将具有可变数量元素的列表的嵌套列表展平到数据框 [英] Flatten nested list of lists with variable numbers of elements to a data frame

问题描述

我有一个嵌套的列表列表，我想将这些列表拼凑成一个带有 id 变量的数据框，以便我知道每个列表元素(和子列表元素)来自哪个列表元素.

I've got a nested list of lists that I'd like to flatten into a dataframe with id variables so I know which list elements (and sub-list elements) each came from.

> str(gc_all)
List of 3
$ 1: num [1:102, 1:2] -74 -73.5 -73 -72.5 -71.9 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:2] "lon" "lat"
$ 2: num [1:102, 1:2] -74 -73.3 -72.5 -71.8 -71 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:2] "lon" "lat"
$ 3:List of 2
..$ : num [1:37, 1:2] -74 -74.4 -74.8 -75.3 -75.8 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : NULL
.. .. ..$ : chr [1:2] "lon" "lat"
..$ : num [1:65, 1:2] 180 169 163 158 154 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : NULL
.. .. ..$ : chr [1:2] "lon" "lat"

我以前使用 plyr::ldply(mylist, rbind) 来展平列表，但由于列表长度可变，我似乎遇到了问题:一些列表元素只包含一个数据框，而其他包含两个数据框的列表.

I've used plyr::ldply(mylist, rbind) for flattening lists before, but I seem to be encountering trouble due to variable list lengths: some list elements contain only one dataframe, whilst others contain a list of two dataframes.

我找到了一个笨拙的解决方案，使用两个 lapply 和一个 ifelse，如下所示:

I've found a clunky solution using two lapplys and an ifelse like so:

# sample latitude-longitude data
df <- data.frame(source_lat = rep(40.7128, 3),
                 source_lon = rep(-74.0059, 3),
                 dest_lat = c(55.7982, 41.0082, -7.2575),
                 dest_lon = c(37.968, 28.9784, 112.7521),
                 id = 1:3)

# split into list
gc_list <- split(df, df$id)

# get great circles between lat-lon for each id; multiple list elements are outputted when the great circle crosses the dateline
gc_all <- lapply(gc_list, function(x) {
  geosphere::gcIntermediate(x[, c("source_lon", "source_lat")],
                 x[, c("dest_lon", "dest_lat")],
                 n = 100, addStartEnd=TRUE, breakAtDateLine=TRUE)
})

gc_fortified <- lapply(1:length(gc_all), function(i) {
  if(class(gc_all[[i]]) == "list") {
    lapply(1:length(gc_all[[i]]), function(j) {
      data.frame(gc_all[[i]][[j]], id = i, section = j)
    }) %>%
      plyr::rbind.fill()
  } else {
    data.frame(gc_all[[i]], id = i, section = 1)
  }
}) %>%
  plyr::rbind.fill()

但我觉得必须有一个更优雅的解决方案，可以作为单线工作，例如dput、data.table?

But I feel like there must be a more elegant solution that works as a one-liner, e.g. dput, data.table?

这是我期望的输出:

> gc_fortified %>% 
    group_by(id, section) %>%
    slice(1)

lon      lat    id section
<dbl>    <dbl> <int>   <dbl>
1 -74.0059 40.71280     1       1
2 -74.0059 40.71280     2       1
3 -74.0059 40.71280     3       1
4 180.0000 79.70115     3       2

推荐答案

首先需要重新设计列表的结构，使其成为一个常规的列表列表，然后我们应用两次map_dfr，使用 .id 参数.

First the structure of the list needs to be reworked so it becomes a regular list of lists, then we apply map_dfr two times, using the .id parameter.

library(purrr)
gc_all_df  <- map(map_if(gc_all,~class(.x)=="matrix",list),~map(.x,as.data.frame))
map_dfr(gc_all_df,~map_dfr(.x,identity,.id="id2"),identity,.id="id1")

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