Pandas 数据帧的元组列表列表? [英] List of LISTS of tuples to Pandas dataframe?

问题描述

我有一个元组列表的列表，其中每个元组的长度相等，我需要将元组转换为Pandas数据框，使数据框的列等于元组的长度，每个元组项目都是跨列的行条目.

I have a list of lists of tuples, where every tuple is of equal length, and I need to convert the tuples to a Pandas dataframe in such a way that the columns of the dataframe are equal to the length of the tuples, and each tuple item is a row entry across the columns.

我已经咨询了有关此主题的其他问题(例如，熊猫数据框的元组列表，

I have consulted other questions on this topic (e.g., Convert a list of lists of tuples to pandas dataframe, List of list of tuples to pandas dataframe, split list of tuples in lists of list of tuples) unsuccessfully.

我最接近的是列表理解，它来自Stack Overflow上的另一个问题:

The closest I get is with list comprehension from a different question on Stack Overflow:

import pandas as pd

tupList = [[('commentID', 'commentText', 'date'), ('123456', 'blahblahblah', '2019')], [('45678', 'hello world', '2018'), ('0', 'text', '2017')]]

# Trying list comprehension from previous stack question:
pd.DataFrame([[y for y in x] for x in tupList])

但这会产生意想不到的结果:

But this yields the unintended result:

    0                                 1
0   (commentID, commentText, date)    (123456, blahblahblah, 2019)
1   (45678, hello world, 2018)        (0, text, 2017)

预期结果如下:

      0            1                 2
0     commentID    commentText       date
1     123456       blahblahblah      2019
2     45678        hello world       2018
3     0            text              2017

总而言之:我需要的列数等于每个元组的长度(在示例中为3)，其中元组中的每个项目都是跨列的行条目.

In sum: I need columns equal to the length of each tuple (in the example, 3), where each item within the tuple is a row entry across the columns.

谢谢！

推荐答案

只需将列表展平到元组列表中(您的初始列表包含元组的子列表):

Just flatten your list into a list of tuples (your initial list contains a sublists of tuples):

In [1251]: tupList = [[('commentID', 'commentText', 'date'), ('123456', 'blahblahblah', '2019')], [('45678', 'hello world', '2018'), ('0', 'text', '2017')]]

In [1252]: pd.DataFrame([t for lst in tupList for t in lst])
Out[1252]: 
           0             1     2
0  commentID   commentText  date
1     123456  blahblahblah  2019
2      45678   hello world  2018
3          0          text  2017

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