如何传播具有重复标识符的列？

Question

A有以下几个:

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

我试图用它tidyr::spread来实现这个目标:

  Female Male
1    NA     21
2    17     NA
3    32     NA
4    NA     29
5    NA     15

我认为spread(gender, age)会工作,但我收到一条错误消息:

Error: Duplicate identifiers for rows (2, 3), (1, 4, 5)

Answer 1

现在你有两个age值Female和三个for Male,没有其他变量可以防止它们折叠成一行,因为spread尝试使用具有相似/没有索引值的值:

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不应用函数来组合多个值(àla dcast),因此必须对行进行索引,以使位置有一个或零值,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

如果您的值未被其他列自然索引,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

如果您想在之后删除它,请添加select(-i).在这种情况下,这不会产生非常有用的data.frame,但在更复杂的重塑中可能非常有用.