问题
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.
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.
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])
But this yields the unintended result:
0 1
0 (commentID, commentText, date) (123456, blahblahblah, 2019)
1 (45678, hello world, 2018) (0, text, 2017)
When the expected result is as follows:
0 1 2
0 commentID commentText date
1 123456 blahblahblah 2019
2 45678 hello world 2018
3 0 text 2017
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.
Thanks!
回答1:
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
回答2:
A shorter code this:
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)
Edit
You can make the list comprehension directly in the from_records
function.
回答3:
tupList = [[('commentID', 'commentText', 'date'), ('123456', 'blahblahblah', '2019')], [('45678', 'hello world', '2018'), ('0', 'text', '2017')]]
print(pd.DataFrame(sum(tupList,[])))
Output
0 1 2
0 commentID commentText date
1 123456 blahblahblah 2019
2 45678 hello world 2018
3 0 text 2017
回答4:
You can do it like this :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()
来源:https://stackoverflow.com/questions/57509968/list-of-lists-of-tuples-to-pandas-dataframe