dplyr:基于变量字符串选择的多个列来改变新列

时间:2017-02-19 19:45:21

标签: r variables select dplyr

鉴于此数据:

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))

我想使用dplyr为所选列的行方向最小值创建一个额外的列min。使用列名称很容易:

df <- df %>% rowwise() %>% mutate(min = min(x2,x5))

但是我有一个带有不同列名的大型df,所以我需要从一些值mycols中匹配它们。现在其他线程告诉我使用select辅助函数,但我必须遗漏一些东西。这是matches

mycols <- c("x2","x5")
df <- df %>% rowwise() %>%
  mutate(min = min(select(matches(mycols))))
Error: is.string(match) is not TRUE

one_of

mycols <- c("x2","x5")
 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')"
In addition: Warning message:
In one_of(c("x2", "x5")) : Unknown variables: `x2`, `x5`

我在俯瞰什么? select_应该有效吗?它不在下面:

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

同样地:

df <- df %>%
  rowwise() %>%
  mutate(min = min(select_(matches(mycols))))
Error: is.string(match) is not TRUE

2 个答案:

答案 0 :(得分:3)

这是另一个解决方案,有点技术性,有来自为函数式编程设计的tidyverse的purrr包的帮助。

来自matches的拳头dplyr助手将正则表达式字符串作为参数而不是向量。这是一种很好的方法,可以找到匹配所有列的正则表达式。 (在你下面的代码中可以使用你想要的dplyr选择助手)

然后,当您了解功能编程的基本方案时,purrr函数与dplyr的效果很好。

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,但在你的情况下它可以正常工作

答案 1 :(得分:1)

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