用 purrr 重组列表

Question

我在使用 R 和 purrr 清理复杂列表时遇到以下问题。示例将清除这个。

需要包裹

library(jsonlite)
library(purrr)
library(tibble)
library(dplyr)

让我们从 NHL json.

获取示例数据

url <- c("https://statsapi.web.nhl.com/api/v1/game/2017010002/feed/live", "https://statsapi.web.nhl.com/api/v1/game/2017010005/feed/live")

data_list <- map(url, fromJSON)

这是一个非常复杂的列表，但最后我想要的是 result。主要思想是我想在其中包含带有 tibble 的列表列，但是 tibble 是从列表的不同部分选择的 vectors/columns 创建的。这是我失败的尝试：

wrong <- tibble(
  gamepk = map_chr(data_list, c("gamePk"), .default = NA),
  home = map_chr(data_list, c("gameData", "teams", "home", "triCode"), .default = NA),
  away = map_chr(data_list, c("gameData", "teams", "away", "triCode"), .default = NA),
  test = list(tibble(event = map(data_list, c("liveData", "plays", "allPlays", "result", "event"), .default = NA),
                     x_cord = map(data_list, c("liveData", "plays", "allPlays", "coordinates", "x"), .default = NA),
                     y_cord = map(data_list, c("liveData", "plays", "allPlays", "coordinates", "y"), .default = NA)))
)

# A tibble: 2 x 4
  gamepk     home  away  test            
  <chr>      <chr> <chr> <list>          
1 2017010002 VAN   VGK   <tibble [2 x 3]>
2 2017010005 BUF   CAR   <tibble [2 x 3]>

这是一种笨拙的方式来完成它并纠正结果。

result <- tibble(
  gamepk = map_chr(data_list, c("gamePk"), .default = NA),
  home = map_chr(data_list, c("gameData", "teams", "home", "triCode"), .default = NA),
  away = map_chr(data_list, c("gameData", "teams", "away", "triCode"), .default = NA))

test1 <- bind_cols(event = data_list[[1]]$liveData$plays$allPlays$result$event, 
          x_cord = data_list[[1]]$liveData$plays$allPlays$coordinates$x,
          y_cord = data_list[[1]]$liveData$plays$allPlays$coordinates$y)

test2 <- bind_cols(event = data_list[[2]]$liveData$plays$allPlays$result$event, 
                  x_cord = data_list[[2]]$liveData$plays$allPlays$coordinates$x,
                  y_cord = data_list[[2]]$liveData$plays$allPlays$coordinates$y)

result$test[1] <- list(test1)
result$test[2] <- list(test2)

result

# A tibble: 2 x 4
  gamepk     home  away  test              
  <chr>      <chr> <chr> <list>            
1 2017010002 VAN   VGK   <tibble [314 x 3]>
2 2017010005 BUF   CAR   <tibble [283 x 3]>

Answer 1

在您已经完成的所有工作的基础上，如果您制作 test 列表列表，您可以使用 tibble() 将其传递给 pmap()。这同时循环遍历嵌套列表的每个元素并将它们 "entibbles" 放在一起，为您留下一个包含两个小标题的列表。

test 的新行如下所示：

test = list(event = map(data_list, c("liveData", "plays", "allPlays", "result", "event"), .default = NA),
                x_cord = map(data_list, c("liveData", "plays", "allPlays", "coordinates", "x"), .default = NA),
                y_cord = map(data_list, c("liveData", "plays", "allPlays", "coordinates", "y"), .default = NA)) %>% 
        pmap(tibble)

将其与您已有的内容相结合，得到 "right" 输出。

right <- tibble(
    gamepk = map_chr(data_list, c("gamePk"), .default = NA),
    home = map_chr(data_list, c("gameData", "teams", "home", "triCode"), .default = NA),
    away = map_chr(data_list, c("gameData", "teams", "away", "triCode"), .default = NA),
    test = list(event = map(data_list, c("liveData", "plays", "allPlays", "result", "event"), .default = NA),
                x_cord = map(data_list, c("liveData", "plays", "allPlays", "coordinates", "x"), .default = NA),
                y_cord = map(data_list, c("liveData", "plays", "allPlays", "coordinates", "y"), .default = NA)) %>% 
        pmap(tibble) )

right

# A tibble: 2 x 4
  gamepk     home  away  test              
  <chr>      <chr> <chr> <list>            
1 2017010002 VAN   VGK   <tibble [314 x 3]>
2 2017010005 BUF   CAR   <tibble [283 x 3]>

用 purrr 重组列表

Restructuring list with purrr

r

list

purrr