[英]apply function across columns using column names within function
我正在嘗試迭代 100 多列以確定單獨列中的變量是否與列名匹配。 我想也許跨 function 可能能夠但無法弄清楚如何在每列上使用變異。 請參見下面的示例。
tst=structure(list(type = c("DOG", "DOG", "DOG", "CAT", "CAT", "CAT",
"MOUSE", "MOUSE", "MOUSE"), CAT = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_), DOG = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_), MOUSE = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_), id = 1:9), row.names = c(NA, -9L
), class = c("tbl_df", "tbl", "data.frame"))
我的表目前具有以下結構。
type CAT DOG MOUSE id
<chr> <chr> <chr> <chr> <int>
1 DOG NA NA NA 1
2 DOG NA NA NA 2
3 DOG NA NA NA 3
4 CAT NA NA NA 4
5 CAT NA NA NA 5
6 CAT NA NA NA 6
7 MOUSE NA NA NA 7
8 MOUSE NA NA NA 8
9 MOUSE NA NA NA 9
我希望最終結果如下所示:
type CAT DOG MOUSE id
<chr> <chr> <chr> <chr> <int>
1 DOG NA TRUE NA 1
2 DOG NA TRUE NA 2
3 DOG NA TRUE NA 3
4 CAT TRUE NA NA 4
5 CAT TRUE NA NA 5
6 CAT TRUE NA NA 6
7 MOUSE NA NA TRUE 7
8 MOUSE NA NA TRUE 8
9 MOUSE NA NA TRUE 9
This works but it is not sufficient for 100 columns.
tst<-tst%>%mutate(CAT=ifelse(type==names(tst[2]),'TRUE',NA))
tst<-tst%>%mutate(DOG=ifelse(type==names(tst[3]),'TRUE',NA))
tst<-tst%>%mutate(MOUSE=ifelse(type==names(tst[4]),'TRUE',NA))
候選解決方案如下(沒有dplyr
)
# initialise list
tmpList <- list()
# iterate over each row
for (i in 1:nrow(tst)) {
tmpList[[i]] <- colnames(tst[-c(1,5)]) %in% tst$type[i]
}
# save as data frame
output <- as.data.frame(do.call(rbind, tmpList))
colnames(output) <- colnames(tst[-c(1,5)])
# cbind with data
output <- cbind(tst[,c(1,5)],output)
如果有更好的解決方案,這給出了您正在尋找的東西。 這不是我容易想到的事情。
最好的!
dplyr
解決方案非常簡潔。 您可以使用 cross( across()
和cur_column()
來獲取當前列的名稱,並將其放入您的公式中:
tst %>%
mutate(across(CAT:MOUSE, ~if_else(type == cur_column(), 'TRUE', .x)))
您可以用任何整潔的 select function 替換CAT:MOUSE
,以獲取 100 多列。
這是完整的代表:
library(tidyverse)
# I like tibbles because they print nicely, but this could just be a plain dataframe
tst <- tibble(
type = c(
"DOG", "DOG", "DOG", "CAT", "CAT", "CAT",
"MOUSE", "MOUSE", "MOUSE"
),
CAT = NA_character_,
DOG = NA_character_,
MOUSE = NA_character_,
id = 1:9
)
# .x here could be NA_character_, if you don't want the value from the existing column
tst %>%
mutate(across(CAT:MOUSE, ~if_else(type == cur_column(), 'TRUE', .x)))
#> # A tibble: 9 x 5
#> type CAT DOG MOUSE id
#> <chr> <chr> <chr> <chr> <int>
#> 1 DOG <NA> TRUE <NA> 1
#> 2 DOG <NA> TRUE <NA> 2
#> 3 DOG <NA> TRUE <NA> 3
#> 4 CAT TRUE <NA> <NA> 4
#> 5 CAT TRUE <NA> <NA> 5
#> 6 CAT TRUE <NA> <NA> 6
#> 7 MOUSE <NA> <NA> TRUE 7
#> 8 MOUSE <NA> <NA> TRUE 8
#> 9 MOUSE <NA> <NA> TRUE 9
由代表 package (v1.0.0) 於 2021 年 4 月 23 日創建
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