pandas 将熊猫系列列表转换为数据框
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Convert pandas series of lists to dataframe
提问by Hatshepsut
I have a series made of lists
我有一系列的列表
import pandas as pd
s = pd.Series([[1, 2, 3], [4, 5, 6]])
and I want a DataFrame with each column a list.
我想要一个每列一个列表的 DataFrame。
None of from_items, from_records, DataFrameSeries.to_frameseem to work.
没有from_items，from_records，DataFrameSeries.to_frame似乎工作。
How to do this?
这该怎么做？
回答by Cleb
You can use from_itemslike this (assuming that your lists are of the same length):
您可以这样使用from_items（假设您的列表长度相同）：
pd.DataFrame.from_items(zip(s.index, s.values))
0 1
0 1 4
1 2 5
2 3 6
or
或者
pd.DataFrame.from_items(zip(s.index, s.values)).T
0 1 2
0 1 2 3
1 4 5 6
depending on your desired output.
取决于您想要的输出。
This can be much faster than using an apply(as used in @Wen's answerwhich, however, does also work for lists of different length):
这可能比使用 an 快得多apply（如@Wen 的回答中使用的那样，但是，它也适用于不同长度的列表）：
%timeit pd.DataFrame.from_items(zip(s.index, s.values))
1000 loops, best of 3: 669 μs per loop
%timeit s.apply(lambda x:pd.Series(x)).T
1000 loops, best of 3: 1.37 ms per loop
and
和
%timeit pd.DataFrame.from_items(zip(s.index, s.values)).T
1000 loops, best of 3: 919 μs per loop
%timeit s.apply(lambda x:pd.Series(x))
1000 loops, best of 3: 1.26 ms per loop
Also @Hatshepsut's answeris quite fast (also works for lists of different length):
另外@Hatshepsut 的回答非常快（也适用于不同长度的列表）：
%timeit pd.DataFrame(item for item in s)
1000 loops, best of 3: 636 μs per loop
and
和
%timeit pd.DataFrame(item for item in s).T
1000 loops, best of 3: 884 μs per loop
Fastest solution seems to be @Abdou's answer(tested for Python 2; also works for lists of different length; use itertools.zip_longestin Python 3.6+):
最快的解决方案似乎是@Abdou 的答案（针对 Python 2 进行了测试；也适用于不同长度的列表；itertools.zip_longest在 Python 3.6+ 中使用）：
%timeit pd.DataFrame.from_records(izip_longest(*s.values))
1000 loops, best of 3: 529 μs per loop
An additional option:
一个额外的选择：
pd.DataFrame(dict(zip(s.index, s.values)))
0 1
0 1 4
1 2 5
2 3 6
回答by Abdou
pd.DataFrame.from_recordsshould also work using itertools.zip_longest:
pd.DataFrame.from_records也应该使用itertools.zip_longest：
from itertools import zip_longest
pd.DataFrame.from_records(zip_longest(*s.values))
# 0 1
# 0 1 4
# 1 2 5
# 2 3 6
回答by Z.Webber
If the length of the series is super high (more than 1m), you can use:
如果系列的长度超高（超过1m），可以使用：
s = pd.Series([[1, 2, 3], [4, 5, 6]])
pd.DataFrame(s.tolist())
回答by Hatshepsut
Iterate over the series like this:
像这样迭代系列：
series = pd.Series([[1, 2, 3], [4, 5, 6]])
pd.DataFrame(item for item in series)
0 1 2
0 1 2 3
1 4 5 6
回答by YOBEN_S
You may looking for
您可能正在寻找
s.apply(lambda x:pd.Series(x))
0 1 2
0 1 2 3
1 4 5 6
Or
或者
s.apply(lambda x:pd.Series(x)).T
Out[133]:
0 1
0 1 4
1 2 5
2 3 6
回答by Dataman
Notethat the from_items()method in the accepted answer is deprecatedin the latest Pandas and from_dict()method should be used instead. Here is how:
请注意，已from_items()接受答案中的方法在最新的 Pandas 中已弃用，from_dict()应改用方法。方法如下：
pd.DataFrame.from_dict(dict(zip(s.index, s.values)))
## OR
pd.DataFrame.from_dict(dict(zip(s.index, s.values))).T
Also note that using from_dict()provides us with the fastestapproach so far:
另请注意， usingfrom_dict()为我们提供了迄今为止最快的方法：
%timeit pd.DataFrame.from_dict(dict(zip(s.index, s.values)))
376 μs ± 14.4 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
## OR
%timeit pd.DataFrame.from_dict(dict(zip(s.index, s.values))).T
487 μs ± 3.5 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
回答by Evan Rosica
Try:
尝试：
import numpy as np, pandas as pd
s = pd.Series([[1, 2, 3], [4, 5, 6]])
pd.DataFrame(np.vstack(s))
