我知道这是一个愚蠢的问题，但仍然。我想 * 理解 * 为什么count()不能在这个非常简单的for循环中工作。

编辑2

我用ch_data$QUnlist <- unlist(Q)解决了这个问题，但我仍然想知道为什么它能工作，是否有更好的方法

编辑1

现在我遇到了与filter()和mutate()相同的问题

• 出现错误的代码：

listDFs <- list()

for (i in 1:2) {
  
## select colums:
  
Q <- names(ch_data[, 1:2][i])

## put into dataframes: 
  
listDFs[[i]] <-  ch_data %>% 
                     dplyr::count(Q) %>% 
                  mutate(prop = round((prop.table(n) * 100), digits = 2),
                  sd = round(sd(prop.table(n)), digits = 2)) %>%
           arrange(Q)
  
}

Error in `dplyr::count()`:
! Must group by variables found in `.data`.
x Column `Q` is not found.

## edit ##

ch_data %>% 
select(Q) %>% 
 filter(!Q %in% 'Não oferta') %>% ## NOT WORKING
 count(!!!syms(Q)) %>% 
  mutate(prop = round((prop.table(n) * 100), digits = 2),
         sd = round(sd(prop.table(n)), digits = 2),
         CH = Q, ### NOT WORKING
         QUESTION = rep('mylabel', length(CH))) %>% select (-Q) %>%
  arrange(CH) %>% 
  relocate(QUESTION, CH, .before = everything())

• 我已经试过了

答：由于我知道count()需要 Dataframe ，所以我执行了Q <- ch_data[, 1:2][i]
B）我想可能是引号的问题，所以我在count()中尝试了str_glue('{noquote(Q)}')，但它也不起作用。
这个问题和this类似，但我还是解不出来。

• 双变量数据

(it有更多的问题，但我想让它简单化）：

dput(ch_data)
structure(list(Q29 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 5L, 1L, 1L, 1L, 1L, 4L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 1L, 5L, 1L, 1L, 1L, 1L, 1L, 1L, 2L), .Label = c("Não oferta", 
"1h - 3h", "Mais de 10h", "50% em LA", "100% em LA"), class = "factor"), 
    Q30 = structure(c(4L, 8L, 3L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 
    1L, 2L, 5L, 1L, 1L, 1L, 7L, 7L, 1L, 1L, 1L, 1L, 5L, 1L, 1L, 
    2L, 1L, 6L, 2L, 8L, 8L, 2L, 8L, 3L, 7L, 7L, 1L, 2L, 2L, 8L, 
    3L, 2L, 1L, 8L, 3L, 1L, 1L, 1L, 1L, 6L, 3L), .Label = c("Não oferta", 
    "1h - 3h", "4h - 5h", "6h - 8h", "9h - 10h", "Mais de 10h", 
    "50% em LA", "100% em LA"), class = "factor")), row.names = c(NA, 
-51L), class = "data.frame")