dplyr错误:组合group_by,mutate和ifelse时的奇怪问题.这是一个错误吗？

Question

我在dplyr以及group_by,mutate和ifelse的组合方面遇到了奇怪的问题.请考虑以下data.frame

> df1
  crawl.id group.id hits.diff
1        1        1        NA
2        1        2        NA
3        2        2         0
4        1        3        NA
5        1        3        NA
6        1        3        NA

当我使用它时,以下代码

library(dplyr)
df1 %>%
  group_by(group.id) %>% 
  mutate( hits.consumed = ifelse(hits.diff<=0,-hits.diff,0) )

出于某种原因,我得到了

Error: incompatible types, expecting a logical vector**

但是,删除任何一个group_by()或ifelse一切按预期工作:

df1 %>%
  mutate( hits.consumed = ifelse(hits.diff<=0,-hits.diff,0) )

crawl.id group.id hits.diff hits.consumed
1        1        1        NA            NA
2        1        2        NA            NA
3        2        2         0             0
4        1        3        NA            NA
5        1        3        NA            NA
6        1        3        NA            NA

df1 %>%
  group_by( group.id ) %>%
  mutate( hits.consumed = -hits.diff )

  crawl.id group.id hits.diff hits.consumed
1        1        1        NA            NA
2        1        2        NA            NA
3        2        2         0             0
4        1        3        NA            NA
5        1        3        NA            NA
6        1        3        NA            NA

这是一个错误还是一个功能？任何人都可以复制这个吗？group_by,mutate和ifelse的特定组合使它失败的特别之处是什么？

我自己的研究在这里引导我:https: //github.com/hadley/dplyr/issues/464 这表明现在应该修复它.

这是dput(df1):

structure(list(crawl.id = c(1, 1, 2, 1, 1, 1), group.id = structure(c(1L, 
2L, 2L, 3L, 3L, 3L), .Label = c("1", "2", "3"), class = "factor"), 
    hits.diff = c(NA, NA, 0, NA, NA, NA)), .Names = c("crawl.id", 
"group.id", "hits.diff"), row.names = c(NA, -6L), class = "data.frame")

Answer 1

将其全部包装as.numeric以强制输出格式,因此NAs logical(默认情况下)不会覆盖输出变量的类:

df1 %>%
  group_by(group.id) %>% 
  mutate( hits.consumed = as.numeric(ifelse(hits.diff<=0,-hits.diff,0)) )

#  crawl.id group.id hits.diff hits.consumed
#1        1        1        NA            NA
#2        1        2        NA            NA
#3        2        2         0             0
#4        1        3        NA            NA
#5        1        3        NA            NA
#6        1        3        NA            NA

很确定这与此处的问题相同:dplyr中的自定义求和函数返回不一致的结果,因此结果表明:

out <- df1[1:2,] %>%  mutate( hits.consumed = ifelse(hits.diff <= 0, -hits.diff, 0))
class(out$hits.consumed)
#[1] "logical"
out <- df1[1:3,] %>%  mutate( hits.consumed = ifelse(hits.diff <= 0, -hits.diff, 0))
class(out$hits.consumed)
#[1] "numeric"