R使用函数式编程将格式更改为数据帧的列

问题描述 投票:1回答:1

情况如下:我有一个数据框列表,对于每个数据框,我都有一个列列表,我需要更改其格式。设置:

df1 <- data.frame(a = c("2020-03-02", "2020-12-22", "2020-07-03"), b = c(4, 5, 6), c = c("2020-03-13", "2019-11-03", "2011-05-02"))

df2 <- data.frame(d = c(1, 2, 3), e = c("2020-05-21", "2014-08-31", "1999-01-21"), f = c(7, 8, 9))

datasets <- list("first" = df1, "second" = df2)

dates <- list("first" = c("a", "c"), "second" = c("e")) 

一个人可以通过以下方法来做到这一点:1.遍历数据框列表,2.遍历每个数据框,遍历要更改的列列表,然后将它们重新分配到位。像这样的东西:

for (i in names(datasets)) {
   for (j in dates[i]) {
      for (k in datasets[[i]][j]) {
         k <- as.Date(k)
      }
   }
} 

这很丑,所以我想尝试使用purrr做同样的事情。我认为这是个好主意:

library(purrr)

walk2(datasets, dates, ~ walk(.x[.y], ~ {.x <- as.Date(.x)}))

但是执行此操作后,数据集仍保持不变。为什么?

r loops functional-programming format purrr
1个回答
0
投票

这里是使用purrr和dplyr的解决方案:

library(purrr)
library(dplyr)

datasets <- datasets %>% 
  imap(~{
    .x %>% 
      mutate_at(vars(dates[[.y]]), as.Date)
  })

str(datasets)
#List of 2
#$ first :'data.frame': 3 obs. of  3 variables:
# ..$ a: Date[1:3], format: "2020-03-02" "2020-12-22" "2020-07-03"
# ..$ b: num [1:3] 4 5 6
# ..$ c: Date[1:3], format: "2020-03-13" "2019-11-03" "2011-05-02"
#$ second:'data.frame': 3 obs. of  3 variables:
# ..$ d: num [1:3] 1 2 3
# ..$ e: Date[1:3], format: "2020-05-21" "2014-08-31" "1999-01-21"
# ..$ f: num [1:3] 7 8 9
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