Pandas 数据帧的元组列表列表？

Question

我有一个元组列表列表，其中每个元组的长度相同，我需要将元组转换为 Pandas 数据帧，使数据帧的列等于元组的长度，并且每个元组项都是跨列的行条目。

我已咨询过有关此主题的其他问题（例如，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）未成功。

我得到的最接近的是 Stack Overflow 上另一个问题的列表理解：

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：

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

当预期结果如下：

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

总而言之：我需要等于每个元组长度的列（在示例中为 3），其中元组中的每个项目都是跨列的行条目。

谢谢！

Answer 1

只需将您的列表展平为一个元组列表（您的初始列表包含一个元组的子列表）：

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

Answer 2

更短的代码：

from itertools import chain
import pandas as pd

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

new_list = [x for x in chain.from_iterable(tupList)]
df = pd.DataFrame.from_records(new_list)

---

编辑

您可以直接在from_records 函数中进行列表推导。

Answer 3

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

输出

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

Answer 4

你可以这样做：D

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

# Trying list comprehension from previous stack question:
df = pd.DataFrame([[y for y in x] for x in tupList])
df_1 = df[0].apply(pd.Series).assign(index= range(0, df.shape[0]*2, 2)).set_index("index")
df_2 = df[1].apply(pd.Series).assign(index= range(1, df.shape[0]*2, 2)).set_index("index")

pd.concat([df_1, df_2], axis=0).sort_index()