这本质上是对@keqiang-li之前的one的后续问题。
dplyr 0.8 group_nest和group_split）。
我本质上想做的是获得另一个列表列，该列为每个政府提供每个以前的政府的列表，其中包含一个表明政党和议席重叠的数据框。
library(tidyverse)
df <- tibble::tribble(
~period, ~party, ~seats,
1, "A", 2,
1, "B", 3,
1, "C", 3,
2, "A", 2,
2, "C", 3,
3, "C", 4,
3, "E", 1,
3, "F", 3
)
df <- bind_rows(AA=df, BB=df, .id="country")
df <- df %>%
group_by(country, period) %>%
group_nest() %>%
#mutate(gov=map(data, "party") %>% map(.,list)) %>%
mutate(prev.govs=map(data, "party") %>%
map(., list) %>%
accumulate(.,union))
df <- df %>%
group_split(country) %>%
map(., ~mutate(., prev.govs.df=map_depth(prev.govs, 2, enframe, value="party")))
df是我的出发点。在失败的尝试之下。
##attempts
df %>%
map(., ~mutate(., df.overlap=map_depth(prev.govs.df, 3, ~map2(., data, inner_join))))
#> Error in UseMethod("inner_join"): nicht anwendbare Methode für 'inner_join' auf Objekt der Klasse "c('integer', 'numeric')" angewendet
df %>%
map(., ~mutate(., df.overlap=map_depth(prev.govs.df, 2, ~map2(., data, inner_join))))
#> Error: Mapped vectors must have consistent lengths:
#> * `.x` has length 2
#> * `.y` has length 3
df %>%
map(., ~mutate(., df.overlap=map2(data, prev.govs.df, ~map2(.x, .y, ~map2(.x, .y, inner_join)))))
#> Error: Mapped vectors must have consistent lengths:
#> * `.x` has length 3
#> * `.y` has length 2
在更具体的水平上，时段3中country AA的解决方案将是3个列表，每个列表都有一个小标题，其中包含来自data的行与与prev.govs.def中的行重叠的行party列（键）
df[[1]][["prev.govs.df"]][[3]]
#> [[1]]
#> # A tibble: 3 x 2
#> name party
#> <int> <chr>
#> 1 1 A
#> 2 2 B
#> 3 3 C
#>
#> [[2]]
#> # A tibble: 2 x 2
#> name party
#> <int> <chr>
#> 1 1 A
#> 2 2 C
#>
#> [[3]]
#> # A tibble: 3 x 2
#> name party
#> <int> <chr>
#> 1 1 C
#> 2 2 E
#> 3 3 F
df[[1]][["data"]][[3]]
#> # A tibble: 3 x 2
#> party seats
#> <chr> <dbl>
#> 1 C 4
#> 2 E 1
#> 3 F 3
先前问题的答案解决了谜语如何使两个列表相交。不幸的是，我无法弄清楚下一步如何拆分数据框并合并嵌套的小对象。
非常有用！
答案 0 :(得分：2)
一个原因是length元素的list中存在差异的问题。我们可以rep组合一个list元素以使长度相同，然后执行inner_join
out <- df %>%
map(., ~ .x %>%
mutate(df.overlap = map2(prev.govs.df, data, ~
map2(rep(list(.y), length(.x)), .x, inner_join))))
-输出
out[[1]]
# A tibble: 3 x 6
# country period data prev.govs prev.govs.df df.overlap
# <chr> <dbl> <list> <list> <list> <list>
#1 AA 1 <tibble [3 × 2]> <list [1]> <list [1]> <list [1]>
#2 AA 2 <tibble [2 × 2]> <list [2]> <list [2]> <list [2]>
#3 AA 3 <tibble [3 × 2]> <list [3]> <list [3]> <list [3]>
# overlap column element
out[[1]]$df.overlap[[3]][[1]]
# A tibble: 1 x 3
# party seats name
# <chr> <dbl> <int>
#1 C 4 3
# input dataset elements used for joining
out[[1]]$data[[3]]
# A tibble: 3 x 2
# party seats
# <chr> <dbl>
#1 C 4
#2 E 1
#3 F 3
out[[1]]$prev.govs.df[[3]][[1]]
# A tibble: 3 x 2
# name party
# <int> <chr>
#1 1 A
#2 2 B
#3 3 C
答案 1 :(得分：1)
OP的第三次尝试实际上已经接近完成。我们只需要像下面这样修改最后一个map：
library(tidyverse)
output <- df %>%
map(~mutate(., df.overlap = map2(data, prev.govs.df, ~map(.y, inner_join, .x))))
输出：
[[1]]
# A tibble: 3 x 6
country period data prev.govs prev.govs.df df.overlap
<chr> <dbl> <list> <list> <list> <list>
1 AA 1 <tibble [3 x 2]> <list [1]> <list [1]> <list [1]>
2 AA 2 <tibble [2 x 2]> <list [2]> <list [2]> <list [2]>
3 AA 3 <tibble [3 x 2]> <list [3]> <list [3]> <list [3]>
[[2]]
# A tibble: 3 x 6
country period data prev.govs prev.govs.df df.overlap
<chr> <dbl> <list> <list> <list> <list>
1 BB 1 <tibble [3 x 2]> <list [3]> <list [3]> <list [3]>
2 BB 2 <tibble [2 x 2]> <list [3]> <list [3]> <list [3]>
3 BB 3 <tibble [3 x 2]> <list [3]> <list [3]> <list [3]>
> output[[1]]$df.overlap[[3]]
[[1]]
# A tibble: 1 x 3
name party seats
<int> <chr> <dbl>
1 3 C 4
[[2]]
# A tibble: 1 x 3
name party seats
<int> <chr> <dbl>
1 2 C 4
[[3]]
# A tibble: 3 x 3
name party seats
<int> <chr> <dbl>
1 1 C 4
2 2 E 1
3 3 F 3