Question

我有一张看起来像的桌子：

> head(test,10)
# A tibble: 10 x 16
   Question_1 Question_2 Question_3 Question_4 Question_5 Question_6 Question_7 Question_8 Question_9
   <chr>      <chr>      <chr>      <chr>      <chr>      <chr>      <chr>      <chr>      <chr>     
 1 B          C          C          E          C          A          C          E          C         
 2 C          C          C          B          C          A          E          D          C         
 3 B          C          C          E          C          A          C          E          C         
 4 C          C          C          D          C          A          C          D          C         
 5 B          B          C          B          A          A          A          D          C         
 6 C          C          C          E          BLANK      A          C          E          C         
 7 C          C          C          E          C          A          E          E          C         
 8 B          C          C          E          C          A          C          D          C         
 9 C          C          C          E          C          A          C          D          C         
10 D          C          E          B          C          A          A          D          C

并想转置，所以我每行有一个问题，在A,B,C,D,E,BLANKS的6个单独列中计数。

Answer 1

我们可以将gather转换为“长”格式，获取“键”，“值”列的count，并spread将其转换为“宽”格式

library(tidyverse)
gather(test) %>% 
    count(key, value) %>%
    spread(value, n, fill = 0)

或使用melt/dcast

library(data.table)
dcast(melt(setDT(test), measure = patterns("^Question")), variable ~ value)

或者在base R中通过复制'test'的列名而无循环，而unlist复制'test'并获得table

table(names(test)[col(test)], unlist(test))       
#              A  B BLANK  C  D  E
#  Question_1  0  4     0  5  1  0
#  Question_2  0  1     0  9  0  0
#  Question_3  0  0     0  9  0  1
#  Question_4  0  3     0  0  1  6
#  Question_5  1  0     1  8  0  0
#  Question_6 10  0     0  0  0  0
#  Question_7  2  0     0  6  0  2
#  Question_8  0  0     0  0  6  4
#  Question_9  0  0     0 10  0  0

注意：无需循环播放

基准

df2 <- test[rep(seq_len(nrow(test)), 1e5), ]

system.time({
vals <- unique(unlist(df2))
t(sapply(df2, function(x) table(factor(x, levels = vals))))


})
# user  system elapsed 
 # 6.987   0.367   7.293 

 system.time({
 table(names(df2)[col(df2)], unlist(df2))  

 })
# user  system elapsed 
#   6.355   0.407   6.720 


system.time({
gather(df2) %>% 
    count(key, value) %>%
    spread(value, n, fill = 0)

})
# user  system elapsed 
# 0.567   0.125   0.695


system.time({
dcast(melt(setDT(df2), measure = patterns("^Question")), variable ~ value)

})
#  user  system elapsed 
#  0.789   0.018   0.195

数据

test <- structure(list(Question_1 = c("B", "C", "B", "C", "B", "C", "C", 
"B", "C", "D"), Question_2 = c("C", "C", "C", "C", "B", "C", 
"C", "C", "C", "C"), Question_3 = c("C", "C", "C", "C", "C", 
"C", "C", "C", "C", "E"), Question_4 = c("E", "B", "E", "D", 
"B", "E", "E", "E", "E", "B"), Question_5 = c("C", "C", "C", 
"C", "A", "BLANK", "C", "C", "C", "C"), Question_6 = c("A", "A", 
"A", "A", "A", "A", "A", "A", "A", "A"), Question_7 = c("C", 
"E", "C", "C", "A", "C", "E", "C", "C", "A"), Question_8 = c("E", 
"D", "E", "D", "D", "E", "E", "D", "D", "D"), Question_9 = c("C", 
"C", "C", "C", "C", "C", "C", "C", "C", "C")), 
class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10"))

Answer 2

R的基本技巧可能是获取数据帧的所有唯一值，并使用sapply并计算列中每个值的频率。

vals <- unique(unlist(test))
t(sapply(test, function(x) table(factor(x, levels = vals))))

#           B  C D E  A BLANK
#Question_1 4  5 1 0  0     0
#Question_2 1  9 0 0  0     0
#Question_3 0  9 0 1  0     0
#Question_4 3  0 1 6  0     0
#Question_5 0  8 0 0  1     1
#Question_6 0  0 0 0 10     0
#Question_7 0  6 0 2  2     0
#Question_8 0  0 6 4  0     0
#Question_9 0 10 0 0  0     0

在R中交换行和列

2 个答案:

基准

数据