我正在努力寻找 dplyr
中跨
列的多个连续操作的正确语法。在此数据中:
df <- structure(list(A1 = c(838.611, 824.048, 668.901, 225.075, 0,
0, 341.291, 0, 101.652, 127.341, 0, 297.092, 0, 0, 0, 0, 0, 764.737,
759.51, 772.21), A2 = c(499.041, 492.997, 486.132, 469.503, 476.782,
464.18, 469.833, 462.317, 455.507, 441.47, 490.147, 430.844,
0, 0, 0, 0, 0, 0, 0, 124.068)), row.names = c(NA, 20L), class = "data.frame")
比如说,我想在 A1
和 A2
列中实现以下更改:
-
- 将
0
替换为NA
- 将
-
- 将离群值设置为
NA
- 将离群值设置为
-
- 插值
NA
- 插值
使用以下语法仅执行更改 1.,但不执行更改 2. 和 3.:
library(dplyr)
library(zoo)
df %>%
mutate(across(starts_with("A"),
~na_if(.,0),
~ifelse(. %in% boxplot(.)$out, NA, .),
~na.approx(., na.rm = FALSE, rule = 2)))
A1 A2
1 838.611 499.041
2 824.048 492.997
3 668.901 486.132
4 225.075 469.503
5 NA 476.782
6 NA 464.180
7 341.291 469.833
8 NA 462.317
9 101.652 455.507
10 127.341 441.470
11 NA 490.147
12 297.092 430.844
13 NA NA
14 NA NA
15 NA NA
16 NA NA
17 NA NA
18 764.737 NA
19 759.510 NA
20 772.210 124.068
编辑: 正确的输出是从这种(重复的)代码类型中获得的(我想避免这种情况):
df %>%
mutate(across(starts_with("A"),
~na_if(.,0))) %>%
mutate(across(starts_with("A"),
~ifelse(. %in% boxplot(.)$out, NA, .))) %>%
mutate(across(starts_with("A"),
~na.approx(., na.rm = FALSE, rule = 2)))
A1 A2
1 838.6110 499.041
2 824.0480 492.997
3 668.9010 486.132
4 225.0750 469.503
5 263.8137 476.782
6 302.5523 464.180
7 341.2910 469.833
8 221.4715 462.317
9 101.6520 455.507
10 127.3410 441.470
11 212.2165 490.147
12 297.0920 430.844
13 375.0328 430.844
14 452.9737 430.844
15 530.9145 430.844
16 608.8553 430.844
17 686.7962 430.844
18 764.7370 430.844
19 759.5100 430.844
20 772.2100 430.844
最佳答案
回答OP在评论中提出的问题。
df %>%
mutate(
across(
starts_with("A"),
list(
~na_if(.,0),
~ifelse(. %in% boxplot(.)$out, NA, .),
~na.approx(., na.rm = FALSE, rule = 2)
)
)
)
A1 A2 A1_1 A1_2 A1_3 A2_1 A2_2 A2_3
1 838.611 499.041 838.611 838.611 838.611 499.041 499.041 499.041
2 824.048 492.997 824.048 824.048 824.048 492.997 492.997 492.997
3 668.901 486.132 668.901 668.901 668.901 486.132 486.132 486.132
4 225.075 469.503 225.075 225.075 225.075 469.503 469.503 469.503
5 0.000 476.782 NA 0.000 0.000 476.782 476.782 476.782
6 0.000 464.180 NA 0.000 0.000 464.180 464.180 464.180
7 341.291 469.833 341.291 341.291 341.291 469.833 469.833 469.833
8 0.000 462.317 NA 0.000 0.000 462.317 462.317 462.317
9 101.652 455.507 101.652 101.652 101.652 455.507 455.507 455.507
10 127.341 441.470 127.341 127.341 127.341 441.470 441.470 441.470
11 0.000 490.147 NA 0.000 0.000 490.147 490.147 490.147
12 297.092 430.844 297.092 297.092 297.092 430.844 430.844 430.844
13 0.000 0.000 NA 0.000 0.000 NA 0.000 0.000
14 0.000 0.000 NA 0.000 0.000 NA 0.000 0.000
15 0.000 0.000 NA 0.000 0.000 NA 0.000 0.000
16 0.000 0.000 NA 0.000 0.000 NA 0.000 0.000
17 0.000 0.000 NA 0.000 0.000 NA 0.000 0.000
18 764.737 0.000 764.737 764.737 764.737 NA 0.000 0.000
19 759.510 0.000 759.510 759.510 759.510 NA 0.000 0.000
20 772.210 124.068 772.210 772.210 772.210 124.068 124.068 124.068
您可以通过(除其他外)命名列表元素来为输出列提供更有意义的名称:
df %>%
mutate(
across(
starts_with("A"),
list(
"Zero"=~na_if(.,0),
"BoxPlot"=~ifelse(. %in% boxplot(.)$out, NA, .),
"Approx"=~na.approx(., na.rm = FALSE, rule = 2)
)
)
)
A1 A2 A1_Zero A1_BoxPlot A1_Approx A2_Zero A2_BoxPlot A2_Approx
1 838.611 499.041 838.611 838.611 838.611 499.041 499.041 499.041
2 824.048 492.997 824.048 824.048 824.048 492.997 492.997 492.997
...
更新以回应下面OP的评论
across()
有一个 .names
参数,允许控制输出列的命名,但这在这里不起作用,因为 across()
对于输入列和函数的每个组合输出一列。我们希望对每个输入列应用多个函数,为每个输入列生成一个输出列。为此,请将处理每一列的函数包装在一个函数中。这与 OP 对原始问题的编辑中的多个 mutate
调用具有相同的效果。
df %>%
mutate(
across(
starts_with("A"),
function(.x) {
.x <- na_if(.x, 0)
.x <- ifelse(.x %in% boxplot(.x)$out, NA, .x)
.x <- na.approx(.x, na.rm = FALSE, rule = 2)
.x
}
)
)
A1 A2
1 838.6110 499.041
2 824.0480 492.997
3 668.9010 486.132
4 225.0750 469.503
5 263.8137 476.782
6 302.5523 464.180
7 341.2910 469.833
8 221.4715 462.317
9 101.6520 455.507
10 127.3410 441.470
11 212.2165 490.147
12 297.0920 430.844
13 375.0328 430.844
14 452.9737 430.844
15 530.9145 430.844
16 608.8553 430.844
17 686.7962 430.844
18 764.7370 430.844
19 759.5100 430.844
20 772.2100 430.844
关于r - `dplyr` 中跨列的多个连续操作的语法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/70707844/