dplyr：根据变量字符串选择的多列来改变新列

Question

挑战如下：我们的任务是使用 dplyr 计算数据框中列子集的按行最小值，但列名称各不相同，并且必须与 mycols 中存储的一组指定值相匹配。让我们用一些无效的代码片段来解决这个问题：

R

# Given data
df=data.frame(
  x1=c(2,0,0,NA,0,1,1,NA,0,1),
  x2=c(3,2,NA,5,3,2,NA,NA,4,5),
  x3=c(0,1,0,1,3,0,NA,NA,0,1),
  x4=c(1,0,NA,3,0,0,NA,0,0,1),
  x5=c(1,1,NA,1,3,4,NA,3,3,1)
)

# Attempt to calculate row-wise minimum using specified columns
mycols <- c("x2","x5")

# Invalid attempt using matches
df <- df %>% rowwise() %>%
  mutate(min = min(select(matches(mycols))))
# Error: is.string(match) is not TRUE

# Another invalid attempt using one_of
df <- df %>%
  rowwise() %>%
  mutate(min = min(select(one_of(mycols))))
# Error: no applicable method for 'select' applied to an object of class "c('integer', 'numeric')"
# Warning message: In one_of(c("x2", "x5")) : Unknown variables: `x2`, `x5`

# Invalid use of select_
df <- df %>%
   rowwise() %>%
   mutate(min = min(select_(mycols)))
# Error: no applicable method for 'select_' applied to an object of class "character"

# Similarly invalid attempt with matches and select_
df <- df %>%
  rowwise() %>%
  mutate(min = min(select_(matches(mycols))))
# Error: is.string(match) is not TRUE

我们似乎在动态选择列的过程中偶然发现了一个挑战。进一步的探索，也许查阅文档可能会揭示前进的道路。

Answer 1

这是另一个有点技术性的解决方案，借助为函数式编程设计的 tidyverse 中的

purrr

包。

来自

matches

 的 

Fist、

dplyr

助手将正则表达式字符串作为参数而不是向量。这是查找与所有列匹配的正则表达式的好方法。 （在下面的代码中，您可以使用

dplyr

选择您想要的帮助器）

然后，当您了解函数式编程的底层方案时，

purrr

函数可以与

dplyr

很好地配合。

解决您的问题：

df=data.frame(
  x1=c(2,0,0,NA,0,1,1,NA,0,1),
  x2=c(3,2,NA,5,3,2,NA,NA,4,5),
  x3=c(0,1,0,1,3,0,NA,NA,0,1),
  x4=c(1,0,NA,3,0,0,NA,0,0,1),
  x5=c(1,1,NA,1,3,4,NA,3,3,1))


# regex to get only x2 and x5 column
mycols <- "x[25]"

library(dplyr)

df %>%
  mutate(min_x2_x5 =
           # select columns that you want in df
           select(., matches(mycols)) %>% 
           # use pmap on this subset to get a vector of min from each row.
           # dataframe is a list so pmap works on each element of the list that is to say each row
           purrr::pmap_dbl(min)
         )
#>    x1 x2 x3 x4 x5 min_x2_x5
#> 1   2  3  0  1  1         1
#> 2   0  2  1  0  1         1
#> 3   0 NA  0 NA NA        NA
#> 4  NA  5  1  3  1         1
#> 5   0  3  3  0  3         3
#> 6   1  2  0  0  4         2
#> 7   1 NA NA NA NA        NA
#> 8  NA NA NA  0  3        NA
#> 9   0  4  0  0  3         3
#> 10  1  5  1  1  1         1

我不会在这里进一步解释

purrr

，但它在你的情况下工作得很好

Answer 2

这有点棘手。在 SE 评估的情况下，您需要将操作作为字符串传递。

mycols <- '(x2,x5)'
f <- paste0('min',mycols)
df %>% rowwise() %>% mutate_(min = f)
df
# A tibble: 10 × 6
#      x1    x2    x3    x4    x5   min
#   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1      2     3     0     1     1     1
#2      0     2     1     0     1     1
#3      0    NA     0    NA    NA    NA
#4     NA     5     1     3     1     1
#5      0     3     3     0     3     3
#6      1     2     0     0     4     2
#7      1    NA    NA    NA    NA    NA
#8     NA    NA    NA     0     3    NA
#9      0     4     0     0     3     3
#10     1     5     1     1     1     1