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Quick estimate of gender distribution of CRAN package maintainers
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library(miniCRAN) | |
library(gender) | |
library(stringr) | |
# Get package description data | |
# This took about an hour to run, so you can load the data directly below | |
# pkgs <- available.packages("http://cran.rstudio.com/src/contrib") | |
# desc <- getCranDescription(pkgs, repos = c(CRAN="http://cran.rstudio.com")) | |
desc <- read.csv("http://www.stat.berkeley.edu/~ledell/data/RStudioCRAN_pkgDesc_20141216.csv") | |
# Grab first name of maintainers | |
# Strip whitespace, split on a single space to get first name and remove punctuation | |
firstname <- function(maintainer) { | |
gsub("[[:punct:]]", "", str_split(str_trim(maintainer), " ", n = 2)[[1]][1]) | |
} | |
maintainers <- sapply(desc$Maintainer, firstname) | |
name_freq <- table(maintainers) | |
# Predicted gender of unique maintainer names (1970 seemed like a good median birth year) | |
# This will be more accurate for U.S. names | |
# This takes a while to run (approx 35 mins) | |
pred_genders <- sapply(names(name_freq), function(x) gender(x, years = c(1970))$gender) | |
# Fix one important miscoding of gender! | |
pred_genders[which(names(pred_genders) == "Hadley")] <- "male" | |
# Estimated gender distribution of CRAN package maintainers | |
gender_table <- table(unlist(sapply(1:length(pred_genders), | |
function(i) rep(pred_genders[i], name_freq[i]))), useNA = "always") | |
#female male <NA> | |
# 539 3731 1835 | |
# Proportions | |
round(gender_table/length(maintainers), 3) | |
#female male <NA> | |
# 0.088 0.611 0.301 |
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