Frequency count of 5 rankings in R

问题

Say I have 5 items A, B, C, D, E in a questionnaire and got respondents to rank them. The data looks like this,

> df
  rank1 rank2 rank3 rank4 rank5
1     A     B     C     D     E
2     A     C     B     D     E
3     C     A     B     E     D
4     B     A     C     D     E
5     A     B     D     C     E

How do I count the frequency of each rank by item so the output looks like this,

  item rank1 rank2 rank3 rank4 rank5
1    A     3     2     0     0     0
2    B     1     2     2     0     0
3    C     1     1     2     1     0
4    D     0     0     1     3     1
5    E     0     0     0     1     4

回答1:

We can use table after converting to factor using base R

lvls <- sort(unique(unlist(df)))
sapply(df, function(x) table(factor(x, levels =lvls)))
#   rank1 rank2 rank3 rank4 rank5
#A     3     2     0     0     0
#B     1     2     2     0     0
#C     1     1     2     1     0
#D     0     0     1     3     1
#E     0     0     0     1     4

Or calling table only once

table(unlist(df), c(col(df)))
#    1 2 3 4 5
#  A 3 2 0 0 0
#  B 1 2 2 0 0
#  C 1 1 2 1 0
#  D 0 0 1 3 1
#  E 0 0 0 1 4

Or compactly with mtabulate from qdapTools

library(qdapTools)
t(mtabulate(df))

data

df <- structure(list(rank1 = c("A", "A", "C", "B", "A"), rank2 = c("B", 
"C", "A", "A", "B"), rank3 = c("C", "B", "B", "C", "D"), rank4 = c("D", 
"D", "E", "D", "C"), rank5 = c("E", "E", "D", "E", "E")), .Names = c("rank1", 
 "rank2", "rank3", "rank4", "rank5"), class = "data.frame", row.names = c("1", 
 "2", "3", "4", "5"))

回答2:

A tidy approach

Here is a way to solve your problem with functions from the tidyverse

library(tidyr)
library(dplyr)

your_data <- tribble(~"rank1", ~"rank2", ~"rank3", ~"rank4", ~"rank5",
 "A",     "B",     "C",     "D",     "E",
 "A",     "C",     "B",     "D",     "E",
 "C",     "A",     "B",     "E",     "D",
 "B",     "A",     "C",     "D",     "E",
 "A",     "B",     "D",     "C",     "E")

your_data %>%  
  gather(key = rank_number, value = rank) %>% 
  count(rank_number, rank) %>% 
  spread(key = rank_number, value = n, fill = 0)
#> # A tibble: 5 x 6
#>   rank  rank1 rank2 rank3 rank4 rank5
#> * <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 A        3.    2.    0.    0.    0.
#> 2 B        1.    2.    2.    0.    0.
#> 3 C        1.    1.    2.    1.    0.
#> 4 D        0.    0.    1.    3.    1.
#> 5 E        0.    0.    0.    1.    4.

来源：https://stackoverflow.com/questions/49830595/frequency-count-of-5-rankings-in-r

标签

frequency

ranking