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So here is a useless fact. There is no letter A in numbers – written as words – from 1 to 100. And of course, we wanted to put this into the test and check if it holds the water.

And sure, there is the result:

Two sets of functions were created in this case (it can be stuffed in single one and super simplified, but it is all about uselessness.

Getting the words for number:

#function
word_a_number <- function(numb){

basLet <- c('one','two','three','four','five','six','seven','eight','nine','ten'
,'eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen'
,'twenty','thirty','forty','fifty','sixty','seventy','eighty','ninety','one hundred')
basNum <- c(1:20,30,40,50,60,70,80,90,100)
df <- data.frame(num = basNum, let = as.character(basLet))

if (numb <= 20) {
im <- df[which(df$num == numb),]$let
print(paste(im, collapse = NULL))
} else {
if (numb %% 10 == 0){
e <- df[which(df$num == numb),]$let
print(paste0(e, collapse=NULL))
} else {
sec <- numb %% 10
fir <- as.integer(numb/10)*10
f_im <- df[which(df$num == fir),]$let
s_im <- df[which(df$num == sec),]$let
res <- paste0(f_im,"-",s_im, collapse = NULL)
print(res)
}
}
}



This functions iterates through a set of numbers, given a boundaries. And since first ten words and and words from eleven to twenty differ significantly, I wrote it in data frame. Next step is to do the increments of ten. And then hundred, thousand, etc.

Once this function is established, it will return the word for given number. Once this is established, a little helper counter would do the job:

#function for  count the frequency
freqLet <- function(x) {
word <- tolower(unlist(strsplit(x,"")))
word_table <- table(word)
ans <- word_table[names(word_table)]
}

getFreq <- function(vect) {
df <- data.frame(word=as.character(), stringsAsFactors = FALSE)
for (i in 1:length(vect)) {
df[i,1] <- as.character(word_a_number(i))
a <<- freqLet(df\$word)
}
return(a)
}


And once this is loaded into environment, we need to run the complete set:

######### Let's check the complete set of numbers

# Automate the function, get a vector of first 100 numbers
vect <- c(1:100)

#Is there A in first 100 words?
getFreq(vect)


And this returns a table of letter frequencies

There are some gotcha moments

It is useless (and that’s the catch) to write down the words of all the numbers. If you only create a dictionary of non-repetitive words (like the one from 11 to 19 or from 0 to 10), you may check only the in these subsets for presence of letter A

Another one is, if you – by some change – decide to write word as hundred and ten (110), you will get high frequency of letter A, unless you decide to write it as hundred-ten.

And the last gotcha moment is – letter A will be present only as in word AND as a helper word for easier pronunciation.

And really last gotcha – the next natural occurrence of letter A is in word gazillion which is a loooot of billions and yet, very informal word.

As always, code is available in at the Github in same Useless_R_function repository.

Happy R-coding!