r - dplyr 0.3.0.9000 如何正确使用 do()

Question

试图在 SO 问题上重现结果:
dplyr: How to apply do() on result of group_by?

这是数据

person = c('Grace', 'Grace', 'Grace', 'Rob', 'Rob', 'Rob')
foods = c('apple', 'banana', 'cucumber', 'spaghetti', 'cucumber', 'banana')
eaten <- data.frame(person, foods, stringsAsFactors = FALSE)

我试图复制的结果是:

[[1]]
     [,1]     [,2]       [,3]      
[1,] "apple"  "apple"    "banana"  
[2,] "banana" "cucumber" "cucumber"

[[2]]
     [,1]        [,2]        [,3]      
[1,] "spaghetti" "spaghetti" "cucumber"
[2,] "cucumber"  "banana"    "banana" 

产生上述结果的原始代码如下，不再有效:

> eaten %>% group_by(person) %>% do(function(x) combn(x$foods, m = 2))
Error: Results are not data frames at positions: 1, 2

尝试了几种使用 do() 函数的方法都无济于事。

> eaten %>% group_by(person) %>% do(combn(.$foods, m = 2))
Error: Results are not data frames at positions: 1, 2

> eaten %>% group_by(person) %>% do(.$foods, combn, m =2)
Error: Arguments to do() must either be all named or all unnamed

> eaten %>% group_by(person) %>% do((combn(.$foods, m=2)))
Error: Results are not data frames at positions: 1, 2

似乎只有下面的一个适用于警告消息:

> eaten %>% group_by(person) %>% do(as.data.frame(combn(.$foods, m = 2)))
#   person        V1        V2       V3
# 1  Grace     apple     apple   banana
# 2  Grace    banana  cucumber cucumber
# 3    Rob spaghetti spaghetti cucumber
# 4    Rob  cucumber    banana   banana
# Warning messages:
# 1: In rbind_all(out[[1]]) : Unequal factor levels: coercing to character
# 2: In rbind_all(out[[1]]) : Unequal factor levels: coercing to character

相信新版本下 do() 的行为必须有所改变。有哪些变化？使用 do() 的正确习惯用法/方法是什么？谢谢。

编辑:安装最新的 dplyr 并运行@hadley 建议的代码

packageVersion("dplyr")
[1] ‘0.3.0.2’

eaten %>% group_by(person) %>% do(x = combn(.$foods, m = 2))
# Source: local data frame [2 x 2]
# Groups: <by row>
#   
#   person          x
# 1  Grace <chr[2,3]>
# 2    Rob <chr[2,3]>

EDIT2:需要按照@hadley 的建议提取列“x”

eaten2 <- eaten %>% group_by(person) %>% do(x = combn(.$foods, m = 2))
eaten2[["x"]]
# [[1]]
# [,1]     [,2]       [,3]      
# [1,] "apple"  "apple"    "banana"  
# [2,] "banana" "cucumber" "cucumber"
# 
# [[2]]
# [,1]        [,2]        [,3]      
# [1,] "spaghetti" "spaghetti" "cucumber"
# [2,] "cucumber"  "banana"    "banana" 

Answer 1

在 Q 中移动 EDIT2 以回答以关闭问题:

最新 dplyr 0.3.0.2+，需要按照@hadley 的建议提取列“x”

eaten2 <- eaten %>% group_by(person) %>% do(x = combn(.$foods, m = 2))
eaten2[["x"]]
# [[1]]
# [,1]     [,2]       [,3]      
# [1,] "apple"  "apple"    "banana"  
# [2,] "banana" "cucumber" "cucumber"
# 
# [[2]]
# [,1]        [,2]        [,3]      
# [1,] "spaghetti" "spaghetti" "cucumber"
# [2,] "cucumber"  "banana"    "banana