r - 在 R 中为多个因子列创建频率表

Question

我是 R 新手。我正在为我的工作编写一份关于常用函数/特性语法的单独手册。我的示例数据框如下:

x.sample <-
structure(list(Q9_A = structure(c(5L, 3L, 5L, 3L, 5L, 3L, 1L, 
5L, 5L, 5L), .Label = c("Impt", "Neutral", "Not Impt at all", 
"Somewhat Impt", "Very Impt"), class = "factor"), Q9_B = structure(c(5L, 
5L, 5L, 3L, 5L, 5L, 3L, 5L, 3L, 3L), .Label = c("Impt", "Neutral", 
"Not Impt at all", "Somewhat Impt", "Very Impt"), class = "factor"), 
Q9_C = structure(c(3L, 5L, 5L, 3L, 5L, 5L, 3L, 5L, 5L, 3L
), .Label = c("Impt", "Neutral", "Not Impt at all", "Somewhat Impt", 
"Very Impt"), class = "factor")), .Names = c("Q9_A", "Q9_B", 
"Q9_C"), row.names = c(NA, 10L), class = "data.frame")

> x.sample
          Q9_A            Q9_B            Q9_C
1        Very Impt       Very Impt Not Impt at all
2  Not Impt at all       Very Impt       Very Impt
3        Very Impt       Very Impt       Very Impt
4  Not Impt at all Not Impt at all Not Impt at all
5        Very Impt       Very Impt       Very Impt
6  Not Impt at all       Very Impt       Very Impt
7             Impt Not Impt at all Not Impt at all
8        Very Impt       Very Impt       Very Impt
9        Very Impt Not Impt at all       Very Impt
10       Very Impt Not Impt at all Not Impt at all

我的原始数据框有 21 列。

如果我想求均值(将其视为序数变量):

> sapply(x.sample,function(x) mean(as.numeric(x), na.rm=TRUE))
Q9_A Q9_B Q9_C 
 4.0  4.2  4.2

我想为我的数据框中的所有变量列出一个频率表。我搜索了互联网和许多论坛，发现最接近的命令是使用 sapply。但当我这样做时，它给出了全 0。

> sapply(x.sample,function(x) table(factor(x.sample, levels=c("Not Impt at all", "Somewhat Impt",            "Neutral", "Impt", "Very Impt"), ordered=TRUE)))
                Q9_A Q9_B Q9_C
Not Impt at all    0    0    0
Somewhat Impt      0    0    0
Neutral            0    0    0
Impt               0    0    0
Very Impt          0    0    0

问题如何利用 sapply 根据上表为数据框中的所有列(即因子)制作频率图表？

PS 很抱歉，如果这看起来很琐碎，但我已经搜索了 2 天但没有找到答案，并尝试了所有可能的组合。可能是我搜索的不够仔细=(

非常感谢。

Answer 1

你就快到了。只要对你的函数做一个小小的改变就可以实现这一点。 function(x) ... 中的 x 需要传递给 table() 调用:

levs <- c("Not Impt at all", "Somewhat Impt", "Neutral", "Impt", "Very Impt")
sapply(x.sample, function(x) table(factor(x, levels=levs, ordered=TRUE)))

对代码进行一些重新调整也可能使其更易于阅读:

sapply(lapply(x.sample,factor,levels=levs,ordered=TRUE), table)

#                Q9_A Q9_B Q9_C
#Not Impt at all    3    4    4
#Somewhat Impt      0    0    0
#Neutral            0    0    0
#Impt               1    0    0
#Very Impt          6    6    6