Frequency of values per column in table

Question

What is a good way to get the independent frequency counts of multiple columns using dplyr? I want to go from a table of values:

# A tibble: 7 x 4
      a     b     c     d
  <int> <int> <int> <int>
1     1     2     1     3
2     1     2     1     3
3     2     2     5     3
4     3     2     4     3
5     3     3     2     3
6     5     3     4     3
7     5     4     2     1

to a frequency table like so:

# A tibble: 5 x 5
      x   a_n   b_n   c_n   d_n
  <int> <int> <int> <int> <int>
1     1     2     0     2     1
2     2     1     4     2     0
3     3     2     2     0     6
4     4     0     1     2     0
5     5     2     0     1     0

I'm still trying to get my head around dplyr, but it seems like this is something it could do. If it is easier to do with an add-on library, that is fine too.

Answer 1

library(dplyr)
library(reshape2)
df %>%
  melt() %>%
  dcast(value ~ variable, fun.aggregate=length)

#   value a b c d
# 1     1 2 0 2 1
# 2     2 1 4 2 0
# 3     3 2 2 0 6
# 4     4 0 1 2 0
# 5     5 2 0 1 0

Data

df <- structure(list(a = c(1L, 1L, 2L, 3L, 3L, 5L, 5L), b = c(2L, 2L, 
2L, 2L, 3L, 3L, 4L), c = c(1L, 1L, 5L, 4L, 2L, 4L, 2L), d = c(3L, 
3L, 3L, 3L, 3L, 3L, 1L)), .Names = c("a", "b", "c", "d"), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7"))

Answer 2

For the same data set that you provided in the question this would be another solution (base-R):

myfreq <- sapply(df, function(x) table(factor(x, levels=unique(unlist(df)), ordered=TRUE)))

Output would be:

> myfreq

#   a b c d 
# 1 2 0 2 1 
# 2 1 4 2 0 
# 3 2 2 0 6 
# 5 2 0 1 0 
# 4 0 1 2 0

Answer 3

library(tidyverse)

dt <- data.frame(a = c(1L, 1L, 2L, 3L, 3L, 5L, 5L), b = c(2L, 2L, 2L, 2L, 3L, 3L, 4L),
                 c = c(1L, 1L, 5L, 4L, 2L, 4L, 2L), d = c(3L, 3L, 3L, 3L, 3L, 3L, 1L))

dt2 <- dt %>%
  mutate(ID = 1:n()) %>%
  gather(Group, x, -ID) %>%
  select(-ID) %>%
  mutate(Group = paste(Group, "n", sep = "_")) %>%
  count(Group, x) %>%
  spread(Group, n, fill = 0L)

Answer 4

Using tabulate in base R:

apply(df,2,function(x) tabulate(x)[min(df):max(df)])

#     a  b c  d
#[1,] 2  0 2  1
#[2,] 1  4 2  0
#[3,] 2  2 0  6
#[4,] 0  1 2 NA
#[5,] 2 NA 1 NA