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ageFemale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/

10-12 18:00