Question

我有一些数据：

df <- data.frame(v1 = c('word',NA,'word','word',NA,'word','word',NA,'word','word'), 
                 v1_open = c('word',NA,'word','word',NA,'word','word',NA,'word','word'),
                 v2 = c('word','word',NA,'word','word',NA,'word','word',NA,'word'), 
                 v2_open = c('word','word',NA,'word','word',NA,'word','word',NA,'word'))

我正在使用apply将包含NA的观察值更改为包含1的其他观察值。

df <- t(apply(df,1,function(x){
  ifelse(is.na(x) ,0,1)
}))

返回

      v1 v1_open v2 v2_open
 [1,]  1       1  1       1
 [2,]  0       0  1       1
 [3,]  1       1  0       0
 [4,]  1       1  1       1
 [5,]  0       0  1       1
 [6,]  1       1  0       0
 [7,]  1       1  1       1
 [8,]  0       0  1       1
 [9,]  1       1  0       0
[10,]  1       1  1       1

我想修改apply函数以排除名称中包含文本'_open'的列，从而导致：

      v1 v1_open v2 v2_open
 [1,]  1    word  1    word  
 [2,]  0    NA    1    word  
 [3,]  1    word  0    NA    
 [4,]  1    word  1    word  
 [5,]  0    NA    1    word  
 [6,]  1    word  0    NA    
 [7,]  1    word  1    word  
 [8,]  0    NA    1    word  
 [9,]  1    word  0    NA    
[10,]  1    word  1    word

这怎么办？

Answer 1

可以做到：

library(dplyr)

df %>%
  mutate_at(vars(-contains("_open")),
            ~ +(!is.na(.)))

输出：

   v1 v1_open v2 v2_open
1   1    word  1    word
2   0    <NA>  1    word
3   1    word  0    <NA>
4   1    word  1    word
5   0    <NA>  1    word
6   1    word  0    <NA>
7   1    word  1    word
8   0    <NA>  1    word
9   1    word  0    <NA>
10  1    word  1    word

Answer 2

我们可以将is.na直接应用于data.frame列的子集，而无需进行任何循环，然后更新列

nm1 <- grep("_open", names(df), value = TRUE, invert = TRUE)
df[nm1] <- +(!is.na(df[nm1]))
df
#   v1 v1_open v2 v2_open
#1   1    word  1    word
#2   0    <NA>  1    word
#3   1    word  0    <NA>
#4   1    word  1    word
#5   0    <NA>  1    word
#6   1    word  0    <NA>
#7   1    word  1    word
#8   0    <NA>  1    word
#9   1    word  0    <NA>
#10  1    word  1    word

Answer 3

如果您的列在.*和.*_open之间交替，那么您可以简单地通过TRUE, FALSE将列子集化，即

df[c(TRUE, FALSE)] <- +(!is.na(df[c(TRUE, FALSE)]))

df
#   v1 v1_open v2 v2_open
#1   1    word  1    word
#2   0    <NA>  1    word
#3   1    word  0    <NA>
#4   1    word  1    word
#5   0    <NA>  1    word
#6   1    word  0    <NA>
#7   1    word  1    word
#8   0    <NA>  1    word
#9   1    word  0    <NA>
#10  1    word  1    word

根据列名称从应用功能中排除列

3 个答案: