r - 如何使用 "contains"和 "ifelse"有条件地改变多个列？

Question

我想改变包含字符串“account”的多个列。具体来说，我希望这些列在满足某个条件时采用“NA”，在不满足条件时采用另一个值。下面我将介绍我的尝试，灵感来自 here和 here 。到目前为止，还没有成功。仍在尝试，但任何帮助将不胜感激。

我的数据

df<-as.data.frame(structure(list(low_account = c(1, 1, 0.5, 0.5, 0.5, 0.5), high_account = c(16, 
16, 56, 56, 56, 56), mid_account_0 = c(8.5, 8.5, 28.25, 28.25, 
28.25, 28.25), mean_account_0 = c(31.174, 30.1922101449275, 30.1922101449275, 
33.3055555555556, 31.174, 33.3055555555556), median_account_0 = c(2.1, 
3.8, 24.2, 24.2, 24.2, 24.2), low_account.1 = c(1, 1, 0.5, 0.5, 0.5, 
0.5), high_account.1 = c(16, 16, 56, 56, 56, 56), row.names = c("A001", "A002", "A003", "A004", "A005", "A006"))))

df
  low_account high_account mid_account_0 mean_account_0 median_account_0 low_account.1 high_account.1 row.names
1         1.0           16          8.50       31.17400              2.1           1.0             16      A001
2         1.0           16          8.50       30.19221              3.8           1.0             16      A002
3         0.5           56         28.25       30.19221             24.2           0.5             56      A003
4         0.5           56         28.25       33.30556             24.2           0.5             56      A004
5         0.5           56         28.25       31.17400             24.2           0.5             56      A005
6         0.5           56         28.25       33.30556             24.2           0.5             56      A006

我的尝试

sample_data<-df%>% mutate_at(select(contains("account") , ifelse(. <= df$low_account&  >= df$high_account, NA, .)))

Error: No tidyselect variables were registered Call rlang::last_error() to see a backtrace

预期输出

df
    low_account high_account mid_account_0 mean_account_0 median_account_0 low_account.1 high_account.1 row.names
    1         1.0           16          8.50       NA                    2.1           1.0             16      A001
    2         1.0           16          8.50       NA                    3.8           1.0             16      A002
    3         0.5           56         28.25       30.19221             24.2           0.5             56      A003
    4         0.5           56         28.25       33.30556             24.2           0.5             56      A004
    5         0.5           56         28.25       31.17400             24.2           0.5             56      A005
    6         0.5           56         28.25       33.30556             24.2           0.5             56      A006

Answer 1

vars(contains('account')) 的问题在于它匹配存在子字符串“account”的所有列，并且当我们进行逻辑比较时，“low_account”会匹配列被转换为 NA 因为它肯定低于或等于 'low_account'，因此只有 NA 替换的列可用。因此，我们可以获取感兴趣的“mid”、“median”、“mean”列，然后进行replace

library(tidyverse)
df %>% 
   mutate_at(vars(matches("(mid|mean|median)_account")),
           ~ replace(., .<= low_account | .>= high_account, NA))
# low_account high_account mid_account_0 mean_account_0 median_account_0 low_account.1 high_account.1 row.names
#1         1.0           16          8.50             NA              2.1           1.0             16      A001
#2         1.0           16          8.50             NA              3.8           1.0             16      A002
#3         0.5           56         28.25       30.19221             24.2           0.5             56      A003
#4         0.5           56         28.25       33.30556             24.2           0.5             56      A004
#5         0.5           56         28.25       31.17400             24.2           0.5             56      A005
#6         0.5           56         28.25       33.30556             24.2           0.5             56      A006