如何传播具有重复标识符的列?

问题描述

A 有以下 tibble:

structure(list(age = c("21", "17", "32", "29", "15"),性别 = 结构(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"))年龄 性别<chr><fctr>1 21 男2 17 女3 32 女4 29 男5 15 男

我正在尝试使用 tidyr::spread 来实现这一点:

 女 男1 不适用 212 17 不适用3 32 不适用4 不适用 295 不适用 15

我认为 spread(gender, age) 会起作用，但我收到一条错误消息:

错误:行 (2, 3), (1, 4, 5) 的标识符重复

解决方案

现在，Female 有两个 age 值，Male 有三个值>，并且没有其他变量阻止它们折叠成一行，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 行...#># 小块:2 x 2#>xy#><chr><int>#>1 一个 1#>2 b 2df %>% spread(x, y) # ...如果每个值只有一个，则变为一.#># 小费:1 x 2#>乙#>* <int><int>#>1 1 2

spread 不应用函数来组合多个值(à la 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 

如果您的值没有被其他列自然索引，您可以添加一个唯一的索引列(例如，通过将行号添加为列)，这将阻止 spread 试图折叠行:

df 

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，但在更复杂的重塑过程中可能非常有用.

A have the following tibble:

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

And I am trying to use tidyr::spread to achieve this:

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

I thought spread(gender, age) would work, but I get an error message saying:

解决方案

Right now you have two age values for Female and three for Male, and no other variables keeping them from being collapsed into a single row, as spread tries to do with values with similar/no index values:

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 doesn't apply a function to combine multiple values (à la dcast), so rows must be indexed so there's one or zero values for a location, e.g.

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

If you your values aren't indexed naturally by the other columns you can add a unique index column (e.g. by adding the row numbers as a column) which will stop spread from trying to collapse the rows:

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

If you want to remove it afterwards, add on select(-i). This doesn't produce a terribly useful data.frame in this case, but can be very useful in the midst of more complicated reshaping.

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