r - 使用 dplyr 在数据帧的多列中计算 "Yes"

Question

假设我有以下数据。 [根据要求，我正在添加数据]

col1 <- c("Team A", "Team A", "Team A", "Team B", "Team B", "Team B", "Team C", "Team C", "Team C", "Team D", "Team D", "Team D")
col2 <- c("High",   "Medium", "Medium", "Low", "Low", "Low", "High", "Medium", "Low", "Medium", "Medium", "Medium")
col3 <- c("Yes", "Yes", "No", "No", "No", "Yes", "No", "Yes", "No", "Yes", "Yes", "Yes")
col4 <- c("No", "Yes", "No", "Yes", "Yes", "No", "No", "Yes", "No", "Yes", "No", "Yes")
df <- data.frame(col1, col2, col3, col4)

# Col1      Col2    Col3    Col4
# Team A    High    Yes     No
# Team A    Medium  Yes     Yes
# Team A    Medium  No      No
# Team B    Low     No      Yes
# Team B    Low     No      Yes
# Team B    Low     Yes     No
# Team C    High    No      No
# Team C    Medium  Yes    Yes
# Team C    Low     No     No 
# Team D    Medium  Yes    Yes
# Team D    Medium  Yes    No
# Team D    Medium  Yes    Yes

我想用dplyr函数得到以下结果。 Status_1 需要是 Col3 中针对每支球队的"is"的数量，而 Status_2 将是 Col4 中针对每支球队的"is"的数量

       High Medium  Low Status_1    Status_2
Team A    1      2    0        2           1
Team B    0      0    3        1           2
Team C    1      1    1        1           1
Team D    0      3    0        3           2

我能够生成正常的摘要，但使用以下语句生成“Status_1”和“Status_2”的最后两列。请问有人可以帮忙吗？

df %>%
  group_by(Col1, Col2) %>%
  summarise(Count = n()) %>%
  spread(Col1, Count, fill = 0)

Answer 1

首先，将数据按 col1 分组统计Yes的数量在 col3和 col4 .然后再次按所有列分组并使用 n() 计算每组中的观察次数.最后，使用 tidyr::pivot_wider将数据从长转换为宽。

df %>%
  group_by(col1) %>%
  mutate_at(vars(col3:col4), ~ sum(. == "Yes")) %>%
  rename(status_1 = col3, status_2 = col4) %>% 
  group_by_all %>%
  summarise(n = n()) %>%
  tidyr::pivot_wider(names_from = col2, values_from = n, values_fill = list(n = 0))

# # A tibble: 4 x 6
#   col1   status_1 status_2  High Medium   Low
#   <fct>     <int>    <int> <int>  <int> <int>
# 1 Team A        2        1     1      2     0
# 2 Team B        1        2     0      0     3
# 3 Team C        1        1     1      1     1
# 4 Team D        3        2     0      3     0