如何在 Python 中将元组元组转换为 pandas.DataFrame？
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How to convert tuple of tuples to pandas.DataFrame in Python?
提问by Oleg Melnikov
No offence, if the questions is too basic. Let me know if you need more information.
没有冒犯，如果问题太基本。如果您需要更多信息，请与我们联系。
I am looking for an idea to convert square-form tuple of tuples to pandas.DataFrame in a clean/efficient/pythonic way, i.e. from
我正在寻找一个想法，以干净/高效/pythonic的方式将元组的方形元组转换为pandas.DataFrame，即从
s =((1,0,0,0,),(2,3,0,0,),(4,5,6,0,),(7,8,9,10,))
to pandas.DataFramelike
到pandas.DataFrame像
1 2 3 4
1 1 0 0 0
2 2 3 0 0
3 4 5 6 0
4 7 8 9 10
Naturally, this list can grow with more zeros appended in the upper-triangular (if we think of s as a tuple of rows).
自然地，这个列表可以随着在上三角形中附加更多的零而增长（如果我们将 s 视为行的元组）。
DataFrame(t)seems to fail.
DataFrame(t)似乎失败了。
采纳答案by furas
import pandas as pd
s = ((1,0,0,0,),(2,3,0,0,),(4,5,6,0,),(7,8,9,10,))
print pd.DataFrame(list(s))
# 0 1 2 3
# 0 1 0 0 0
# 1 2 3 0 0
# 2 4 5 6 0
# 3 7 8 9 10
print pd.DataFrame(list(s), columns=[1,2,3,4], index=[1,2,3,4])
# 1 2 3 4
# 1 1 0 0 0
# 2 2 3 0 0
# 3 4 5 6 0
# 4 7 8 9 10
回答by unutbu
Pass a listof tuples instead of a tuple of tuples:
传递元组列表而不是元组元组：
In [13]: pd.DataFrame(list(s))
Out[13]:
0 1 2 3
0 1 0 0 0
1 2 3 0 0
2 4 5 6 0
3 7 8 9 10
pd.DataFrame(data)follows different code paths when datais a tuple as opposed to a list.
pd.DataFrame(data)当data是元组而不是列表时，遵循不同的代码路径。
Pandas developer Jeff Reback explains:
Pandas 开发人员Jeff Reback 解释说：
list-of-tuples is the specified type, tuple-of-tuple is not allowed as I think it can signify nested types that would require more parsing.
list-of-tuples 是指定的类型，tuple-of-tuple 是不允许的，因为我认为它可以表示需要更多解析的嵌套类型。
