如何基于“长”数据集创建新比率

Question

这是一个示例数据集：

structure(list(LD_wday = c(6, 2, 6, 1, 4, 4, 7, 6, 1, 3, 1, 3, 
6, 1, 6, 4, 7, 7, 6, 2, 7, 1, 5, 2, 2, 2, 3, 3, 5, 1, 2, 5, 1, 
6, 3, 4, 3, 4, 1, 6, 3, 6, 2, 6, 5, 5, 4, 3, 5, 6), status = c("successful", 
"failed", "live", "successful", "failed", "successful", "failed", 
"successful", "successful", "successful", "live", "successful", 
"successful", "failed", "failed", "successful", "failed", "live", 
"successful", "successful", "failed", "live", "successful", "successful", 
"failed", "successful", "successful", "successful", "failed", 
"failed", "failed", "failed", "failed", "successful", "live", 
"failed", "live", "successful", "successful", "successful", "successful", 
"failed", "failed", "live", "successful", "failed", "successful", 
"failed", "failed", "successful")), row.names = c(NA, -50L), class = c("tbl_df", 
"tbl", "data.frame"))

因此，根据下图，我试图找出如何创建成功/失败比率。

我一直使用group_by和summary，但最终得到类似下面的输出。如何从提供的数据集中创建成功/失败比率？

sample %>%
  filter(status == "failed" | status == "successful") %>%
  group_by(LD_wday, status) %>%
  summarize(count = n())

OUTPUT:
# A tibble: 13 x 3
# Groups:   LD_wday [7]
   LD_wday status     count
     <dbl> <chr>      <int>
 1       1 failed         3
 2       1 successful     3
 3       2 failed         4
 4       2 successful     3
 5       3 failed         1
 6       3 successful     5
 7       4 failed         2
 8       4 successful     4
 9       5 failed         4
10       5 successful     2
11       6 failed         2
12       6 successful     7
13       7 failed         3

任何帮助将不胜感激，对于表达问题有困难，我深表歉意。

Answer 1

如果我们想创建两者的比率，则可以除以{count}的sum，因为它已经被'LD_wday'分组了

library(dplyr)
sample %>%
  filter(status == "failed" | status == "successful") %>%
  group_by(LD_wday, status) %>%
  summarize(count = n()) %>%
   mutate(status = count/sum(count))