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I'm eager to know,

  • how many package names on CRAN have two, three, N characters?
  • which combinations have not yet been used ("unpoppler")
  • how many package names use full-caps, or camelCase?
  • how many package names end in 2?

I think it might reveal some interesting facts.

Edit: bonus points for animated graphics showing the time-evolution of CRAN packages.

baptiste
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  • It's an interesting question, but I'm not sure if it's really SO-style. Should just be a matter of scraping the names off http://cran.r-project.org/web/packages/available_packages_by_name.html and running a few regexes on them, though. – Owen Sep 11 '11 at 23:28
  • @Owen I agree on both points, but curiosity's sake won me over. – baptiste Sep 11 '11 at 23:36
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    Scraping is overkill: just look at `myList[,"Package"]` where `myList <- available.packages()`. This list is subject to change every day. – Iterator Sep 12 '11 at 00:03
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    As for the time evolution, here is a database with API: https://github.com/metacran/crandb (Disclaimer: I am the author of it.) It has some incorrect data, in particular dates of archivals are often wrong. Some of these I can and will fix, but for some, there is just no information available AFAIK. – Gabor Csardi Sep 09 '14 at 02:00
  • you're doing an outstanding work with metacran – baptiste Sep 09 '14 at 10:55

4 Answers4

14

A better way than scraping a web page to get the names of packages is to use the available.packages() function and process those results. available.packages() returns a matrix contains details of all packages available (but is filtered by default — see the Details section of ?available.packages for more).

pkgs <- available.packages(filters = "duplicates")
nameCount <- unname(nchar(pkgs[, "Package"]))
table(nameCount)

> table(nameCount)
nameCount
  2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21 
 32 311 374 360 434 445 368 277 199 132  99  56  56  43  22  19  18   2  12   8 
 22  24  25  31 
  5   2   1   1

Using nameCount we can select packages with names containing any number of characters without needing to resort to regexp etc:

> unname(pkgs[which(nameCount == 2), "Package"])
 [1] "BB" "bs" "ca" "cg" "dr" "ez" "FD" "ff" "HH" "HI" "iv" "JM" "ks" "M3" "mi"
[16] "np" "oc" "oz" "PK" "PP" "qp" "QT" "RC" "rv" "Rz" "sm" "sn" "sp" "st" "SV"
[31] "tm" "wq"
Gavin Simpson
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10

here's one shot based on various suggestions.

 packages <- available.packages()[,'Package']

 ggplot(data.frame(n = nchar(packages))) +
   geom_histogram(aes(n), binwidth=1)

histogram

 all <- length(packages)
 ## 3168
 up <- sum(toupper(packages) == packages)
 ## 262
 low <- sum(tolower(packages) == packages)
 ## 1697
 pie(c(up, low, all-up-low), labels=c("UPPERCASE","lowercase","cAmElCaSe"))

pie

 let <- sapply(sapply(letters, grep, tolower(packages)), length)
 barplot(let)

barplot 

 length(packages[grep("2$", packages, perl=TRUE)])
 # 29
baptiste
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5

Here is a short piece of code to answer some questions. I will keep adding to my answer when I find time.

library(XML); library(ggplot2);

url = 'http://cran.r-project.org/web/packages/available_packages_by_name.html'
packages = readHTMLTable(url, stringsAsFactors = F)[[1]][-1,]

# histogram of number of characters in package name
qplot(nchar(V1), data = packages)
Ramnath
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  • Nice, `RthroughExcelWorkbooksInstaller2` anyone? :) – baptiste Sep 11 '11 at 23:52
  • +1, though this might slightly overstate the number with two characters as it is picking up some `NA` values (e.g. `packages[128,]`) when the first letter changes, perhaps from html section names. – Henry Sep 12 '11 at 06:49
  • yes. but i think the `available.packages` solution is more elegant and robust. – Ramnath Sep 12 '11 at 13:47
1

Make a vector of all the packages using

myList <- available.packages()[,'Package']

Then you can analyze however you want. For example, a list of packages with just two character names

myList[grep('^..$', myList)]
adamleerich
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