我正在尝试迭代 100 多列以确定单独列中的变量是否与列名匹配。我想也许 cross 函数可能能够但无法弄清楚如何在每一列上使用 mutate。请参见下面的示例。
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结构。
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
我希望最终结果如下所示:
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))