如何根据R中具有相同字符串的两列的计算添加具有相同逻辑的多个新列答案

【问题标题】：How to add multiple new columns with the same logic based on calculation of two columns with the same string in R如何根据R中具有相同字符串的两列的计算添加具有相同逻辑的多个新列
【发布时间】：2021-02-03 18:46:00
【问题描述】：

我有一张如下所示的表格

data <- data.frame(A = c(1,2,3,4,5),
                   B = c(1,2,3,4,5),
                   C = c(11,12,13,14,15),
                   trans_A = c(NA,1,2,3,4),
                   trans_B = c(NA,1,2,3,4),
                   trans_C = c(NA,11,12,13,14))

我需要使用 difference_A = A - trans_A、difference_B = B - trans_B 等逻辑在此数据框中创建/添加多个新列，新名称为 difference_A、difference_B 等。我该如何在 R 中执行此操作？

我试过了：

new_df[paste("difference", cols, sep = "_")]<- c(NA, new_df[cols]-new_df[paste("trans",cols,sep = "_")], na.rm=TRUE)

它给了我错误消息“列的重复下标”

【问题讨论】：

标签： r

【解决方案1】：

dplyr 和 purrr 的一个解决方案可能是：

map(.x = names(select(data, -starts_with("trans_"))),
    ~ data %>%
     mutate(across(all_of(.x), .names = "difference_{col}") - across(all_of(paste0("trans_", .x))))) %>%
 reduce(full_join)

  A B  C trans_A trans_B trans_C difference_A difference_B difference_C
1 1 1 11      NA      NA      NA           NA           NA           NA
2 2 2 12       1       1      11            1            1            1
3 3 3 13       2       2      12            1            1            1
4 4 4 14       3       3      13            1            1            1
5 5 5 15       4       4      14            1            1            1

【讨论】：

【解决方案2】：

我们可以使用pivot_longer/pivot_wider

library(dplyr)
library(tidyr)
library(stringr)
data %>% 
    rename_at(vars(which(!str_detect(names(.), '_'))), ~ str_c("orig_", .)) %>% 
    mutate(rn = row_number()) %>% 
    pivot_longer(cols = -rn, names_to = c(".value", "group"), names_sep = "_") %>%
    transmute(group = str_c('difference_', group), rn, difference = orig - trans) %>%
    pivot_wider(names_from = group, values_from = difference ) %>% 
    select(-rn) %>% 
    bind_cols(data, .)
#  A B  C trans_A trans_B trans_C difference_A difference_B difference_C
#1 1 1 11      NA      NA      NA           NA           NA           NA
#2 2 2 12       1       1      11            1            1            1
#3 3 3 13       2       2      12            1            1            1
#4 4 4 14       3       3      13            1            1            1
#5 5 5 15       4       4      14            1            1            1

【讨论】：

【解决方案3】：

基本 R 选项

cbind(
  data,
  lapply(
    split.default(
      data,
      gsub(".*(\\w)$", "difference_\\1", names(data))
    ), function(v) do.call("-", v)
  )
)

给予

  A B  C trans_A trans_B trans_C difference_A difference_B difference_C
1 1 1 11      NA      NA      NA           NA           NA           NA
2 2 2 12       1       1      11            1            1            1
3 3 3 13       2       2      12            1            1            1
4 4 4 14       3       3      13            1            1            1
5 5 5 15       4       4      14            1            1            1

【讨论】：

【解决方案4】：

另一个命题：

df <- read.table(header = TRUE, text = "
  A  B  C trans_A trans_B trans_C
1 1  1 11      NA      NA      NA
2 2  2 12       1       1      11
3 3  3 13       2       2      12
4 4  4 14       3       3      13
5 5  5 15       4       4      14
")

for (x in c("A","B","C")) {
  varname <- paste0("difference_",x)
  df[[varname]] <- with(df, get(x) - get(paste0("trans_",x)))
}

df
#>   A B  C trans_A trans_B trans_C difference_A difference_B difference_C
#> 1 1 1 11      NA      NA      NA           NA           NA           NA
#> 2 2 2 12       1       1      11            1            1            1
#> 3 3 3 13       2       2      12            1            1            1
#> 4 4 4 14       3       3      13            1            1            1
#> 5 5 5 15       4       4      14            1            1            1

# Created on 2021-02-03 by the reprex package (v0.3.0.9001

或

for (x in c("A","B","C")) {
  df <- within (df, {
    assign(paste0("difference_",x), get(x) - get(paste0("trans_",x)))
  })
}

df
#>   A B  C trans_A trans_B trans_C difference_A difference_B difference_C
#> 1 1 1 11      NA      NA      NA           NA           NA           NA
#> 2 2 2 12       1       1      11            1            1            1
#> 3 3 3 13       2       2      12            1            1            1
#> 4 4 4 14       3       3      13            1            1            1
#> 5 5 5 15       4       4      14            1            1            1

# Created on 2021-02-03 by the reprex package (v0.3.0.9001)

（与Use dynamic variable names in `dplyr`相关）

问候，

【讨论】：