Summarize distribution of factors in R data frame - r

Question

Say I have a data.frame like this:
X1 X2 X3
1 A B A
2 A C B
3 B A B
4 A A C
I would like to count the occurrences of A, B, C, etc. in each column, and return the result as
A_count B_count C_count
X1 3 1 0
X2 2 1 1
X3 1 2 1
I'm sure this question has a thousand duplicates, but I can't seem to find an answer that works for me :(
By running
apply(mydata, 2, table)
I get something like
$X1
B A
1 3
$X2
A C B
2 1 1
But it's not exactly what I want and if I try to build it back into a data frame, it doesn't work because I don't get the same number of columns for every row (like $X1 above where there are no C's).
What am I missing?
Many thanks!

Answer 1

You can refactor to include the factor levels common to each column, then tabulate. I would also recommend using lapply() instead of apply(), as apply() is for matrices.
df <- read.table(text = "X1 X2 X3
1 A B A
2 A C B
3 B A B
4 A A C", h=T)
do.call(
rbind,
lapply(df, function(x) table(factor(x, levels=levels(unlist(df)))))
)
# A B C
# X1 3 1 0
# X2 2 1 1
# X3 1 2 1

Answer 2

Suppose your data frame is x, I would simply do:
do.call(rbind, tapply(unlist(x, use.names = FALSE),
rep(1:ncol(x), each = nrow(x)),
table))
# A B C
#1 3 1 0
#2 2 1 1
#3 1 2 1
Benchmarking
# a function to generate toy data
# `k` factor levels
# `n` row
# `p` columns
datsim <- function(n, p, k) {
as.data.frame(replicate(p, sample(LETTERS[1:k], n, TRUE), simplify = FALSE),
col.names = paste0("X",1:p), stringsAsFactors = TRUE)
}
# try `n = 100`, `p = 500` and `k = 3`
x <- datsim(100, 500, 3)
## DirtySockSniffer's answer
system.time(do.call(rbind, lapply(x, function(u) table(factor(u, levels=levels(unlist(x)))))))
# user system elapsed
# 21.240 0.068 21.365
## my answer
system.time(do.call(rbind, tapply(unlist(x, use.names = FALSE), rep(1:ncol(x), each = nrow(x)), table)))
# user system elapsed
# 0.108 0.000 0.111
Dirty's answer can be improved, by:
## improved DirtySockSniffer's answer
system.time({clevels <- levels(unlist(x, use.names = FALSE));
do.call(rbind, lapply(x, function(u) table(factor(u, levels=clevels))))})
# user system elapsed
# 0.108 0.000 0.108
Also consider user20650's answer:
## Let's try a large `n`, `p`, `k`
x <- datsim(200, 5000, 5)
system.time(t(table(stack(lapply(x, as.character)))))
# user system elapsed
# 0.592 0.052 0.646
While my answer does:
system.time(do.call(rbind, tapply(unlist(x, use.names = FALSE), rep(1:ncol(x), each = nrow(x)), table)))
# user system elapsed
# 1.844 0.056 1.904
Improved Dirty's answer does:
system.time({clevels <- levels(unlist(x, use.names = FALSE));
do.call(rbind, lapply(x, function(u) table(factor(u, levels=clevels))))})
# user system elapsed
# 1.240 0.012 1.263