Sheldon Cooper's favorite number

      If you are a fan of the television series "The Big Bang Theory", then you know Sheldon often wears a shirt with 73 on it.

      In the tenth episode of the fourth season of The Bing Bang Theory, aired December 9, 2010, https://www.imdb.com/title/tt1632225/characters/nm1433588 , Sheldon asks what everyone’s favorite number is.

      Raj suggests 5318008. Flip this number upside-down to see what Raj has in mind.

      Sheldon declares the best number is 73 because it satisfies the following two properties: a. the product property, and b. the mirror property:

1. p=73 is the i = 21st prime number. The product of the digits of p, 7 and 3, equals i.
2. The reverse of the digits of 73 (its mirror) is 37. 37 is the 12th prime number. 12 is the reverse of 21, and 21 is the product of the digits of 73.

      We will call a prime number satisfying both the product property and mirror property a “Sheldon Prime”.

      Sheldon adds that 73 in base two, 1001001, is a palindrome number. However, many base two numbers are palindrome numbers; 5 in base two is 101.

      This episode was the 73rd episode of Bing Bang. Actor Jim Parsons was born in 1973, and he was 37 years old when the episode aired. Coincidences?

      The 73rd day of a non-leap year is March 14, also known as Pi Day.

      73 is an emirp. An emirp (prime spelled backwards) is a prime number that results in a different prime when its digits are reversed.

      Mathematicians Pomerance and Spicer prove that 73 is the only prime satisfying both the product property and mirror property. https://math.dartmouth.edu/~carlp/sheldon02132019.pdf . They also state that the only primes less than 10^10 satisfying the product property are p=17, i=7; p=73, i=21; and p=2475989, i=181440.

      The following R code: a. verifies that 73 is the only Sheldon Prime within the first 10000 primes. b. verifies that 73 in base two, 1001001, is a palindrome number. c. flips Raj’s 5318008 number upside-down.

# a. 73 is the only Sheldon prime in first 10000 primes:

library(RcppAlgos)
# Function to calculate the product of digits
digit_product <- function(x) {
  digits <- as.numeric(strsplit(as.character(x), NULL)[[1]])
  digits <- digits[!is.na(digits)]
  if (length(digits) == 0) return(NA)
  prod(digits)
}

# Function to reverse digits of a number
reverse_digits <- function(x) {
  as.numeric(paste(rev(strsplit(as.character(x), NULL)[[1]]), collapse = ""))
}

# Function to check if a number is prime
is_prime <- function(num) {
  if (num <= 1) return(FALSE)
  if (num == 2) return(TRUE)
  if (num %% 2 == 0) return(FALSE)
  for (i in 3:sqrt(num)) {
   if (num %% i == 0) return(FALSE)
  }
  return(TRUE)
}

# Generate the first 10000 primes # stopped here for memory & speed issues
primes <- primeSieve(104730) # Generates first 10000 primes; the 10,000th prime is 104729
results <- list()

for (i in seq_along(primes)) {
  if (digit_product(primes[i]) == i) { # product property
   reversed_prime <- reverse_digits(primes[i])
   if (is_prime(reversed_prime)) { # mirror property
    reversed_index <- which(primes == reversed_prime)
    if (reverse_digits(reversed_index) == digit_product(primes[i])){
     results <- append(results, list(c(prime = primes[i], index = i, reversed_prime = reversed_prime,     reversed_index = reversed_index)))
    }
   }
  }
}

if (length(results) > 0) {
  print(results)
} else {
  print("No such prime found.")
}

# b. verify that 73 in base two, 1001001, is a palindrome number.

library(gtools)
x <- 73
v <- baseOf(x, base = 2)
w <- rev(v)
if (all(v == w)) {
  cat(v, "is a palindrome number")
} else {
  cat(v, " is not a palindrome number")
}

# c. flips Raj’s 5318008 number upside-down.

flip_string <- function(x) {
  flip <- c('0' = '0', '1' = 'I', '2' = '2', '3' = 'E', '4' = '4', '5' = 'S', '6' = '9', '7' = '7', '8' = 'B', '9' = '6')
  paste(rev(sapply(unlist(strsplit(as.character(x), NULL)), function(ch) flip[ch])), collapse = "")
}

Raj <- '5318008'
flipped_Raj <- flip_string(Raj)
flipped_Raj

# End

Rene Descartes walks into a bar

Rene Descartes walks into a bar, by Jerry Tuttle

I recently told the old Rene Descartes joke to a math class: Rene Descartes walks into a bar. The bartender asks, "Would you like a beer?" Descartes pauses for a moment and then replies, "I think not." Then poof - he disappears.

Of course I naively assumed my students had been exposed to the Descartes quote, "I think, therefore I am." Philosopher and mathematician Rene Descartes wrote this in French in 1637, and later in Latin as cogito, ergo sum."

After explaining the Descartes quote, I think the students understood the joke. Well, maybe it's not that funny.

But perhaps funnier to math people than you realize, is: this joke is logically flawed because the punchline is the inverse to the original conditional statement, and an inverse is not logically equivalent to the original statement.

Let P and Q be declarative sentences that can be definitively classified as either true or false. Then define:

• Conditional Statement: If P, then Q.
• Converse: If Q, then P.
• Inverse: If not P, then not Q.
• Contrapositive: If not Q, then not P

Two conditional statements are defined as logically equivalent when their truth values are identical for every possible combination of truth values for their individual declarative sentences.

P	Q	statement	converse	inverse	contrapositive
TRUE	TRUE	TRUE	TRUE	TRUE	TRUE
TRUE	FALSE	FALSE	TRUE	TRUE	FALSE
FALSE	TRUE	TRUE	FALSE	FALSE	TRUE
FALSE	FALSE	TRUE	TRUE	TRUE	TRUE

The above table shows statement and contrapositive have the same truth values in columns 3 and 6, and so are logically equivalent. Statement and inverse are not logically equivalent.

The Descartes quote is, "If I think, therefore I am", or "If P then Q". The punchline is, "If I don't think, therefore I am not", or "If not P, then not Q". The punchline is the inverse, and is not logically equivalent to the quote. If P is false, then "if P then Q" is true regardless of the value of Q. So Q can be either true or false.

Occasionally on television someone, often a police detective, will make a statement where they confuse a statement with its converse or inverse, and I have been known to yell at the television.

Descartes is known for developing analytic geometry, which uses algebra to describe geometry. Descartes' rule of signs counts the roots of a polynomial by examining sign changes in its coefficients.

And before someone else feels the need to say this, I will: "Don't put Descartes before the horse." This is perhaps the punchline to changing the original joke to "A horse walks into a bar ... "

The following is R code to create truth tables. Logical is a variable type in R. Conditional statements in R are created using the fact that “If P then Q” is equivalent to “Not P or Q”. I am defining the logic rules for statement, converse, inverse, contrapositive, but I could have defined the rules for more complicated statements as well.

# Define the possible values for P and Q
P <- c(TRUE, TRUE, FALSE, FALSE)
Q <- c(TRUE, FALSE, TRUE, FALSE)

# Calculate the 4 logical rules: statement, converse, inverse, contrapositive
# (Note that “if P then Q” is equivalent to “Not P or Q”.)
P_implies_Q <- !P | Q
Q_implies_P <- !Q | P
not_P_implies_not_Q <- P | !Q
not_Q_implies_not_P <- Q | !P

# expand.grid(P, Q) would also be a good start, but I wanted a specific ordering
# Create a data frame to display the truth table
truth_table <- data.frame(
P = P,
Q = Q,
`P -> Q` = P_implies_Q,
`Q -> P` = Q_implies_P,
`!P -> !Q` = not_P_implies_not_Q,
`!Q -> !P` = not_Q_implies_not_P
)

# Print the truth table
colnames(truth_table) <- c("P", "Q", "statement", "converse", "inverse", "contrapositive")
print(truth_table)

P_variable <- "I think"
Q_variable <- "I am"

colnames(truth_table) <- c(P_variable, Q_variable, "statement", "converse", "inverse", "contrapositive")
print(truth_table)

End

Mind reader game, and Unicode symbols

Mind reader game, and Unicode symbols, by Jerry Tuttle

Perhaps you've seen this Mind Reader game? Think of a two-digit positive whole number, such as 54. Subtract each of the two digits from your number, such as 54 - 5 - 4 = 45, and call 45 the RESULT. Examine the table of symbols below and find the SYMBOL that corresponds with your RESULT. Concentrate on the SYMBOL, and remember it. Then scroll down below, and I will read your mind to predict your SYMBOL.

Before I get to my prediction, let's talk about Unicode symbols.

Unicode symbols

Unicode is a standard international system that assigns a unique Unicode number to each letter, digit, or symbol. This permits thousands of symbols to be written across different platforms, programs, and languages. An example of a Unicode number is "U+" followed by a hex number, such as U+1F499.

The table of symbols in the Mind Reader game is based on plotting a variety of Unicode symbols by printing their Unicode numbers.

A Unicode symbol is printed in R with its Unicode number, but beginning with a backslash to escape the U, and omitting the plus sign. For example, here is how to print a heart symbol.

print("\U1F499")
"💙"

Example with Hebrew letters

Let's spell out the Hebrew word shalom , letter by letter, with each letter's Unicode number, and then use the R paste command to paste the letters together.

    shin <- "\U05E9"     # "ש"
    kamatz <- "\U05B7"     # vowel as two perpendicular lines "ַ"
    lamed <- "\U05DC"     # "ל"
    vav <- "\U05D5"     # "ו"
    final_mem <- "\U05DD"     # "ם"
    shin_with_kamatz <- "\U05E9\U05B7"     # "שַ"
    paste(shin_with_kamatz, lamed, vav, final_mem, sep = "")

"שַלום"

Note that the letters are entered in the paste statement in order of first Hebrew letter, second Hebrew letter, etc., but they are printed in Hebrew right-to-left. Also, a better choice than vav is cholam, which is vav with a dot above it, "\U05BA", but this doesn't print for me.

My prediction

Here is my prediction of your symbol:

Want to play again? Think of another two-digit positive whole number, such as 54. Subtract each of the two digits from your number, such as 54 - 5 - 4 = 45, and call 45 the RESULT. Examine the table of symbols below and find the SYMBOL that corresponds with your RESULT. Concentrate on the SYMBOL, and remember it. Then scroll down below, and I will read your mind to predict your SYMBOL.

Plotting with Unicode characters

Tired of plotting points with those boring 25 pch symbols? You can use Unicode symbols, but you can't simply use pch = . Here for no good reason I use a heart and a thumbs up.

library(ggplot2)
x <- seq(from=0,to=4, by=1)
y <- x^2
z <- exp(x)
df1 <- data.frame(x, y)
df2 <- data.frame(x, z)
heart <- "\U1F499"
thumbs_up <- "\U1F44D"
# Create a custom function to convert Unicode values to GeomPoint
custom_points <- function(data, mapping, ..., shape = heart) {
    ggplot2::geom_point(data = data, mapping = mapping, ..., shape = shape)
}
ggplot() +
    custom_points(data = df1, aes(x = x, y = y, color = "y = x^2"), shape = heart, size=5) +
    geom_line(data = df1, aes(x = x, y = y, color = "y = x^2")) +
    custom_points(data = df2, aes(x = x, y = z, color = "z = exp(x)"), shape = thumbs_up, size=5) +
    geom_line(data = df2, aes(x = x, y = z, color = "z = exp(x)")) +
    ggtitle("Plot with Unicode points") +
    labs(color = "Function") +
    theme(legend.position = c(0.15, 0.85),
      plot.title = element_text(color="black", size=14, face="bold"),
      legend.text = element_text(color="black", size=10, face="bold"))

My second prediction

Here is my second prediction:

Here is the code for the mind reader:

# start with arbitrary set of unicode symbols
description =
    c("a_bengali","a_gurmukhi","approximately_equal","biohazard","black_diamond",
    "black_heart","black_scissors","mercury",
    "mushroom","nya_gujarati",
    "section","snowflake",
    "snowman","teardrop_spoked_asterisk","thunderstorm",
    "umbrella_raindrops","white_cross","white_florette",
    "zhe_cyrillic", "airplane",
   "black_right_arrow","black_telephone","blue_heart",
    "two_xs","hot_beverage","green apple","pill",
    "trophy","thumbs_up")

unicode=
    c("\U0986","\U0A05","\U2248","\U2623","\U25C6","\U2665","\U2702","\U263F",
    "\U1F344","\U0A9E","\U00A7","\U2746",
    "\U26C4","\U273B","\U2608","\U2614","\U271E","\U2740",
    "\U04DC","\U2708","\U27A4","\U260E","\U1F499","\U1F9E0","\U2615","\U1F34F",
   "\U1F48A","\U1F3C6","\U1F44D")

df_uni <- data.frame(cbind(description, unicode))
n <- nrow(df_uni)
x <- seq(1,n,1)
s <- sample(x, 1)
diag <- unicode[s]
diag
x <- x[-x[s]]
y <- sample(x, 100, replace = TRUE)     # randomly choose 100 symbols

df <- data.frame(matrix(ncol = 20, nrow = 10))

for (i in seq(1, 19, by = 2)) {
    df[, i] <- seq(99, 9, -10) - (i - 1) / 2 # row i, odd columns
    df[, i + 1] <- y[((i - 1) / 2 * 10 + 1):((i - 1) / 2 * 10 + 10)] # row i, even columns
    df[, i + 1] <- unicode[df[, i + 1]]
}

for (i in 1:10) {
    df[i, 20 - 2*(i - 1)] <- diag
}

op <- par(bg = "thistle")
plot(x = c(0, 50*20), y = c(0, 50*10), type = "n", xlab = "", ylab = "", xaxt = 'n',
    yaxt = 'n', main = "Mind Reader # 1", cex=2, font=2)

# Loop through each cell of the dataframe to draw rectangles and add text
for (i in 1:20) {
    for (j in 1:10) {
      # Determine the color based on whether the column index is odd or even
      fill_color <- ifelse(i %% 2 == 1, "cornsilk", "lightblue")

      # Draw rectangle with the determined color
      rect(50*(i-1), 50*(10-j), 50*i, 50*(10-j+1), col = fill_color, border = "blue")

      # Add text in the center of the rectangle
      text(50*(i-1) + 25, 50*(10-j) + 25, df[j, i], col = "navyblue", cex = 1, font = 2)
      }
}

# Restore original graphics parameters
par(op)

Hint on prediction

The prediction relies on a little algebra. As a hint, your original two-digit whole number is of the form, 10*T + U. What happens when you subtract the digits?

End

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Text analysis of 2024 US Presidential convention speeches

Text analysis of 2024 US Presidential convention speeches, by Jerry Tuttle

      Can an actuary / mathematician / data analyst say anything objective and data-oriented about the 2024 US presidential campaign?

      Yes, if I confine my remarks to a numerical text analysis of the candidates' words, rather than attempt to comment on the candidates' political and economic views. This analysis is based solely on data that I collected from the two convention speeches.

      I performed a sentiment analysis of the candidates' emotional words. Here is a graphical summary, discussion to follow:

Sentiment analysis

      I also calculated the candidates' most frequent words, excluding stop words like "a", "and", "the" and so on that don't carry much useful information:

Most frequent words

      I calculated statistics of things like average sentence length and average word length:

Summary statistics

      This blog is my attempt at some objective and data-oriented analysis of the two presidential candidates from their convention sppeches. This is a text analysis of their speeches. Text analysis is a branch of data analysis that takes unstructured text documents, breaks them into sentences and words, and attempts to find some meaning. Text analysis is used by marketers to evaluate customer opinions, by police departments to monitor potential or actual criminal activites, by researchers to evaluate things like whether Shakespeare really wrote all the plays attributed to him and which Beatles songs were mostly written by John versus which by Paul, and by many others.

      This is intended as an objective analysis. I am not trying to make either candidate look good, or bad. I have used these same text analysis techniques in other projects such as analyzing Hamlet, analying short stories, and analyzing the Twitter tweets of a radio talk show host.

      For each of Trump and Harris, I started with a transcript of their convention sppeches. I believe the transcripts are their spoken transcripts, not their written transcripts, based on their very first few sentences. I used various computer packages in R such as tm and tidytext to tokenize the documents into individual sentences, words, and characters. I was guided by the works of Silge and Robinson in Text Mining with R and Williams in Data Science Desktop Survival Guide.

      A summary and a comparison of the of the tokenization of the speeches is the following, repeated from before.

      Obviously Trump spoke for much longer than Harris, so he had many more total words. Harris had a larger average sentence length (18.06 words per sentence versus 10.41), but the two were close in average characters per word and average number of syllables per word. Their numerical Flesch scores were about equal. Flesch score is a measure of the grade level required to understand the text, and we would expect a low grade level in a political speech to the public, in contrast to say a student college paper. Harris had a larger percentage of unique or different words in her speech than Trump (22% to 12%). Trump had a larger percentage of negative words than Harris (43% to 37%), which I will discuss below.

      Sentiment analysis ia analyzing text to determine its emotional tone. Linguistics experts have built dictionaries that associate a large list of words with eight emotions (anger, anticipation, disgust, fear, joy, sadness, surprise, and trust) and also negative and positive sentiments. For example, the word "ache" is associated with sadness. Some words have more than one emotion; for example "admirable" is associated with both joy and trust. Further, some words have either negative or positive associations.

      There are some limitations in sentiment analysis. The sentiment dictionary does not recognize sarcasm, and I am limiting my analysis to single words so I am not picking up negation (such as "not expensive") or other instances where the emotion requires several words. A conclusion from the sentiment distribution graph is that the candidates are surprisingly similar in most of these emotions. The biggest differences are that Trump has a greater portion of his words categorized as anger and negative than Harris has.

Most frequent positive and megative words

      Trump's frequent positive words include "love", adjectives "beautiful" and "wonderful", and also "money". His negative words include "tax", "invasion", "inflation", "war", "illegal", and "crime".

      Harris's frequent positive words include "freedom", "love", "opportunity", and "forward". Her negative words include "tax", "enforcement", "abuse", and "abortion". Interestingly "mother" is both a positive word and a negative word, as is "vote". (I rechecked the sentiment dictionary to confirm these.)

      Trump has a larger percentage of negative words (negative divided by positive plus negative) than Harris (43% to 37%). These positive and negative lists seem consistent with my memory of their speeches.

Most frequent words

      Word frequency provides some clue on what is important to each candidate. Not surprisingly, both candidates frequently used words like "America", "Americans", "country", "nation", and "border". Trump used adjectives like "beautiful" and "incredible". Harris spent considerable time talking about Trump and also about her mother. Harris frequently used "middle", "class", "women", and "law".

Distribution of word sizes

Distribution of syllables per word

      The distributions of size of words and of syllables per word are pretty similar for both candidtaes.

Final thoughts

      It is hard to be indifferent about the 2024 US presidential election. You have your opinion, and I have mine. Much of what the candidates say is their opinion, or their plan if elected, and these things are not things we can verify.

      Some things the candidates say as if they are facts, are stated in a way that is open to interpretation. A good example is that "You are better (or worse) off financially today than four years ago." I can choose one measure, collect some data, and show I am better off; or I can choose a very different measure, collect some data, and show I am worse off.

      Some things the candidates say as facts ARE verifiable. I am in no position to do such verifying, but a number of third parties do this. Here are a few links. I can not vouch for their reliability or bias.

• AP News
• FACT Check (Annenberg Public Policy Center)
• Reuters
• Snopes
• Verify This (TEGNA)

      Text analysis is a popular data analysis technique, and I hope I have added some insight into the candidates in an objective manner.

      The following is my R code:

# Trump convention speech, July 19, 2024
# https://www.nytimes.com/2024/07/19/us/politics/trump-rnc-speech-transcript.html

# Harris convention speech, August 23, 2024
# https://singjupost.com/full-transcript-kamala-harriss-2024-dnc-speech/?singlepage=1

library(tidytext)
library(tm)
library(dplyr)
library(nsyllable)
library(SnowballC)
library(ggplot2)
library(forcats)
library(ggpubr)

speaker <- readline(prompt = "Enter Trump, or enter Harris: ")
if (speaker == "Trump" | speaker == "Harris") print(speaker) else print("Invalid input")
trump_file <- "C:/Users/Jerry/Desktop/Harris_Trump/trump_convention_speech.txt"
harris_file <- "C:/Users/Jerry/Desktop/Harris_Trump/harris_convention_speech.txt"
textfile <- ifelse(speaker=="Trump", trump_file, harris_file)
textfile <- readLines(textfile)
text_df <- data.frame(line = 1:length(textfile), text=textfile)
names(text_df)[2] <- "text"

docs <- Corpus(VectorSource(text_df$text))
docs <- tm_map(docs, content_transformer(tolower))
docs <- tm_map(docs, removeNumbers)
docs <- tm_map(docs, removePunctuation, ucp=TRUE)
docs <- tm_map(docs, stripWhitespace)
inspect(docs[1:8])

custom_colors <- c("#1f77b4", "#ff7f0e", "#2ca02c", "#d62728", "#9467bd",
      "#8c564b", "#e377c2", "#7f7f7f", "#bcbd22", "#17becf",
      "#6a3d9a", "#ff9e1b", "#f6c6c7", "#8dd3c7", "#ffffb3",
      "#bebada", "#fb8072", "#80b1d3", "#fdb462", "#b3e2cd",
      "#ccebc5")

common_theme <- theme(
      legend.position="NULL",
      plot.title = element_text(size=15, face="bold"),
      plot.subtitle = element_text(size=12.5, face="bold"),
      axis.title = element_text(size=15, face="bold"),
      axis.text = element_text(size=15, face="bold"),
      legend.title = element_text(size=15, face="bold"),
      legend.text = element_text(size=15, face="bold"))

docs_df <- data.frame(text = sapply(docs, as.character)) # Convert Corpus to data.frame
wordfile <- unnest_tokens(docs_df, word, text, token = "words")
wordfile %>% count(word, sort=TRUE)
wordcountfile <- mutate(wordfile, numbchars = nchar(word)) # characters per word
long1 <- wordcountfile[which(wordcountfile$numbchars == max(wordcountfile$numbchars)),1] # longest word long2 <- wordcountfile[which(wordcountfile$numbchars == max(wordcountfile$numbchars)),2]
numberchars <- sum(wordcountfile$numbchars)
numberwords <- sum(count(wordcountfile, word, sort=TRUE)$n) # no. words
avgcharperword <- round(numberchars / numberwords, digits=2)
sentencefile <- unnest_tokens(text_df, sentence, text, token = "sentences")
sentencecount <- sum(count(sentencefile, sentence, sort=TRUE)$n)
avgwordpersent <- round(numberwords/sentencecount,2)

wordcountdist <- wordcountfile %>% count(numbchars)
wordcountdist$numbchars <- as.factor(wordcountdist$numbchars)
title <- paste(speaker, "- Distribution of Word Size")
subtitle <- paste("Longest word: ", long1, long2, "characters")
ggplot(wordcountdist, aes(numbchars, n, fill=numbchars)) +
      geom_bar(stat="identity", position = "dodge", width=0.5) +
      labs(title=title, subtitle=subtitle) +
      xlab("number of characters per word") + ylab("") +
      scale_fill_manual(values = custom_colors) +
      theme(legend.position = "none") +
      common_theme

syls <- nsyllable(wordfile$word, language = "en")
syls[which(wordfile$word == "jd")] <- 2 # used because nsyllable generated error here
syls[which(wordfile$word == "nd")] <- 1 # used because nsyllable generated error here
syls[which(wordfile$word == "st")] <- 3 # s/b 21st; used because nsyllable generated error here
syls[which(wordfile$word == "gasolinepowered")] <- 5 # used because nsyllable erred here
long2 <- min(syls[(syls == max(syls, na.rm=TRUE))], na.rm=TRUE)
w <- min(which(syls == long2))
long1 <- wordfile$word[w]
avgsylperword <- round(sum(syls)/numberwords, digits = 2)
avgsylperword
syls <- data.frame(syls) %>% count(syls)
syls$syls <- as.factor(syls$syls)
colnames(syls) <- c("syllables", "n")
title <- paste(speaker, "- Distribution of No. Syllables per Word")
subtitle <- paste("Most syllables: ", long1, long2, "syllables")
ggplot(syls, aes(syllables, n, fill = syllables)) +
      geom_bar(stat="identity", position = "dodge", width=0.5) +
      labs(title=title, subtitle=subtitle) +
      xlab("number of syllables per word") + ylab("") +
      scale_fill_manual(values = custom_colors) +
      theme(legend.position = "none") +
      common_theme

# Flesch-Kincaid reading ease formula
flesch <- round(206.835 - 1.015*(numberwords/sentencecount) - 84.6*(sum(syls$n)/numberwords),2) # Flesch reading ease
flesch
flesch_df <- data.frame(score = c(0,30,50,60,70,80,90,100),
      grade = c("College graduate","College","10th - 12th grade","8th - 9th grade",
      "7th grade","6th grade","5th grade","below 5th grade"))

# Function to find the grade based on score; vlookup
find_grade <- function(score, flesch_df) {
  idx <- findInterval(score, flesch_df$score)
  if (idx == 0) {
   return("below 5th grade") # Handle case where score is below the minimum
  } else {
   return(flesch_df$grade[idx])
  }
}

score_to_find <- flesch
flesch_grade <- find_grade(score_to_find, flesch_df)
flesch_grade

# delete stop words
docs_df <- data.frame(text = sapply(docs, as.character)) # Convert Corpus to data.frame
wordfile <- unnest_tokens(docs_df, word, text, token = "words")
stop_words <- data.frame(tidytext::stop_words) # more words than tm
my_stop_words <- data.frame(word = c("theyre", "hes", "dont",
   "didnt","youre","cant", "im","whats", "weve", "theyve", "youve",
   "couldnt", "wont", "youd"))
wordfile <- anti_join(wordfile, stop_words)
wordfile <- anti_join(wordfile, my_stop_words)

wordfreq <- wordfile
wordfreq <- count(wordfreq, word, sort=TRUE) # word frequency excl stop words
wordfreqdf <- data.frame(wordfreq)

unique_words <- nrow(wordfreq)
portion_unique_words <- round(unique_words / numberwords, digits=2)
wordfreqdf20 <- wordfreqdf[1:21,] # Think about threshold
wordfreqdf20

graphtitle <- paste(speaker, "Word Frequency")
wordfreqdf20$word <- fct_reorder(wordfreqdf20$word, wordfreqdf20$n, .desc = FALSE)
ggplot(data=wordfreqdf20, aes(x=word, y=n, fill=word)) +
      geom_bar(stat="identity", position = "dodge", width=0.5) +
      coord_flip() +
      common_theme +
      xlab("") + ylab("Frequency") +
      ggtitle(graphtitle) +
      scale_fill_manual(values = custom_colors) +
      theme(legend.position = "none")

# sentiments; note mother is both positive and negative!
df1 <- data.frame(wordfile)
colnames(df1) <- "word"
df2 <- get_sentiments("nrc")
df3 <- merge(x=df1, y=df2, by="word", all.x=TRUE, stringsAsFactors=FALSE)
df3 <- subset(df3, !is.na(sentiment))
table(df3$sentiment)
w <- data.frame(table(df3$sentiment))
colnames(w) <- c("sentiment", "n")

sentiment_colors <- c(
   "Anger" = "red",
   "Anticipation" = "green",
   "Disgust" = "brown",
   "Fear" = "purple",
   "Joy" = "yellow",
   "Negative" = "gray",
   "Positive" = "lightblue",
   "Sadness" = "blue",
   "Surprise" = "pink",
   "Trust" = "turquoise")

title <- paste(speaker, "- Sentiment Plot")
ggplot(w, aes(sentiment, n)) +
      geom_bar(stat = "identity", position = "dodge", width = 0.5, fill = sentiment_colors) +
      ggtitle(title) +
      ylab("") +
      common_theme +
      theme(axis.text.x = element_text(angle = 45, hjust=1))

df4 <- df3 %>% filter(sentiment == "positive" | sentiment == "negative")
w <- with(df4, table(sentiment))
neg <- w[1]
pos <- w[2]
neg_ratio <- round(w[1] / (w[1] + w[2]), digits=2)
df5 <- df4 %>% group_by(sentiment) %>% count(word, sort=TRUE)
pos_freq <- df5 %>% filter(sentiment=="positive") %>% top_n(10, wt = n) %>% slice_head(n = 10)
neg_freq <- df5 %>% filter(sentiment=="negative") %>% top_n(10, wt = n) %>% slice_head(n = 10) # ties
pos_freq$word <- fct_reorder(pos_freq$word, pos_freq$n, .desc = FALSE)
neg_freq$word <- fct_reorder(neg_freq$word, neg_freq$n, .desc = FALSE)

title <- paste(speaker, "- Most Frequent Positive and Negative Words")
p1 <- ggplot(pos_freq, aes(word, n)) +
      geom_bar(stat="identity", position = "dodge", width=0.5, fill="darkgreen") +
      ggtitle("Positves") +
      common_theme +
      xlab("") +
      coord_flip()
p2 <- ggplot(neg_freq, aes(word, n)) +
      geom_bar(stat="identity", position = "dodge", width=0.5, fill="red") +
      ggtitle("Negatives") +
      common_theme +
      xlab("") +
      coord_flip()
plot <- ggarrange(p1,p2, ncol=2, nrow=1, legend=NULL)
annotate_figure(plot, top = text_grob(title,
    color = "black", face = "bold", size = 14))

if (speaker == "Trump"){
   t <- data_frame(speaker, numberwords, avgwordpersent, avgcharperword, avgsylperword, flesch, flesch_grade, portion_unique_words, neg_ratio)
   print(t)
   } else {h <- data_frame(speaker, numberwords, avgwordpersent, avgcharperword, avgsylperword, flesch, flesch_grade, portion_unique_words, neg_ratio)
    conclusion <- data.frame(rbind(t,h))
    conclusion <- t(conclusion)
    colnames(conclusion) <- c("Trump", "Harris")
    conclusion <- conclusion[-1,]
    print(conclusion)
}

# the results of stemming and lemmatizing were not used in the report
# stemming
wordfile <- wordfile %>%
    mutate(stem = wordStem(word)) %>%
    count(stem, sort = TRUE)

# lemmatize
df1 <- wordfile # df1 has col named stem
url <- "https://raw.githubusercontent.com/michmech/lemmatization-lists/master/lemmatization-en.txt"
df2 <- read.table(url, header = FALSE, sep = "\t", quote = "", stringsAsFactors = FALSE)
names(df2) <- c("stem", "word")
df3 <- merge(x = df1, y = df2, by = "stem", all.x = TRUE, stringsAsFactors=FALSE)

# if word is not in dictionary, then leave word as is; otherwise, use stemmed word.
df3$word <- ifelse(is.na(df3$word), df3$stem, df3$stem)

# End

Is the Mona Lisa thinking about irrational numbers?

Is the Mona Lisa thinking about irrational numbers? by Jerry Tuttle

      As a math teacher I sometimes share the following problem-solving strategy: If you are really stuck on a problem, let it sit, come back the next day, and maybe you will have a fresh insight. This often works for me.

      58 years later ...

      I recently applied this strategy with a problem posed by my beloved Bryant High School math teacher Anthony Holman. However, he posed it in 1966, so it took me 58 years before I had any insight. Unfortunately, Mr. Holman passed away in 1985, so I never had the opportunity to discuss this with him.

      In calculus class Mr. Holman taught us about e, sometimes called Euler’s number, which is the base of the natural log function. He told us e, like π , would appear in many unlikely places in math, and of course he was right. Then he hypothesized that e and π and other famous irrational numbers were probably part of the great works of art like the Mona Lisa.

      That was the only time I remember him making that statement. I had no idea how to approach it. This was 1966, long before computers, even mainframe computers, were readily available to the public. I forgot about his statement for about 55 years.

      More recently I have been programming in R, and I have learned a little about computer images. A computer image can be represented as an array of many small elements called pixels, and every pixel is a base 16 number of the hex code of that pixel's color. I decided to interpret Mr. Holman’s hypothesis as whether a finite sequence of the digits of a number like π were contained within the pixel hex code of the Mona Lisa. I decided to separately test four famous irrational numbers: π, e, φ, and √2.

      A quick note about φ: This is the Golden Ratio phi, defined as the ratio of a/b such that φ = a/b = (a + b)/a. The Golden Ratio is well-known to appear in numerous artworks including the Mona Lisa, where various measures of the subject’s face appear in Golden ratio proportions. I am assuming this is not what Mr. Holman intended in his hypothesis.

      I am making a few assumptions here. An irrational number is an infinite, non-repeating decimal. I can not test whether an infinite sequence of digits appears within some other sequence of numbers. So I am truncating these irrational numbers to ten decimal places each, which I think is sufficient for this exercise. The tenth decimal place is truncated, not rounded. Of course the test may pass at ten decimal places and then fail at the eleventh.

      Another assumption is that I have a sufficiently high-quality image of the original Mona Lisa. The painting was completed in approximately 1517. It has aged and there has been some physical restoration. A computer image is the result of a photograph, and these photographs also contain some digital retouching. So the result may not be equivalent to the original. Finally, there is a variety of available resolutions including one that has a size of 90 MB. I do not have sufficient computer memory to handle that file size and the calculations of a file that large.

      I am also assuming it is sufficient to do this problem with just the Mona Lisa.

      In my first attempt, I was able to get the base 16 number of the hex code of every pixel of my Mona Lisa file. The file looks like this for the first six pixels. I show each pixel's base 16 hex code and its base 10 decimal equivalent.

      I converted each of the four irrational numbers from base 10 to base 16. Each irrational number is nine characters in base 16. I created one very long string of the pixel hex codes, and I searched whether each irrational number was contained in the hex string. The string has 1.7 million characters, and I am searching for a nine character sequence. However, the result was none of the irrationals was contained in the string.

      I decided the first attempt was faulty. The Mona Lisa was created in a base 10 world, and it didn’t make sense to force the irrationals into an artificial base 16 comparison.

      In my second attempt, I converted each base 16 pixel to base 10, to compare against base 10 irrationals. I think this is a more natural comparison, and if an irrational is going to be contained in a larger sequence then it would be as a base ten irrational.

      Unfortunately this also failed for each of the four irrationals. So sadly, I can not confirm Mr. Holman’s hypothesis. Maybe I should let this problem sit for a few more years and come back when I have a fresh idea (or when I buy a computer with more memory). Maybe Mr. Holman has solved it in Heaven. Nevertheless, I’m sure he is smiling at my efforts.

      The following is my R code:

# Mona Lisa problem matching base 10 codes
library(imager)
# url1 is huge: 89.9 MB; url2 is 70 KB.
# use url2 unless you have a lot of memory.

url1 <- "https://upload.wikimedia.org/wikipedia/commons/e/ec/Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg"
url2 <- "https://upload.wikimedia.org/wikipedia/commons/thumb/e/ec/Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg/402px-Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg"
img <- load.image(url2)
plot(img)

img_data <- as.data.frame(img)
head(img_data) # x, y, cc(color channel: r, g, b, transparent), intensity (normalized 0,1)
dim(img_data) # 722,394 x 4
library(dplyr)
img_data <- arrange(img_data, x, y, cc)

# Pivot the data to wide format so each pixel has its R, G, B values in separate columns
library(tidyr)
img_wide <- pivot_wider(img_data, names_from = cc, values_from = value, values_fill = list(value = 0)) # # If there are missing values, fill them with 0
dim(img_wide) # 240,798 x 5

# Convert normalized values to 0-255 range
img_wide$R <- img_wide$`1` * 255
img_wide$G <- img_wide$`2` * 255
img_wide$B <- img_wide$`3` * 255

# Convert RGB values to hexadecimal format
img_wide$hex <- rgb(img_wide$R, img_wide$G, img_wide$B, maxColorValue = 255)
img_wide$hex <- gsub('#', '', img_wide$hex)

# Convert hex values to decimal format
library(Rmpfr) # extended floating point computations
options(scipen = 999)
img_wide$dec <- as.numeric(mpfr(img_wide$hex, base=16))

# Check the structure of img_wide
head(img_wide)
# Concatenate dec values into a single string
dec_string <- paste(img_wide$dec, collapse = "")
nchar(dec_string) # 1701600

test_value1 <- "04111665813" # starts in position 2
pi_base10 <- "31415926535" # https://www.angio.net/pi/digits/50.txt 10 decimal places, truncated not rounded
e_base10 <- "27182818284" # https://www.i4cy.com/euler/
phi_base10 <- "16180339887" # https://nerdparadise.com/math/reference/phi10000
root_two_base10 <- "14142135623" # https://nerdparadise.com/math/reference/2sqrt10000

# Check if the dec string contains the digits of pi, e, phi, root 2

check_irrational <- function(value, name) {
if (grepl(value, dec_string)) {
cat("The first ten digits of", name, "are present in the pixel data.\n")
start_position <- regexpr(value, dec_string)
cat("The match starts at position:", start_position, "\n")
} else {
cat("The first ten digits of", name, "are not present in the pixel data.")
}
}

check_irrational(test_value1, "test1")
check_irrational(pi_base10, "π")
check_irrational(e_base10, "e")
check_irrational(phi_base10, "φ")
check_irrational(root_two_base10, "√2")

options(scipen = 0)

# END

#######################################

The distribution has changed; and pretty tables in base R

The distribution has changed; and pretty tables in base R, by Jerry Tuttle

      So you spent hours, or maybe days, cranking out thousands of numbers, you submit it to your boss just at the deadline, your boss quickly peruses your exhibit of numbers, points to a single number and says, "This number doesn't look right." Bosses have an uncanny ability to do this.

      Your boss is pointing to something like this: Your company sells property insurance on both personal and commercial properties. The average personal property premium increased 10% in 2024. The average commercial property premium increased 10% in 2024. But you say the combined average property premium decreased 3% in 2024. You realize that negative 3% does not look right.

      You might have made an input error or a calculation error, but you don't want to admit to that. So you blurt out, "That's because the distribution has changed." And to your relief, the boss buys into that.

      "The distribution has changed" is probably a pretty good answer in more instances than you realize. A more common example is if your investment portfolio starts at 90% stocks and 10% bonds, but you have a good year with stocks, at year-end the distribution of your portfolio has changed to 94% stocks and 6% bonds, and you may want to rebalance your portfolio. How about the distribution of population by state? It has definitely changed since the prior Census. It might be hard to think of a non-trivial example of something where the distribution has not changed.

      The key calculations are the 2023 average premium is (2000 * 90 + 10000 * 10) / (90 + 10) = 2800, the 2024 average premium is (2200 * 110 + 11000 * 7) / (110 + 7) = 2726, and so the average premium change is 2726 / 2800 = .97 (rounded) = - 3%. And the distribution between personal and commercial DID change, as measured either by the distribution of number of policies (was 90% personal, now 94%) or by the distribution of premium (was 64% personal, now 76%). So because the distribution has changed towards more smaller personal policies, this pulls the combined average premium down, even though the average premium for each of the two separate subgroups has increased. (There are alternatives to the key calculations, such as weighting the percentage changes, instead of weighting the average premiums.)

      Years ago I posed a dilemma like this during job interviews for actuarial trainees to see how well they would respond to a sort of non-routine problem, but I decided it was too difficult.

      I did the above exhibit in Excel because it was quick. It was also pretty easy to add custom colors to alternating rows, which I did via FONT > FILL > MORE COLORS > CUSTOM > enter HEX CODE. Here I chose cornsilk #fff8dc and cyan #00ffff for fun.

      Then I wondered how easy it would be to make a pretty table in R. If you google something like "pretty tables in R", you will find a number of R packages that create HTML type code that can be saved as an HTML file, a PDF file, or another file format. Much has been written about these packages, but they seem a little complicated for basic work, and further, I like the idea of staying exclusively within the R environment. When I realized a table is just a collection of rectangles, it occurred to me that the base R commands of rectangle and text are pretty much all I need.

      Here is a table of sample rectangles with text, written in R. The rectangle syntax is rect(xleft, ybottom, xright, ytop, col, border) and the text syntax is text(x, y, labels, col, cex, font). The numerical axis is helpful when first defining the rectangles, but can be deleted by adding axes = FALSE to the plot function for the final table.

# rectangle syntax: rect(xleft, ybottom, xright, ytop)
plot(x = c(0, 500), y = c(0, 500), type= "n", xlab = "", ylab = "", main = "Sample rectangles with text", cex=2, font=2)

rect(0, 0, 250, 250, col = "#E41A1C", border = "blue")
text(125, 125, "red rectangle, white font", col="white", cex=1.15, font=2)

rect(250, 0, 500, 250, col = "yellow", border = "blue")
text(375, 125, "yellow rectangle, blue font", col="navyblue", cex=1.15, font=2)

rect(0, 250, 250, 500, col = "cornsilk", border = "blue")
text(125, 375, "cornsilk rectangle, black font", col="black", cex=1., font=2)

rect(250, 250, 500, 500, col = "cyan", border = "blue")
text(375, 375, "cyan rectangle, purple font", col="purple", cex=1.15, font=2)

      The R equivalent of the Excel exhibit is the following. Note that all code is in base R.

title <- c("Subgroups increase, but the average decreases")
df <- data.frame(Personal = c(2000, 90, 2200, 110, 10),
Commercial = c(10000, 10, 11000, 7, 10),
Weighted = c(2800, 100, 2726, 117, -3))

rownames(df) = c("2023 Avg Prem", "2023 No. Policies","2024 Avg Prem", "2024 No. Policies", "Avg Prem % Change")
df

# rectangle syntax: rect(xleft, ybottom, xright, ytop)
op <- par(bg = "thistle")
col1 = "cornsilk"
col2 = "cyan"

plot(x = c(0, 500), y = c(0, 700), type= "n", xlab = "", ylab = "", axes = FALSE)
height = 100

rect(0, 6*height, 500, 7*height, col = col1, border = "blue")
text(250, 6.5*height, title, col="black", cex=1.25, font=2)

rect(0, 5*height, 200, 6*height, col = col2, border = "blue")
rect(200, 5*height, 300, 6*height, col = col2, border = "blue")
rect(300, 5*height, 400, 6*height, col = col2, border = "blue")
rect(400, 5*height, 500, 6*height, col = col2, border = "blue")

text(100, 5.5*height, "", col="blue")
text(250, 5.5*height, colnames(df)[1], col="blue")
text(350, 5.5*height, colnames(df)[2], col="blue")
text(450, 5.5*height, colnames(df)[3], col="blue")

rect(0, 4*height, 200, 5*height, col = col1, border = "blue")
rect(200, 4*height, 300, 5*height, col = col1, border = "blue")
rect(300, 4*height, 400, 5*height, col = col1, border = "blue")
rect(400, 4*height, 500, 5*height, col = col1, border = "blue")

text(100, 4.5*height, rownames(df)[1], col="blue")
text(250, 4.5*height, df[1,1], col="blue")
text(350, 4.5*height, df[1,2], col="blue")
text(450, 4.5*height, df[1,3], col="blue")

rect(0, 3*height, 200, 4*height, col = col2, border = "blue")
rect(200, 3*height, 300, 4*height, col = col2, border = "blue")
rect(300, 3*height, 400, 4*height, col = col2, border = "blue")
rect(400, 3*height, 500, 4*height, col = col2, border = "blue")

text(100, 3.5*height, rownames(df)[2], col="blue")
text(250, 3.5*height, df[2,1], col="blue")
text(350, 3.5*height, df[2,2], col="blue")
text(450, 3.5*height, df[2,3], col="blue")

rect(0, 2*height, 200, 3*height, col = col1, border = "blue")
rect(200, 2*height, 300, 3*height, col = col1, border = "blue")
rect(300, 2*height, 400, 3*height, col = col1, border = "blue")
rect(400, 2*height, 500, 3*height, col = col1, border = "blue")

text(100, 2.5*height, rownames(df)[3], col="blue")
text(250, 2.5*height, df[3,1], col="blue")
text(350, 2.5*height, df[3,2], col="blue")
text(450, 2.5*height, df[3,3], col="blue")

rect(0, height, 200, 2*height, col = col2, border = "blue")
rect(200, height, 300, 2*height, col = col2, border = "blue")
rect(300, height, 400, 2*height, col = col2, border = "blue")
rect(400, height, 500, 2*height, col = col2, border = "blue")

text(100, 1.5*height, rownames(df)[4], col="blue")
text(250, 1.5*height, df[4,1], col="blue")
text(350, 1.5*height, df[4,2], col="blue")
text(450, 1.5*height, df[4,3], col="blue")

rect(0, 0, 200, height, col = col1, border = "blue")
rect(200, 0, 300, height, col = col1, border = "blue")
rect(300, 0, 400, height, col = col1, border = "blue")
rect(400, 0, 500, height, col = col1, border = "blue")

text(100, .5*height, rownames(df)[5], col="blue")
text(250, .5*height, paste(df[5,1], "%"), col="black", cex=1.5)
text(350, .5*height, paste(df[5,2], "%"), col="black", cex=1.5)
text(450, .5*height, paste(df[5,3], "%"), col="black", cex=2, font=2)

par(op)

# END

#######################################

Radar charts and five-tool baseball players

Radar charts and five-tool baseball players by Jerry Tuttle

      I was looking for an opportunity to practice with radar charts and I came across an article on five-tool baseball players, so this seemed like a perfect application for this kind of chart.

      A radar chart is an alternative to a column chart to display three or more quantitative variables. The chart graphs the values in a circular manner around a center point.

      The five tools in baseball are: (1) hitting for average; (2) hitting for power; (3) defense; (4) throwing; and (5) speed. A five-tool player excels in all five of these.

      Among current players, Mike Trout is considered a five-tool player. The measurement of Trout’s five tools can be displayed in the following radar chart:

      Trout is rated at 80 for hitting for average, 70 for hitting for power, and his lowest scores are 60 for defense, throwing and speed. This is based on a 20-to-80 rating system, where 80 is elite, 70 is plus-plus, and 60 is plus. Sorry - I could not get the points to line up with the concentric pentagons.

      For comparison, here is a display of Aaron Judge's ratings.

      Judge is rated at 80 for hitting for power, 70 for hitting for average, 60 for defense, 70 for throwing, and 50 for speed, where the 50 is average at the major league level.

      The results of several players can be displayed in a single radar chart, but this becomes hard to read. Three players are probably the maximum for readability.

      The alternative to visualizing several players is either to create several individual radar charts or else to create a bar (horizontal) chart or a column (vertical) chart.

      Each of the five tools is generally rated on a 20-to-80 scale, where 50 is average (for a major leaguer), 80 is elite, and every 10 points is supposed to represent one standard deviation. I suspect the standard deviation concept is more judgmental than mathematical. There is not a single rating system; some use traditional baseball statistics, and others use modern motion tracking data.

      The numerical data above was obtained from an article by Jake Mintz in 2022 for Fox Sports https://www.foxsports.com/stories/mlb/trout-betts-rodriguez-the-definition-of-mlbs-five-tool-players . In Mintz's data, all numbers are shown rounded to the nearest 10. Mintz only has five current players as five-tool players: Mike Trout, Mookie Betts, Trea Turner, Byron Buxton, and Julio Rodriguez. I tried graphing all five players in a single radar chart, but this was too hard to read. Mintz thinks a true five-tool player should have a grade of at least 60 in each of the five categories. By this measure, Aaron Judge is not quite a five-tool player due to a 50 in speed, and a number of elite major leaguers have at least one 50. Note that each category is considered separately. If there were some sort of weighting system, many people would weigh hitting with power as most important, followed by hitting for average, although perhaps the weights should vary by position with higher weights for defense and throwing for catcher, middle infielders, and center fielder. Pitchers have a different grading system.

      What about Shohei Ohtani? At the time of his article, Mintz did not have sufficient data on Ohtani.

      Mintz observes that Mike Trout worked one winter to improve his throwing, and Julio Rodriguez worked to increase his speed. This suggests that the ratings probably change over the life of a player and are dependent on when they are measured.

      Other authors suggest that there is a sixth tool of exceptional players such as mental makeup and character. Another tool might be situational game awareness.

      Modern motion tracking data by Statcast and others did not exist until fairly recently. Willie Mays is generally considered the greatest five-tool player. Using statistical measures, author Herm Krabbenhoft suggests Tris Speaker, Ty Cobb, and Honus Wagner should be considered as five tool players, although Krabbenhoft measures hitting for power with SLG (slugging percentage) and ISO (isolated power), not home runs https://sabr.org/journal/article/honus-wagner-baseballs-prototypical-five-tooler/ . A very different measure of hitting with power would be something like home run distance greater than 425 feet or launch angle and velocity.

      What about Babe Ruth? We know Babe Ruth's career numbers are .342 batting average and 714 home runs. I have not read anything about his defense, throwing, or speed. He did steal 123 bases, including home 10 times; maybe he was faster than we realize. He is remembered for getting thrown out stealing second to end the 1926 World Series, but perhaps the hit-and-run play was on, and Bob Meusel, the batter, swung and missed the pitch? See https://baseballegg.com/2019/10/30/babe-ruths-failed-stolen-base-attempt-ended-the-1926-world-series-or-is-that-what-really-happened/ . Ruth had 204 assists as an outfielder, which sounds like a lot. I wonder how he would have ranked in defense, throwing, and speed?

      Here is my R code. I do like radar charts for comparing one to three observations over five variables, as a change of pace from column charts. I used the fmsb library for the radar charts. There is also a ggradar library, but I did not like its visualization. One of the quirks of fmsb is that the axis for each variable can have its own scale. Originally I used each variable's max and min values, but the axes were out of sync, so I replaced this with the grand max and min. Also, I could not get the values, which are all multiples of ten, to line up on the concentric pentagons.

library(fmsb)
library(scales)

group = c("Hit_avg", "Hit_power", "Defense", "Throwing", "Speed")
player_names = c("Trout","Betts","Judge")
players <- data.frame(
   row.names = player_names,
   Hit_avg = c(80, 70, 70),
   Hit_power = c(70,60,80),
   Defense = c(60,70,60),
   Throwing = c(60,80,70),
   Speed = c(60,70,50))
players

# The row 1 should contain the maximum values for each variable
# The row 2 should contain the minimum values for each variable
# Data for cases or individuals should be given starting from row 3
# Define the variable ranges: maximum and minimum; however, want axes to have equal scales

max_min <- data.frame(
   Hit_avg = c(max(players), min(players)),
   Hit_power = c(max(players), min(players)),
   Defense = c(max(players), min(players)),
   Throwing = c(max(players), min(players)),
   Speed = c(max(players), min(players)))

rownames(max_min) <- c("Max", "Min") # row 1 has max's, row 2 has min's.
df <- rbind(max_min, players)
df

player1_data <- df[c("Max", "Min", player_names[1]), ]
player2_data <- df[c("Max", "Min", player_names[2]), ]
player3_data <- df[c("Max", "Min", player_names[3]), ]

chart <- function(data, color, title){
     radarchart(data, axistype = 0,
       pcol = color, pfcol = scales::alpha(color, 0.5), plwd = 2, plty = 1,
        vlabels = colnames(data), vlcex = 1.5,
       cglcol = "black", cglty = 1, cglwd = 0.8,
        caxislabels = NULL,
       title = title)
}

# Plot the data for players 1, 2, and 3 separately
chart(data=player1_data, color="#00AFBB", title="MIKE TROUT 5 Tools")
chart(data=player2_data, color="#E7B800", title="MOOKIE BETTS 5 Tools")
chart(data=player3_data, color="#FC4E07", title="AARON JUDGE 5 Tools")

# Plot the data for three players
chart(data=df, color=c("#00AFBB", "#E7B800", "#FC4E07"), # blue-green, red-green, red-green
     title="TROUT, BETTS, JUDGE 5 Tools")
legend(
     x = "bottom", legend = rownames(df[-c(1,2),]), horiz = FALSE,
     bty = "n", pch = 20 , col = c("#00AFBB", "#E7B800", "#FC4E07"),
     text.col = "black", cex = 1.25, pt.cex = 1.5)

###########################################

# column graphs

library(tibble)
library(tidyr)
library(ggplot2)
# Reshape data to long format
players_long <- players %>%
     rownames_to_column("player") %>%
     pivot_longer(cols = -player, names_to = "group", values_to = "value")

# Common theme for graphs
common_theme <- theme(
   legend.position="right",
   plot.title = element_text(size=15, face="bold"),
   axis.title = element_text(size=15, face="bold"),
   axis.text = element_text(size=15, face="bold"),
   legend.title = element_text(size=15, face="bold"),
   legend.text = element_text(size=15, face="bold"))

# Create column graph: Tool Ratings by Player
ggplot(players_long, aes(x = player, y = value, fill = group, title = "Tool Ratings by Player")) +
   geom_col(position = "dodge") +
   labs(x = "Player", y = "Rating", fill = "Group") +
   common_theme

# Create the column graph: Player Ratings for each Tool
ggplot(players_long, aes(x = group, y = value, fill = player)) +
   geom_col(position = "dodge") +
   labs(x = "Group", y = "Rating", fill = "Player", title = "Player Ratings for each Tool") +
   common_theme

### END

##################################################################################

Something a llttle different: Hexbin maps

Something a llttle different: Hexbin maps by Jerry Tuttle

      I recently became acquainted with hexbin maps, so I thought I would experiment with one. In a hexbin map, each geographical region is represented by an equally sized regular hexagon. (A regular hexagon can tessellate a plane, one of only three regular polygons that can do so besides an equilateral triangle and a square.) A hexbin map can be used to illustrate the distribution of a categorical variable by using colors, at a cost that geographical areas are distorted by size, shape, and orientation. Here is a hexbin map, followed by the same data in a conventional map. The four categories represent which of the two political parties had the larger popular vote in the 2020 presidential election and by what margin over the other party.

# Hexbin map

# Conventional map

      A hexbin is based on a geoJson file that provides the hexagon boundaries as two-dimensional coordinates. I downloaded a file from https://team.carto.com/u/andrew/tables/andrew.us_states_hexgrid/public/map in geo json format containing the boundaries of the US states. I read it in with read_sf from the R sf (simple features spatial vector data) library. I used a datafile of 2020 presidential election popular votes by state scraped from https://www.presidency.ucsb.edu/statistics/elections/2020 . I binned percent Democraric votes to total votes minus percent Republican votes to total votes into four bins and joined to the sf file. I used ggplot to plot the hexbin map, where label = iso3166_2 provides a two-character state abbreviation.

      For the conventional map, I used the R usmap library.

      A hexbin map is a nice alternative to a conventional map. With a conventional map, it's hard to read those small states. Howwever, it is not the right visualization tool if the data is quantitative. It is also not the right tool if the size or geography is important. Utah is not the same size as Califirnia in either area or number of voters, and South Carolina is not east of North Carolina.

      An alternative to hexagonal areas is to make them squares, but hexagons seem more interesting.

      Here is the R code I used:

# plot the conventional map first:
library(usmap)
# Color maps with data; must have fips column
library(ggplot2)
library(readxl)
data <- read_excel("C:/Users/Jerry/Desktop/R_files/presidential_2020.xlsx")
data$party <- ifelse(data$Dem_Minus_Rep >=0 & data$Dem_Minus_Rep <= 0.10, 'Dem 0 to 10%',
      ifelse(data$Dem_Minus_Rep > 0.10, 'Dem > 10%',
      ifelse(data$Dem_Minus_Rep <0 & data$Dem_Minus_Rep >= -0.10, 'Rep 0 to 10%',
      ifelse(data$Dem_Minus_Rep < -0.10, 'Rep > 10%', "Error"))))
#
party_colors <- c("blue", "lightskyblue", "red", "#FF66CC")
mytitle="2020 Presidential Election Popular Vote by State"
#
plot_usmap(data = data, values = "party", labels = TRUE, label_color = "white") +
    labs(title=mytitle) +
    scale_fill_manual(values=party_colors) +
    theme(
      legend.position="bottom",
      plot.title = element_text(size=15, face="bold"),
      legend.title = element_text(size=12, face="bold"),
      legend.text = element_text(size=12, face="bold"))
#
# plot the hexbin map:
# Download coordinates from https://team.carto.com/u/andrew/tables/andrew.us_states_hexgrid/public/map
# in geo json format
my_sf <- read_sf("C:/Users/Jerry/Desktop/R_files/us_states_hexgrid.geojson")
my_sf <- my_sf %>%
    mutate(google_name = gsub(" \\(United States\\)", "", google_name))
#
# join sf file with data file
my_sf_data <- my_sf %>%
    left_join(data, by = c("google_name" = "STATE"))
#
ggplot(my_sf_data) +
    geom_sf(aes(fill = party), linewidth = 0, alpha = 0.9) +
    geom_sf_text(aes(label = iso3166_2), color = "white", size = 4, alpha = 1) +
    theme_void() +
    scale_fill_manual(values=party_colors) +
    ggtitle(mytitle) +
    theme(
      legend.position="bottom",
      plot.title = element_text(size=15, face="bold"),
      legend.title = element_text(size=12, face="bold"),
      legend.text = element_text(size=12, face="bold"))
# End

##################################################################################

Taylor Swift and Data Analysis

Taylor Swift and Data Analysis. by Jerry Tuttle

Who will be the most talked-about celebrity before, during, and after the Super Bowl?