汇总多个回答问题 [英] Tabulating multiple response questions

问题描述

想象一下，我有一个问题，其中有四个选项，而受访者可以选择零或这四个选项的任意组合.这些变量分别命名为A，B，C和D，并且响应存储在data.frame中，如下所示.

Imagine that I have a question for which there are four options, and a respondent can select zero or any combination of the four. The variables are named A, B, C, and D and the responses are stored in a data.frame as below.

set.seed(1)
dat = data.frame(A = sample(c(0, 1), 20, replace=TRUE), 
                 B = sample(c(0, 1), 20, replace=TRUE), 
                 C = sample(c(0, 1), 20, replace=TRUE),
                 D = sample(c(0, 1), 20, replace=TRUE))

我可以通过以下方法将响应的组合制成表格(例如，单独使用A或A + B或C + D等等，有多少响应):

I can tabulate the combination of responses (for example, how many responded with A alone, or A+B, or C+D, and so on) by doing the following:

data.frame(table(dat))
#    A B C D Freq
# 1  0 0 0 0    2
# 2  1 0 0 0    2
# 3  0 1 0 0    0
# 4  1 1 0 0    1
# 5  0 0 1 0    1
# 6  1 0 1 0    3
# 7  0 1 1 0    0
# 8  1 1 1 0    2
# 9  0 0 0 1    0
# 10 1 0 0 1    2
# 11 0 1 0 1    1
# 12 1 1 0 1    1
# 13 0 0 1 1    2
# 14 1 0 1 1    0
# 15 0 1 1 1    3
# 16 1 1 1 1    0

我现在想创建一个新列，以显示此输出表示的字母组合.例如，第4行代表A + B响应的计数，第14行代表A + C + D响应的计数.

I would like to now create a new column that shows the letter combination that is being represented by this output. For example, row 4 represents the count of A+B responses, and row 14 represents the count of A+C+D responses.

我认为apply函数之一在这里很有用，但是我不确定如何进行.

I think that one of the apply functions would be useful here, but I'm not sure how to proceed.

推荐答案

dat.t <- data.frame(table(dat))    
dat.t$combn <- apply(dat.t[,1:4] == 1, 1, function(x) paste(names(dat)[x], collapse=' + '))

> dat.t
   A B C D Freq         combn
1  0 0 0 0    2              
2  1 0 0 0    2             A
3  0 1 0 0    0             B
4  1 1 0 0    1         A + B
5  0 0 1 0    1             C
6  1 0 1 0    3         A + C
7  0 1 1 0    0         B + C
8  1 1 1 0    2     A + B + C
9  0 0 0 1    0             D
10 1 0 0 1    2         A + D
11 0 1 0 1    1         B + D
12 1 1 0 1    1     A + B + D
13 0 0 1 1    2         C + D
14 1 0 1 1    0     A + C + D
15 0 1 1 1    3     B + C + D
16 1 1 1 1    0 A + B + C + D
> 

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