Code

twirdle <- function(tweets) {
  
  g   <- "\U1F7E9"
  o   <- "\U1F7E7"
  y   <- "\U1F7E8"
  blu <- "\U1F7E6"
  bla <- "\U2B1B"
  w   <- "\U2B1C"
  
  rx_all   <- paste(g, o, y, blu, bla, w, sep = "|")
  rx_right <- paste(g, o, sep = "|")
  rx_place <- paste(y, blu, sep = "|")
  rx_wrong <- paste(bla, w, sep = "|")
  rx_color <- paste(o, blu, sep = "|")

  tweets$meta <- regexpr(
    "Wordle \\d{1,} [\\d{1}|X]/\\d{1}",
    tweets$text,
    perl = TRUE
  )
  
  tweets$meta <- setNames(
    tweets$meta, 
    ifelse(tweets$meta < 0, FALSE, TRUE)
  )
  
  tweets$meta <- ifelse(
    names(tweets$meta),
    regmatches(tweets$text, tweets$meta),
    NA_character_
  )
  
  tweets <- tweets[!is.na(tweets$meta), ]
  
  tweets$edition <- as.numeric(
    regmatches(
      tweets$meta, 
      regexpr("\\d{1,}", tweets$meta)
    )
  )
  
  tweets$attempts <- regmatches(
    tweets$meta,
    regexpr("[\\d{1}|X](?=/)", tweets$meta, perl = TRUE)
  )
  
  tweets$attempts <- ifelse(
    tweets$attempts == "X",
    NA_character_,
    tweets$attempts
  )
  
  tweets$attempts <- as.numeric(tweets$attempts)
  
  tweets$allowed <- as.numeric(
    regmatches(
      tweets$meta, 
      regexpr("(?<=/)\\d{1}", tweets$meta, perl = TRUE)
    )
  )
  
  tweets$grid <- regmatches(
    tweets$text, 
    gregexpr(rx_all, tweets$text) 
  )
  
  tweets$grid <- lapply(
    tweets$grid, 
    function(x) paste(x, collapse = "")
  )
  
  tweets$colorblind <- ifelse(
    grepl(rx_color, tweets$grid), TRUE, FALSE
  )
  
  tweets$mode <- ifelse(
    grepl(bla, tweets$grid), "dark",
    ifelse(grepl(w, tweets$grid), "light", "unknown")
  )
  
  tweets$grid <- gsub(rx_right, "G", tweets$grid)
  tweets$grid <- gsub(rx_place, "Y", tweets$grid)
  tweets$grid <- gsub(rx_wrong, "-", tweets$grid)
  
  tweets[, c("edition", "attempts", "allowed", "mode",
             "colorblind", "grid", "status_id")]
  
}

I also put the code for the function in a GitHub Gist.

Example

To give an example of twirdle() in action, let’s first grab a small number of tweets using the {rtweet} package by Mike Kearney. I think it’s best to supply the query string of search_tweets() with the word ‘Wordle’ and at least one white or black emoji (signifying an incorrect letter).⁴

tweets <- rtweet::search_tweets(
  q = "Wordle \U2B1B OR \U2B1C",
  n = 10,  # return 10 tweets
  include_rts = FALSE  # no retweets
)

And now we can pass the returned dataframe of tweets to twirdle(). It outputs a row per tweet, but there may be fewer tweets than we asked for because the content doesn’t conform to the output provided by Wordle. Sometimes people add their own comments into the results, disrupting the expected format. There’s also a Spanish version that has ‘(ES)’ in the meta information that we’re going to exclude for our purposes.

Note that anyone who didn’t complete the puzzle in six tries gets a score of ‘X/6’, which is returned as NA_real_ in the attempts column.

twirdle(tweets)

## # A tibble: 7 × 7
##   edition attempts allowed mode  colorblind grid                 status_id      
##     <dbl>    <dbl>   <dbl> <chr> <lgl>      <chr>                <chr>          
## 1     213        5       6 light FALSE      GY--YGGG--GGG--GGG-… 14834445690229…
## 2     213        3       6 dark  FALSE      Y-----YY--GGGGG      14834445680161…
## 3     213        6       6 dark  FALSE      --Y-----YG-G--G--G-… 14834445678988…
## 4     213        6       6 light FALSE      -----Y------Y-G-G--… 14834445674540…
## 5     213       NA       6 light FALSE      ---Y---G-YGGG--GGG-… 14834445662083…
## 6     213        5       6 dark  FALSE      -G------YYGGG--GGG-… 14834445661120…
## 7     213        3       6 dark  FALSE      ----Y--GY--GG--GGG-… 14834445648075…

You could use this to do a number of things, like calculate the mean number of attempts for each day’s puzzle, look for guess patterns at scale, or maybe see whether dark-mode users are more skilled than light-mode users.

Code

eldrow <- function(guess, answer, words = wordle::wordle_dict){
  
  answer <- tolower(answer)
  guess <- toupper(guess)
  
  guess_chars <- strsplit(guess, "")[[1]]
  answer_chars <- strsplit(answer, "")[[1]]
  
  exact <- ifelse(guess_chars == "G", answer_chars, ".") |>
    paste0(collapse = "")
  
  wrong_spot <- gsub(
    "\\.", "", ifelse(guess_chars == "Y", answer_chars, ".")
  )
  
  exact_chars <- regmatches(exact, gregexpr("\\w", exact))[[1]]
  correct_chars_table <- table(
    c(exact_chars, wrong_spot[wrong_spot != ""])
  )
  min_count <- as.vector(correct_chars_table)
  names(min_count) <- names(correct_chars_table)
  
  possibles <- wordle::filter_words(
    words, exact, wrong_spot, min_count, min_count
  )
  
  possibles <- possibles[which(possibles != answer)]
  possibles[order(possibles)]
  
}

I also put the code for this in a GitHub Gist.

Example

So, let’s say someone took three attempts at the word ‘shirt’⁷ and their first and second guesses gave encodings of "YGG--" then "-GGG-".

We can start by passing the answer and the encoding for the prior guess.

guess_2 <- eldrow(guess = "-GGG-", answer = "shirt")
guess_2

##  [1] "chirk" "chirl" "chirm" "chiro" "chirp" "chirt" "chiru" "shire" "shirk"
## [10] "shirs" "third" "thirl" "whirl" "whirs"

Okay, so logicically they could have guessed any of these 14 options before their final, correct guess.

Naturally, we can extrapolate one step further back and infer the earlier potential guesses.

You can iterate over these 14 possible words as the answer argument to eldrow(), setting the guess argument to the encoding for the previous attempt (i.e. "YGG--" was the first guess in our example).

guesses_1_2 <- lapply(
  guess_2,
  \(x) eldrow("YGG--", x)
) |> 
  lapply(\(x) x[which(x != "shirt")]) |>
  setNames(guess_2)

str(guesses_1_2)

## List of 14
##  $ chirk: chr "thick"
##  $ chirl: chr "thick"
##  $ chirm: chr "thick"
##  $ chiro: chr "thick"
##  $ chirp: chr "thick"
##  $ chirt: chr "thick"
##  $ chiru: chr "thick"
##  $ shire: chr [1:22] "chias" "chibs" "chics" "chiks" ...
##  $ shirk: chr [1:22] "chias" "chibs" "chics" "chiks" ...
##  $ shirs: chr [1:22] "chias" "chibs" "chics" "chiks" ...
##  $ third: chr [1:13] "ahint" "chirt" "chits" "shift" ...
##  $ thirl: chr [1:13] "ahint" "chirt" "chits" "shift" ...
##  $ whirl: chr(0) 
##  $ whirs: chr(0)

The output is a list with elements containing the potential first guesses and is named for each of the potential second guesses. So, logically, a possible set of guesses by this imaginary person was ‘shift’ then ‘third’ then ‘shirts’.

Of course, you can eliminate any potential second guesses that failed to yield a potential first guess, like ‘whirl’ in this example.

You’ll also notice a number of the potential first guesses are the same word. The more they appear, the more likely that word is to have been the actual starting guess, I suppose?

words <- c()
for (i in guesses_1_2) {
  words <- c(words, i)
}

counts <- as.data.frame(
  table(words),
  responseName = "n"
)
counts$total <- sum(counts$n)
counts$percent <- round(100 * (counts$n / counts$total), 1)

likeliest <- counts[order(-counts$percent), ]
rownames(likeliest) <- NULL
head(likeliest, 3)

##   words n total percent
## 1 thick 7    99     7.1
## 2 chits 5    99     5.1
## 3 whist 5    99     5.1

So, purely on the basis of this frequency, ‘thick’ was most likely to be the first guess in this contrived example. You might want to consider how likely someone is to actually submit some of these words, given their obscurity.⁸