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我正在尝试构建一个辅助函数来提取参数中给出的列中的数字。我可以在里面使用我的函数mutate（并对所有感兴趣的列重复它），但它似乎在里面不起作用mutate_at。

这是我的数据的示例：

> set.seed(20190928)
> evalYr <- 2018
> n <- 5
> (df <- data.frame(
+     AY = sample(2016:2019, n, replace = T),
+     Pay00 = rgamma(n, 2, 1/1000),
+     Pay01 = rgamma(n, 2, 1/1000),
+     Pay02 = rgamma(n, 2, 1/1000),
+     Pay03 = rgamma(n, 2, 1/1000)
+ ))
    AY     Pay00     Pay01     Pay02     Pay03
1 2018 2520.3772 2338.9490  919.8245  629.1657
2 2016  259.7804 1543.4450  661.6488 2382.7916
3 2018 2446.3075  312.5143 2297.9717  942.5627
4 2017 1386.6288 4179.0352 2370.2669 1846.5838
5 2018  541.8261 2104.4589 2622.1758 2606.0694

所以我已经构建（使用dplyr语法）这个助手来改变PayXX我拥有的每一列：

# Helper function to get the number inside column `PayXX` name
f1 <- function(pmt) enquo(pmt) %>% quo_name() %>% str_extract('(\\d)+') %>% as.numeric()

此功能适用于dplyr::mutate：

> df %>% mutate(Pay00_numcol = f1(Pay00),
+               Pay01_numcol = f1(Pay01),
+               Pay02_numcol = f1(Pay02),
+               Pay03_numcol = f1(Pay03))
    AY     Pay00     Pay01     Pay02     Pay03 Pay00_numcol Pay01_numcol Pay02_numcol Pay03_numcol
1 2018 2520.3772 2338.9490  919.8245  629.1657            0            1            2            3
2 2016  259.7804 1543.4450  661.6488 2382.7916            0            1            2            3
3 2018 2446.3075  312.5143 2297.9717  942.5627            0            1            2            3
4 2017 1386.6288 4179.0352 2370.2669 1846.5838            0            1            2            3
5 2018  541.8261 2104.4589 2622.1758 2606.0694            0            1            2            3

但是当我尝试在内部使用相同的函数时mutate_at，它会返回 NA：

> df %>% mutate_at(vars(starts_with('Pay')), list(numcol = ~f1(.)))
    AY     Pay00     Pay01     Pay02     Pay03 Pay00_numcol Pay01_numcol Pay02_numcol Pay03_numcol
1 2018 2520.3772 2338.9490  919.8245  629.1657           NA           NA           NA           NA
2 2016  259.7804 1543.4450  661.6488 2382.7916           NA           NA           NA           NA
3 2018 2446.3075  312.5143 2297.9717  942.5627           NA           NA           NA           NA
4 2017 1386.6288 4179.0352 2370.2669 1846.5838           NA           NA           NA           NA
5 2018  541.8261 2104.4589 2622.1758 2606.0694           NA           NA           NA           NA

有人遇到过类似的问题吗？在这种情况下，我该如何处理该mutate_at功能？

谢谢，

可复制的例子

library(tidyverse)
library(stringr)
set.seed(20190928)
evalYr <- 2018
n <- 5
(df <- data.frame(
    AY = sample(2016:2019, n, replace = T),
    Pay00 = rgamma(n, 2, 1/1000),
    Pay01 = rgamma(n, 2, 1/1000),
    Pay02 = rgamma(n, 2, 1/1000),
    Pay03 = rgamma(n, 2, 1/1000)
))

# Helper function to get the number inside column `PayXX` name
f1 <- function(pmt) enquo(pmt) %>% quo_name() %>% str_extract('(\\d)+') %>% as.numeric()

# Working
df %>% mutate(Pay00_numcol = f1(Pay00),
              Pay01_numcol = f1(Pay01),
              Pay02_numcol = f1(Pay02),
              Pay03_numcol = f1(Pay03))

# Not working
df %>% mutate_at(vars(starts_with('Pay')), list(numcol = ~f1(.)))
4

1 回答 1

0

我想到的第一种方法是重塑数据可能更容易。但是，tidyr要得到1）“Pay00”，“Pay01”等列；2）提取数字；3）操纵，所以你可以tidyr::spread用来回到宽形；和 4) 传播并删除我添加的“_value”位。

我相信最近版本的 有更好的方法来做到这一点tidyr，因为新pivot_wider功能应该能够将多个列作为value. 我根本没有搞砸这个，但也许其他人可以写出来。

library(tidyverse)

df %>%
  rowid_to_column() %>%
  gather(key, value, -AY, -rowid) %>%
  mutate(numcol = as.numeric(str_extract(key, "\\d+$"))) %>%
  gather(key = coltype, value, value, numcol) %>%
  unite(key, key, coltype) %>%
  spread(key, value) %>%
  select(AY, ends_with("value"), ends_with("numcol")) %>%
  rename_all(str_remove, "_value")
#>     AY     Pay00     Pay01     Pay02     Pay03 Pay00_numcol Pay01_numcol
#> 1 2018 2520.3772 2338.9490  919.8245  629.1657            0            1
#> 2 2016  259.7804 1543.4450  661.6488 2382.7916            0            1
#> 3 2018 2446.3075  312.5143 2297.9717  942.5627            0            1
#> 4 2017 1386.6288 4179.0352 2370.2669 1846.5838            0            1
#> 5 2018  541.8261 2104.4589 2622.1758 2606.0694            0            1
#>   Pay02_numcol Pay03_numcol
#> 1            2            3
#> 2            2            3
#> 3            2            3
#> 4            2            3
#> 5            2            3

或者，如果您想坚持使用 tidyeval 方法：获取要调用函数的列的名称。请注意，如果您使用list(numcol = ~f1(.))符号，所有这些 quosures 都会出现.

f1 <- function(pmt) {
  str_extract(rlang::as_name(enquo(pmt)), "\\d+$") %>%
    as.numeric()
}

df %>%
  mutate_at(vars(starts_with("Pay")), list(numcol = f1))
# same output as prev
于 2019-09-27T17:59:15.650 回答