我们可以使用across，但名称会发生一些变化：

df %>%
  mutate(
    across(ends_with("_SAC"),
           ~ .x - get(sub("_SAC$", "_OVX", cur_column())),
           .names = "chg_{sub('_SAC$','',.col)}")
  )
# # A tibble: 5 × 12
#   Leptin_OVX Leptin_SAC MIP1A_OVX MIP1A_SAC IL4_OVX IL4_SAC IL1B_OVX IL1B_SAC chg_Leptin chg_MIP1A chg_IL4 chg_IL1B
#        <dbl>      <dbl>     <dbl>     <dbl>   <dbl>   <dbl>    <dbl>    <dbl>      <dbl>     <dbl>   <dbl>    <dbl>
# 1    101537.     19786.    199.      153.      84.4       0     20.5      0      -81751.    -46.4    -84.4    -20.5
# 2     34184.    124224.      9.39      0        0         0      0       19.0     90040.     -9.39     0       19.0
# 3     54567.     89540.      6.31      6.31     0         0      0       15.0     34973.      0        0       15.0
# 4     29559.     29335.     31.4      12.1     84.4       0      0        0        -223.    -19.4    -84.4      0  
# 5     40930.     49086.     61.7      56.3      0         0     27.0      0        8156.     -5.38     0      -27.0

考虑简单的矩阵算术减去整个列块：

stubs <- unique(gsub("_.*", "", colnames(df)))

df[paste0("chg_", stubs)] <- (
    df[paste0(stubs, "_SAC")] - df[paste0(stubs, "_OVX")]
)

MyCompiler.io Demo

**Solution1 map_dfr：**一个优雅的方法来完成这样的任务是视觉上分开要减去的对：

split.default(df, sub("\\_.*", "", names(df)))

我们得到了所有的对：

$IL1B
# A tibble: 5 × 2
  IL1B_OVX IL1B_SAC
     <dbl>    <dbl>
1     20.5      0  
2      0       19.0
3      0       15.0
4      0        0  
5     27.0      0  

$IL4
# A tibble: 5 × 2
  IL4_OVX IL4_SAC
    <dbl>   <dbl>
1    84.4       0
2     0         0
3     0         0
4    84.4       0
5     0         0

$Leptin
# A tibble: 5 × 2
  Leptin_OVX Leptin_SAC
       <dbl>      <dbl>
1    101537.     19786.
2     34184.    124224.
3     54567.     89540.
4     29559.     29335.
5     40930.     49086.

$MIP1A
# A tibble: 5 × 2
  MIP1A_OVX MIP1A_SAC
      <dbl>     <dbl>
1    199.      153.  
2      9.39      0   
3      6.31      6.31
4     31.4      12.1 
5     61.7      56.3

对于purrr s map_dfr和reduce，我们将-（减法）应用于所有对，重命名并与原始 Dataframe 绑定：

library(purrr)
library(stringr)
library(dplyr)

split.default(df, sub("\\_.*", "", names(df))) %>% 
  map_dfr(reduce, `-`) %>% 
  rename_all(~ str_c('diff_', .)) %>% 
  bind_cols(df, .)

**解决方案2整形：**如果事情太复杂（特别是初学者），长格式的整形将有助于掌握整个过程：

下面是一个长格式的解决方案：

library(dplyr)
library(tidyr)
library(stringr)

df %>%
  mutate(rn = row_number()) %>%
  pivot_longer(cols = -rn, names_to = c( "grp", '.value'), names_sep = "\\_") %>%
  transmute(rn, grp = str_c('diff_', grp), diff = SAC - OVX) %>% 
  pivot_wider(names_from = grp, values_from = diff) %>%
  bind_cols(df, .)

解决方案3 map2_df：

library(purrr)
library(dplyr)

map2_df(
  df %>% select(ends_with("OVX")), 
  df %>% select(ends_with("SAC")), 
  `-`
) %>% 
  rename_with(~ str_c('diff_', .)) %>% 
  bind_cols(df, .)

Leptin_OVX Leptin_SAC MIP1A_OVX MIP1A_SAC IL4_OVX IL4_SAC IL1B_OVX IL1B_SAC    rn diff_Leptin diff_MIP1A diff_IL4
       <dbl>      <dbl>     <dbl>     <dbl>   <dbl>   <dbl>    <dbl>    <dbl> <int>       <dbl>      <dbl>    <dbl>
1    101537.     19786.    199.      153.      84.4       0     20.5      0       1     -81751.     -46.4     -84.4
2     34184.    124224.      9.39      0        0         0      0       19.0     2      90040.      -9.39      0  
3     54567.     89540.      6.31      6.31     0         0      0       15.0     3      34973.       0         0  
4     29559.     29335.     31.4      12.1     84.4       0      0        0       4       -223.     -19.4     -84.4
5     40930.     49086.     61.7      56.3      0         0     27.0      0       5       8156.      -5.38      0  
# ℹ 1 more variable: diff_IL1B <dbl>