鉴于这种情况
id = "a" df <- tibble( id = c("a", "b", "c"), value = c(1, 2, 3) ) df %>% dplyr::filter(id == id)
我期望最后一行具有相同的输出,df %>% dplyr::filter(id == "a")但它仍然引用id为 df 列而不是变量。我可以强制它寻找变量id吗?
df %>% dplyr::filter(id == "a")
id
r dplyr
dplyr ×1
r ×1