如何将键值对的列表列表转换为数据框？

huangapple
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英文:

How to convert list of lists of key value pairs to dataframe?

问题

注意，我正在处理的输入形状与类似问题中的不同：
https://stackoverflow.com/questions/29674661/r-list-of-lists-to-data-frame; https://stackoverflow.com/questions/59928743/converting-a-list-of-lists-to-a-dataframe-in-r-the-tidyverse-way

我正在使用返回以下形状的Web API 工作

  1. listOfListsOfKeyVals <- lapply(1:5, function(i){
  2.     list(
  3.       col1 = i,
  4.       col2 = runif(1)
  5.     )
  6. })

你可能认为以下方法会起作用

  1. do.call(rbind, listOfListsOfKeyVals)

但仔细检查后，结果实际上是一个列表的数据框

  1. do.call(rbind, listOfListsOfKeyVals) |> tibble()
  3. # A tibble: 5 × 1
  4.   `do.call(rbind, listOfListsOfKeyVals)`[,"col1"] [,"col2"]
  5.   <list>                                          <list>   
  6. 1 <int [1]>                                       <dbl [1]>
  7. 2 <int [1]>                                       <dbl [1]>
  8. 3 <int [1]>                                       <dbl [1]>
  9. 4 <int [1]>                                       <dbl [1]>
  10. 5 <int [1]>                                       <dbl [1]>

我提出了以下解决方案

  1. foreach(x = listOfListsOfKeyVals, .combine = rbind) %do% {
  2.   as.data.frame(x)
  3. } |> tibble()

但对于大型数据集来说速度非常慢。有更好的方法吗？

英文:

Note the input shape I am dealing with is not the same as in these similar questions:
https://stackoverflow.com/questions/29674661/r-list-of-lists-to-data-frame; https://stackoverflow.com/questions/59928743/converting-a-list-of-lists-to-a-dataframe-in-r-the-tidyverse-way

I am working with a web api which returns results in this shape

  1. listOfListsOfKeyVals <- lapply(1:5, function(i){
  2.     list(
  3.       col1 = i,
  4.       col2 = runif(1)
  5.     )
  6. })

You might think that the following will work

  1. do.call(rbind, listOfListsOfKeyVals)

But on closer inspection the result is actually a dataframe of lists

  1. do.call(rbind, listOfListsOfKeyVals) |> tibble()
  3. # A tibble: 5 × 1
  4.   `do.call(rbind, listOfListsOfKeyVals)`[,"col1"] [,"col2"]
  5.   <list>                                          <list>   
  6. 1 <int [1]>                                       <dbl [1]>
  7. 2 <int [1]>                                       <dbl [1]>
  8. 3 <int [1]>                                       <dbl [1]>
  9. 4 <int [1]>                                       <dbl [1]>
  10. 5 <int [1]>                                       <dbl [1]>

I have come up with the following solution

  1. foreach(x = listOfListsOfKeyVals, .combine = rbind) %do% {
  2.   as.data.frame(x)
  3. } |> tibble()

But it is painfully slow for large data sets. Is there a better way?

答案1

得分: 4

我认为你正在寻找 dplyr::bind_rows。

  1. library(dplyr)
  3. set.seed(12)
  4. listOfListsOfKeyVals <- lapply(1:5, function(i){
  5.   list(
  6.     col1 = i,
  7.     col2 = runif(1)
  8.   )
  9. })
  11. bind_rows(listOfListsOfKeyVals)

输出：

  1. # A tibble: 5 × 2
  2.    col1   col2
  3.   <int>  <dbl>
  4. 1     1 0.0694
  5. 2     2 0.818 
  6. 3     3 0.943 
  7. 4     4 0.269 
  8. 5     5 0.169
英文:

I think you are looking for dplyr::bind_rows.

  1. library(dplyr)
  3. set.seed(12)
  4. listOfListsOfKeyVals &lt;- lapply(1:5, function(i){
  5.   list(
  6.     col1 = i,
  7.     col2 = runif(1)
  8.   )
  9. })
  11. bind_rows(listOfListsOfKeyVals)

Output:

  1. # A tibble: 5 &#215; 2
  2.    col1   col2
  3.   &lt;int&gt;  &lt;dbl&gt;
  4. 1     1 0.0694
  5. 2     2 0.818 
  6. 3     3 0.943 
  7. 4     4 0.269 
  8. 5     5 0.169

答案2

得分: 1

另一个选项是 rrapply：

  1. rrapply::rrapply(listOfListsOfKeyVals, how = "bind")
  3. #   col1       col2
  4. # 1    1 0.21794909
  5. # 2    2 0.81600287
  6. # 3    3 0.04631368
  7. # 4    4 0.10518273
  8. # 5    5 0.46489659

还有一个基础的 R 选项，使用 matrix：

  1. l <- listOfListsOfKeyVals
  2. data.frame(matrix(unlist(l), nrow = length(l), byrow = TRUE)) ->
  3.   setNames(names(l[[1]]))
英文:

Another option is rrapply:

  1. rrapply::rrapply(listOfListsOfKeyVals, how = &quot;bind&quot;)
  3. #   col1       col2
  4. # 1    1 0.21794909
  5. # 2    2 0.81600287
  6. # 3    3 0.04631368
  7. # 4    4 0.10518273
  8. # 5    5 0.46489659

And a base R option with matrix:

  1. l &lt;- listOfListsOfKeyVals
  2. data.frame(matrix(unlist(l), nrow = length(l), byrow = TRUE)) |&gt;
  3.   setNames(names(l[[1]]))

huangapple
  • 本文由 发表于 2023年6月8日 13:47:27
  • 转载请务必保留本文链接：https://go.coder-hub.com/76428910.html
  • dataframe
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