在mutate_at中使用dplyr quosure自定义函数

Question

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mutate中使用我的函数(并对所有感兴趣的列重复该函数)，但它似乎不能在mutate_at中工作。

下面是我的数据的一个示例：

> 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

因此，我(使用dplyr语法)构建了这个帮助器，以便对我拥有的每个PayXX列进行变异：

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

此函数可以很好地与dplyr::mutate配合使用：

> 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mutate_at中使用相同的函数时，它返回NA的：

> 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，我如何处理mutate_at函数？

谢谢,

可重现的例子

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(.)))

Answer 1

我想到的第一个方法是，通过重塑数据，这可能会更容易。然而，仍然需要一堆tidyr函数才能得到1)一列"Pay00“、"Pay01”等；2)提取数字；3)进行操作以便可以使用tidyr::spread返回宽形；4)展开并删除我附加的"_value“位。

我相信在最新版本的tidyr中有一种更好的方法来实现这一点，因为新的pivot_wider函数应该能够接受多个列作为value。我根本没有弄乱这个，但也许其他人可以把它写下来。

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，如果您想继续使用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(.))表示法，那么所有这些查询都会显示为.

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