Question

我有以下df

df <- structure(list(ID = c(1, 1, 1, 1, 1, 2, 2, 2, 2), value = c("p", 
"p", "p1", "p2", "p3", "a", "b", "c", "d"), i1 = c(1, 1, 1, 1, 
1, 1, 1, 1, 1)), row.names = c(NA, -9L), class = c("tbl_df", 
"tbl", "data.frame"))

     ID value    i1
  <dbl> <chr> <dbl>
1     1 p         1
2     1 p         1
3     1 p1        1
4     1 p2        1
5     1 p3        1
6     2 a         1
7     2 b         1
8     2 c         1
9     2 d         1

当我尝试旋转时，出现一个错误，说有重复项。

df %>% pivot_wider(names_from = value, values_from = i1, values_fill = list(i1 = 0))

Warning message:
Values in `i1` are not uniquely identified; output will contain list-cols.
* Use `values_fn = list(i1 = list)` to suppress this warning.
* Use `values_fn = list(i1 = length)` to identify where the duplicates arise
* Use `values_fn = list(i1 = summary_fun)` to summarise duplicates

我想确定每个唯一ID重复的值，以便进行过滤。或者，也许我可以在pivot_wider（）步骤中删除重复项。源代码具有我设置为“唯一”的name_repair。没用！

理想的输出是：

    p  p1  p2  p3  a  b  c  d
1   1   1   1   1  0  0  0  0
2   0   0   0   0  1  1  1  1

Answer 1

我认为，在OP的尝试中，他们试图做的是删除重复项，然后转移可以使用distinct和pivot_wider完成的数据。

library(dplyr)
library(tidyr)

df %>%
 distinct() %>%
 pivot_wider(names_from = value, values_from = i1, values_fill = list(i1 = 0))

# A tibble: 2 x 9
#     ID     p    p1    p2    p3     a     b     c     d
#  <dbl> <int> <int> <int> <int> <int> <int> <int> <int>
#1     1     1     1     1     1     0     0     0     0
#2     2     0     0     0     0     1     1     1     1

我们还可以使用count和pivot_wider

df %>% 
  count(ID, value) %>%
  mutate(n = +(n > 0)) %>%
  pivot_wider(names_from = value, values_from = n, values_fill = list(n = 0))

具有重复项的pivot_wider

1 个答案: