有以下小标题:
structure(list(age = c("21", "17", "32", "29", "15"),
gender = structure(c(2L, 1L, 1L, 2L, 2L), .Label = c("Female", "Male"), class = "factor")),
row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"), .Names = c("age", "gender"))
age gender
<chr> <fctr>
1 21 Male
2 17 Female
3 32 Female
4 29 Male
5 15 Male
我正在尝试使用
tidyr::spread
实现此目的: Female Male
1 NA 21
2 17 NA
3 32 NA
4 NA 29
5 NA 15
我以为
spread(gender, age)
可以用,但是我收到一条错误消息:最佳答案
现在,您有两个用于age
的Female
值和三个用于Male
的值,并且没有其他变量可以防止它们折叠成一行,因为spread
试图处理具有相似/无索引值的值:
library(tidyverse)
df <- data_frame(x = c('a', 'b'), y = 1:2)
df # 2 rows...
#> # A tibble: 2 x 2
#> x y
#> <chr> <int>
#> 1 a 1
#> 2 b 2
df %>% spread(x, y) # ...become one if there's only one value for each.
#> # A tibble: 1 x 2
#> a b
#> * <int> <int>
#> 1 1 2
spread
没有应用组合多个值的函数(àla dcast
),因此必须对行进行索引,以便某个位置有一个或零个值,例如df <- data_frame(i = c(1, 1, 2, 2, 3, 3),
x = c('a', 'b', 'a', 'b', 'a', 'b'),
y = 1:6)
df # the two rows with each `i` value here...
#> # A tibble: 6 x 3
#> i x y
#> <dbl> <chr> <int>
#> 1 1 a 1
#> 2 1 b 2
#> 3 2 a 3
#> 4 2 b 4
#> 5 3 a 5
#> 6 3 b 6
df %>% spread(x, y) # ...become one row here.
#> # A tibble: 3 x 3
#> i a b
#> * <dbl> <int> <int>
#> 1 1 1 2
#> 2 2 3 4
#> 3 3 5 6
如果您的值没有被其他列自然索引,则可以添加唯一索引列(例如,通过将行号添加为列),这将阻止
spread
尝试折叠行:df <- structure(list(age = c("21", "17", "32", "29", "15"),
gender = structure(c(2L, 1L, 1L, 2L, 2L),
.Label = c("Female", "Male"), class = "factor")),
row.names = c(NA, -5L),
class = c("tbl_df", "tbl", "data.frame"),
.Names = c("age", "gender"))
df %>% mutate(i = row_number()) %>% spread(gender, age)
#> # A tibble: 5 x 3
#> i Female Male
#> * <int> <chr> <chr>
#> 1 1 <NA> 21
#> 2 2 17 <NA>
#> 3 3 32 <NA>
#> 4 4 <NA> 29
#> 5 5 <NA> 15
如果要在以后删除它,请添加
select(-i)
。在这种情况下,这不会产生非常有用的data.frame,但在更复杂的重塑过程中可能非常有用。关于r - 如何传播具有重复标识符的列?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/45898614/