使用purrr从列表列表中提取数据到自己的data.frame中

Question

代表性样本数据（列表列表）：

l <- list(structure(list(a = -1.54676469632688, b = "s", c = "T", 
d = structure(list(id = 5L, label = "Utah", link = "Asia/Anadyr", 
    score = -0.21104594634643), .Names = c("id", "label", 
"link", "score")), e = 49.1279871269422), .Names = c("a", 
"b", "c", "d", "e")), structure(list(a = -0.934821052832427, 
b = "k", c = "T", d = list(structure(list(id = 8L, label = "South Carolina", 
    link = "Pacific/Wallis", score = 0.526540892113734, externalId = -6.74354377676955), .Names = c("id", 
"label", "link", "score", "externalId")), structure(list(
    id = 9L, label = "Nebraska", link = "America/Scoresbysund", 
    score = 0.250895465294041, externalId = 16.4257470807879), .Names = c("id", 
"label", "link", "score", "externalId"))), e = 52.3161400117052), .Names = c("a", 
"b", "c", "d", "e")), structure(list(a = -0.27261485993069, b = "f", 
c = "P", d = list(structure(list(id = 8L, label = "Georgia", 
    link = "America/Nome", score = 0.526494135483816, externalId = 7.91583574935589), .Names = c("id", 
"label", "link", "score", "externalId")), structure(list(
    id = 2L, label = "Washington", link = "America/Shiprock", 
    score = -0.555186440792989, externalId = 15.0686663219837), .Names = c("id", 
"label", "link", "score", "externalId")), structure(list(
    id = 6L, label = "North Dakota", link = "Universal", 
    score = 1.03168296038975), .Names = c("id", "label", 
"link", "score")), structure(list(id = 1L, label = "New Hampshire", 
    link = "America/Cordoba", score = 1.21582056168681, externalId = 9.7276418869132), .Names = c("id", 
"label", "link", "score", "externalId")), structure(list(
    id = 1L, label = "Alaska", link = "Asia/Istanbul", score = -0.23183264861979), .Names = c("id", 
"label", "link", "score")), structure(list(id = 4L, label = "Pennsylvania", 
    link = "Africa/Dar_es_Salaam", score = 0.590245339334121), .Names = c("id", 
"label", "link", "score"))), e = 132.1153538536), .Names = c("a", 
"b", "c", "d", "e")), structure(list(a = 0.202685974077313, b = "x", 
c = "O", d = structure(list(id = 3L, label = "Delaware", 
    link = "Asia/Samarkand", score = 0.695577130634724, externalId = 15.2364820698193), .Names = c("id", 
"label", "link", "score", "externalId")), e = 97.9908914452971), .Names = c("a", 
"b", "c", "d", "e")), structure(list(a = -0.396243444741009, 
b = "z", c = "P", d = list(structure(list(id = 4L, label = "North Dakota", 
    link = "America/Tortola", score = 1.03060272795705, externalId = -7.21666936522344), .Names = c("id", 
"label", "link", "score", "externalId")), structure(list(
    id = 9L, label = "Nebraska", link = "America/Ojinaga", 
    score = -1.11397997280413, externalId = -8.45145052697411), .Names = c("id", 
"label", "link", "score", "externalId"))), e = 123.597945533926), .Names = c("a", 
"b", "c", "d", "e")))

我有一个列表列表，借助JSON数据下载。

该列表包含176个元素，每个元素包含33个嵌套元素，其中一些也是不同长度的列表。

我对分析特定嵌套列表中包含的数据感兴趣，该列表的长度为〜150，其中每个176个元素包含4个或5个元素-有些包含4个元素，有些包含5个元素。我正在尝试提取此嵌套的列表感兴趣的列表并将其转换为data.frame以便能够执行一些分析。

在上面的代表性示例数据中，我对l的5个元素中的每个元素的嵌套列表d感兴趣。 因此，所需的data.frame如下所示：

id           label            link       score  externalId
 5            Utah     Asia/Anadyr  -0.2110459          NA
 8  South Carolina  Pacific/Wallis   0.5265409   -6.743544
 .
 .

我一直在尝试使用purrr ，它似乎在处理列表中的数据时具有合理且一致的流程，但是我遇到了无法完全理解原因的错误-很有可能是因为我没有正确理解purrr或列表（可能两者）的命令/逻辑。 这是我一直在尝试的代码，但会引发相关的错误：

df <- map_df(l, "d", ~as.data.frame(.))
Error: incompatible sizes (5 != 4)

我相信这与每个组件的d长度不同有关，或者与包含的数据不同（有时4个元素，有时是5个）有关，或者我在这里使用的函数未正确指定-实际上我不确定。

我通过使用for循环解决了这个问题，我知道它效率低下，因此我在SO上提出了问题。

这是我目前使用的for循环：

df <- data.frame(id = integer(), label = character(), score = numeric(), externalId = numeric())
for(i in seq_along(l)){
    df_temp <- l[[i]][[4]] %>% map_df(~as.data.frame(.))
    df <- rbind(df, df_temp)
}

一些援助最好用purrr或者某个版本的- apply ，因为这仍是优于我的for循环-将不胜感激。 另外，如果有上述资源，我想了解，而不仅仅是找到正确的代码。

Answer 1

您可以在三个步骤做到这一点，首先拉出d ，然后中各个要素结合的行d ，然后结合一切都变成一个单一的对象。

我将bind_rows的bind_rows用于列表内行绑定。 map_df执行最后的行绑定。

library(purrr)
library(dplyr)

l %>%
    map("d") %>%
    map_df(bind_rows)

这也等效：

map_df(l, ~bind_rows(.x[["d"]] ) )

结果看起来像：

# A tibble: 12 x 5
      id          label                 link      score externalId
   <int>          <chr>                <chr>      <dbl>      <dbl>
 1     5           Utah          Asia/Anadyr -0.2110459         NA
 2     8 South Carolina       Pacific/Wallis  0.5265409  -6.743544
 3     9       Nebraska America/Scoresbysund  0.2508955  16.425747
 4     8        Georgia         America/Nome  0.5264941   7.915836
 5     2     Washington     America/Shiprock -0.5551864  15.068666
 6     6   North Dakota            Universal  1.0316830         NA
 7     1  New Hampshire      America/Cordoba  1.2158206   9.727642
 8     1         Alaska        Asia/Istanbul -0.2318326         NA
 9     4   Pennsylvania Africa/Dar_es_Salaam  0.5902453         NA
10     3       Delaware       Asia/Samarkand  0.6955771  15.236482
11     4   North Dakota      America/Tortola  1.0306027  -7.216669
12     9       Nebraska      America/Ojinaga -1.1139800  -8.451451

Answer 2

有关purrr的更多信息，我推荐Grolemund和Wickham的“ R for Data Science” http://r4ds.had.co.nz/

我认为您面临的一个问题是l$d中的某些项目是变量列表，每个变量都有一个观察值，可以转换为数据框，而其他项目是此类列表的列表。

但是我本人不是很擅长Purrr。 这是我的处理方式：

l <- lapply(l, function(x){x$d}) ## work with the data you need.

list_of_observations <- Filter(function(x) {!is.null(names(x))},l)

list_of_lists <- Filter(function(x) {is.null(names(x))}, l)

another_list_of_observations <- unlist(list_of_lists, recursive=FALSE)

df <- lapply(c(list_of_observations, another_list_of_observations),
             as.data.frame) %>% bind_rows