使用 mutate 和 case_when 遍歷列

Question

我想遍歷許多列並在某些條件下替換值。 例如，如果 disease=0 且treatment=1，則將處理單元替換為 99。

數據：

df <- data.frame(id=1:5,
                 disease1=c(1,1,0,0,0),
                 treatment1=c(1,0,1,0,0),
                 outcome1=c("survived", "died", "survived", NA,NA),
                 disease2=c(1,1,0,0,0),
                 treatment2=c(1,0,1,0,0),
                 outcome2=c("survived", "died", "survived", NA,NA))

> df
  id disease1 treatment1 outcome1 disease2 treatment2 outcome2
1  1        1          1 survived        1          1 survived
2  2        1          0     died        1          0     died
3  3        0          1 survived        0          1 survived
4  4        0          0     <NA>        0          0     <NA>
5  5        0          0     <NA>        0          0     <NA>

對於單個列， case_when 效果很好：

df %>% mutate(treatment=case_when((disease1!=1&treatment1==1)~99, TRUE~treatment1))

對於多列，以下在基礎 R 中有效：

for(i in 1:2) {
  df[,paste0("treatment",i)] <- ifelse(df[,paste0("disease",i)]!=1&df[,paste0("treatment",i)]==1,99, df[,paste0("treatment",i)])
}

我正在尋找一種在 tidyverse 中完成這一切的方法，但我很難找到正確的食譜。 先感謝您。

Answer 1

也許考慮使用pivot_longer輸入長格式，然后在多個列之間進行mutate會更容易。 如果所有疾病都應放在一個列中（並且對於 1 列中的治療和 1 列中的結果相同），這將是一種“更整潔”的方法。

library(tidyverse)

df %>%
  pivot_longer(cols = -id, names_to = c(".value", "number"), names_pattern = "(\\w+)(\\d+)") %>%
  mutate(treatment = ifelse(disease == 0 & treatment == 1, 99, treatment))

Answer 2

帶有names_sep的pivot_longer中的case_when選項

library(dplyr)
library(tidyr)
pivot_longer(df, cols = -id, names_to = c('.value', 'number'), 
     names_sep="(?<=[a-z])(?=[0-9])") %>% 
    mutate(treatment = replace(treatment, !disease & treatment == 1, 99))
# A tibble: 10 x 5
#      id number disease treatment outcome 
#   <int> <chr>    <dbl>     <dbl> <chr>   
# 1     1 1            1         1 survived
# 2     1 2            1         1 survived
# 3     2 1            1         0 died    
# 4     2 2            1         0 died    
# 5     3 1            0        99 survived
# 6     3 2            0        99 survived
# 7     4 1            0         0 <NA>    
# 8     4 2            0         0 <NA>    
# 9     5 1            0         0 <NA>    
#10     5 2            0         0 <NA>