Visualizing 2-letter combinations

Question

The answers to this question on SO returned a set of approximately 125 one- to two-letter names: http://stackoverflow.com/questions/6979630/what-1-2-letter-object-names-conflict-with-existing-r-objects

  [1] "Ad" "am" "ar" "as" "bc" "bd" "bp" "br" "BR" "bs" "by" "c"  "C" 
 [14] "cc" "cd" "ch" "ci" "CJ" "ck" "Cl" "cm" "cn" "cq" "cs" "Cs" "cv"
 [27] "d"  "D"  "dc" "dd" "de" "df" "dg" "dn" "do" "ds" "dt" "e"  "E" 
 [40] "el" "ES" "F"  "FF" "fn" "gc" "gl" "go" "H"  "Hi" "hm" "I"  "ic"
 [53] "id" "ID" "if" "IJ" "Im" "In" "ip" "is" "J"  "lh" "ll" "lm" "lo"
 [66] "Lo" "ls" "lu" "m"  "MH" "mn" "ms" "N"  "nc" "nd" "nn" "ns" "on"
 [79] "Op" "P"  "pa" "pf" "pi" "Pi" "pm" "pp" "ps" "pt" "q"  "qf" "qq"
 [92] "qr" "qt" "r"  "Re" "rf" "rk" "rl" "rm" "rt" "s"  "sc" "sd" "SJ"
[105] "sn" "sp" "ss" "t"  "T"  "te" "tr" "ts" "tt" "tz" "ug" "UG" "UN"
[118] "V"  "VA" "Vd" "vi" "Vo" "w"  "W"  "y"

And R import code:

nms <- c("Ad","am","ar","as","bc","bd","bp","br","BR","bs","by","c","C","cc","cd","ch","ci","CJ","ck","Cl","cm","cn","cq","cs","Cs","cv","d","D","dc","dd","de","df","dg","dn","do","ds","dt","e","E","el","ES","F","FF","fn","gc","gl","go","H","Hi","hm","I","ic","id","ID","if","IJ","Im","In","ip","is","J","lh","ll","lm","lo","Lo","ls","lu","m","MH","mn","ms","N","nc","nd","nn","ns","on","Op","P","pa","pf","pi","Pi","pm","pp","ps","pt","q","qf","qq","qr","qt","r","Re","rf","rk","rl","rm","rt","s","sc","sd","SJ","sn","sp","ss","t","T","te","tr","ts","tt","tz","ug","UG","UN","V","VA","Vd","vi","Vo","w","W","y")

Since the point of the question was to come up with a memorable list of object names to avoid, and most humans are not so good at making sense out of a solid block of text, I would like to visualize this.

Unfortunately I'm not exactly certain of the best way to do this. I had thought of something like a stem-and-leaf plot, only since there are no repeated values each "leaf" was placed in the appropriate column rather than being left justified. Or a wordcloud-style adaptation where letters are sized according to its prevalence.

How might this be most clearly and efficiently be visualized?

Visualizations which do either of the following fit in the spirit of this question:

• Primary goal: Enhance the memorizability of the set of names by revealing patterns in the data

• Alternate goal: Highlight interesting features of the set of names (e.g. which help visualize the distribution, most common letters, etc.)

Answers in R are preferred, but all interesting ideas are welcome.

Ignoring the single-letter names is allowed, since those are easier to just give as a separate list.

Answer 1

Here is a start: visualize these on a grid of first and second letters:

combi <- c("Ad", "am", "ar", "as", "bc", "bd", "bp", "br", "BR", "bs", 
"by", "c",  "C",  "cc", "cd", "ch", "ci", "CJ", "ck", "Cl", "cm", "cn", 
"cq", "cs", "Cs", "cv", "d",  "D",  "dc", "dd", "de", "df", "dg", "dn", 
"do", "ds", "dt", "e",  "E",  "el", "ES", "F",  "FF", "fn", "gc", "gl", 
"go", "H",  "Hi", "hm", "I",  "ic", "id", "ID", "if", "IJ", "Im", "In", 
"ip", "is", "J",  "lh", "ll", "lm", "lo", "Lo", "ls", "lu", "m",  "MH", 
"mn", "ms", "N",  "nc", "nd", "nn", "ns", "on", "Op", "P",  "pa", "pf", 
"pi", "Pi", "pm", "pp", "ps", "pt", "q",  "qf", "qq", "qr", "qt", "r",  
"Re", "rf", "rk", "rl", "rm", "rt", "s",  "sc", "sd", "SJ", "sn", "sp", 
"ss", "t",  "T",  "te", "tr", "ts", "tt", "tz", "ug", "UG", "UN", "V",  
"VA", "Vd", "vi", "Vo", "w",  "W",  "y")

df <- data.frame (first = factor (gsub ("^(.).", "\\1", combi), 
                                  levels = c (LETTERS, letters)),
                  second = factor (gsub ("^.", "", combi), 
                                  levels = c (LETTERS, letters)),
                  combi = combi))

library(ggplot2)
ggplot (data = df, aes (x = first, y = second)) + 
   geom_text (aes (label = combi), size = 3) + 
   ## geom_point () +
   geom_vline (x = 26.5, col = "grey") + 
   geom_hline (y = 26.5, col = "grey")

(was: )

ggplot (data = df, aes (x = second)) + geom_histogram ()

ggplot (data = df, aes (x = first)) + geom_histogram ()

I gather:

• of the one letter names,

  • fortunately i, j, k, and l are available (so I can index up to 4d arrays)
  • unfortunately t (time), c (concentration) are gone. So are m (mass), V (volume) and F (force). No radius r nor diameter d.
  • I can have pressure (p), amount of substance (n), and length l, though.

  • Maybe I'll have to change to greek names: ε is OK, but then shouldn't

    π <- pi
    

    ?

• I can have whatever lowerUPPER name I want.

• In general, starting with an upper case letter is a safer bet than lower case.

• don't start with c or d

Answer 2

Ok, here's my very quick take on a "periodic table"-like visualization, based on the SO question and the comments of the others. The main problem is the big difference in number of variables between packages, which kind of hinders the visualization... I realize this is very rough, so please feel free to change it as you wish.

Here is the current output (from my package list)

And the code

# Load all the installed packages
lapply(rownames(installed.packages()), require, 
       character.only = TRUE)
# Find variables of length 1 or 2
one_or_two <- unique(apropos("^[a-zA-Z]{1,2}$"))
# Find which package they come from
packages <- lapply(one_or_two, find)
# Some of the variables may belong to multiple packages, so determine the length 
# of each entry in packages and duplicate the names accordingly
lengths <- unlist(lapply(packages, length))
var.data <- data.frame(var = rep(one_or_two, lengths), 
                   package = unlist(packages))

Now, we have a data frame like this:

> head(var.data, 10)
   var           package
1   ar     package:stats
2   as   package:methods
3   BD    package:fields
4   bs      package:VGAM
5   bs   package:splines
6   by      package:base
7    c      package:base
8    C     package:stats
9   cm package:grDevices
10   D     package:stats

We can now split the data by package

 data.split <- split(var.data, var.data$package)

We can see that most variables come from the base and stats package

> unlist(lapply(data.split, nrow))
     package:base  package:datasets    package:fields 
               16                 1                 2 
  package:ggplot2 package:grDevices  package:gWidgets 
                2                 1                 1 
  package:lattice      package:MASS    package:Matrix 
                1                 1                 3 
  package:methods      package:mgcv      package:plyr 
                3                 2                 1 
     package:spam   package:splines     package:stats 
                1                 2                14 
 package:survival     package:utils      package:VGAM 
                1                 2                 4 

Finally, the drawing routine

plot(0, 0, "n", xlim=c(0, 100), ylim=c(0, 120), 
     xaxt="n", yaxt="n", xlab="", ylab="")

side.len.x <- 100 / length(data.split)
side.len.y <- 100 / max(unlist(lapply(data.split, nrow)))
colors <- rainbow(length(data.split), start=0.2, end=0.6)    

for (xcnt in 1:length(data.split))
    {
    posx <- side.len.x * (xcnt-1)

    # Remove "package :" in front of the package name
    pkg <- unlist(strsplit(as.character(data.split[[xcnt]]$package[1]), ":"))
    pkg <- pkg[2]

    # Write the package name
    text(posx + side.len.x/2, 102, pkg, srt=90, cex=0.95, adj=c(0, 0))

    for (ycnt in 1:nrow(data.split[[xcnt]]))
        {
        posy <- side.len.y * (ycnt-1)
        rect(posx, posy, posx+side.len.x*0.85, posy+side.len.y*0.9, col = colors[xcnt])
        text(posx+side.len.x/2, posy+side.len.y/2, data.split[[xcnt]]$var[ycnt], cex=0.7)
        }
    }

Answer 3

Here's a letter-based histogram. Considered sizing the first letters by number, but decided against since that's already encoded in the vertical component.

# "Load" data
nms <- c("Ad","am","ar","as","bc","bd","bp","br","BR","bs","by","c","C","cc","cd","ch","ci","CJ","ck","Cl","cm","cn","cq","cs","Cs","cv","d","D","dc","dd","de","df","dg","dn","do","ds","dt","e","E","el","ES","F","FF","fn","gc","gl","go","H","Hi","hm","I","ic","id","ID","if","IJ","Im","In","ip","is","J","lh","ll","lm","lo","Lo","ls","lu","m","MH","mn","ms","N","nc","nd","nn","ns","on","Op","P","pa","pf","pi","Pi","pm","pp","ps","pt","q","qf","qq","qr","qt","r","Re","rf","rk","rl","rm","rt","s","sc","sd","SJ","sn","sp","ss","t","T","te","tr","ts","tt","tz","ug","UG","UN","V","VA","Vd","vi","Vo","w","W","y") #all names
two_in_base <- c("ar", "as", "by", "cm", "de", "df", "dt", "el", "gc", "gl", "if", "Im", "is", "lh", "lm", "ls", "pf", "pi", "pt", "qf", "qr", "qt", "Re", "rf", "rm", "rt", "sd", "ts", "vi") # 2-letter names in base R
vowels <- c("a","e","i","o","u")
vowels <- c( vowels, toupper(vowels) )

# Constants
yoffset.singles <- 3

# Define a function to give us consistent X coordinates
returnX <- function(vec) {
  sapply(vec, function(x) seq(length(all.letters))[ x == all.letters ] )
}

# Make df of 2-letter names
combi <- nms[ sapply( nms, function(x) nchar(x)==2 ) ]
combidf <- data.frame( first = substr(combi,1,1), second=substr(combi,2,2) )
library(plyr)
combidf <- arrange(combidf,first,second)

# Add vowels
combidf$first.vwl <- (combidf$first %in% vowels)
combidf$second.vwl <- (combidf$second %in% vowels)

# Flag items only in base R
combidf$in_base <- paste(combidf$first,combidf$second,sep="") %in% two_in_base

# Create a data.frame to hold our plotting information for the first letters
combilist <- dlply(combidf,.(first),function(x) x$second)
combi.first <- data.frame( first = names(combilist), n = sapply(combilist,length) ,stringsAsFactors=FALSE )
combi.first$y <- 0
all.letters <-  c(letters,LETTERS) # arrange(combi.first,desc(n))$first to go in order of prevalence (which may break the one-letter name display)
combi.first$x <- returnX( combi.first$first )

# Create a data.frame to hold plotting information for the second letters
combidf$x <- returnX( combidf$first )
combidf$y <- unlist( by( combidf$second, combidf$first, seq_along ) )

# Make df of 1-letter names
sngldf <- data.frame( sngl = nms[ sapply( nms, function(x) nchar(x)==1 ) ] )
singles.y <- max(combidf$y) + yoffset.singles
sngldf$y <- singles.y
sngldf$x <- returnX( sngldf$sngl )

# Plot
library(ggplot2)
ggplot(data=combidf, aes(x=x,y=y) ) +
  geom_text(aes( label=second, size=3, colour=combidf$in_base ), position=position_jitter(w=0,h=.25)) +
  geom_text( data=combi.first, aes( label=first, x=x, y=y, size=4 ) ) +
  geom_text( data=sngldf, aes( label=sngl, x=x, y=y, size=4 ) ) +
  scale_size(name="Order (2-letter names)",limits=c(1,4),breaks=c(1,2),labels=c("Second","First")) +
  scale_x_continuous("",breaks=c(13,39),labels=c("lower","UPPER")) +
  scale_y_continuous("",breaks=c(0,5,singles.y),labels=c("First letter of two-letter names","Second letter of two-letter names","One-letter names") ) +
  coord_equal(1.5) +
  labs( colour="In base R" )

Answer 4

Periodic Table for 100, Alex. I don't have code for it, though. :(

One might think that a "periodic table" package might already exist in CRAN. The idea of a coloring scheme and layout of such data could be interesting and useful.

These could be colored by package and sorted vertically by frequency, e.g. in a sample of code on CRAN or as they appear in one's local codebase.

Answer 5

The first two pages in chapter 2 of MacKay's ITILA has nice diagrams showing the conditional probabilities of all character pairings in the English language. You may find it of use.

I'm embarrassed to say that I don't remember what program was used to produce them.