Convert pandas series of lists to dataframe

Question

I have a series made of lists

import pandas as pd
s = pd.Series([[1, 2, 3], [4, 5, 6]])

and I want a DataFrame with each column a list.

Answer 1

As @Hatshepsut pointed out in the comments, from_items is deprecated as of version 0.23. The link suggests to use from_dict instead, so the old answer can be modified to:

pd.DataFrame.from_dict(dict(zip(s.index, s.values)))

--------------------------------------------------OLD ANSWER-------------------------------------------------------------

You can use from_items like this (assuming that your lists are of the same length):

pd.DataFrame.from_items(zip(s.index, s.values))

   0  1
0  1  4
1  2  5
2  3  6

or

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.

This can be much faster than using an apply (as used in @Wen's answer which, however, does also work for lists of different length):

%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

%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 answer is quite fast (also works for lists of different length):

%timeit pd.DataFrame(item for item in s)
1000 loops, best of 3: 636 µs per loop

and

%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_longest in Python 3.6+):

%timeit pd.DataFrame.from_records(izip_longest(*s.values))
1000 loops, best of 3: 529 µs per loop

An additional option:

pd.DataFrame(dict(zip(s.index, s.values)))

   0  1
0  1  4
1  2  5
2  3  6

Answer 2

If the length of the series is super high (more than 1m), you can use:

s = pd.Series([[1, 2, 3], [4, 5, 6]])
pd.DataFrame(s.tolist())

Answer 3

Note that the from_items() method in the accepted answer is deprecated in the latest Pandas and from_dict() method should be used instead. Here is how:

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 fastest approach so far:

%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)

Answer 4

pd.DataFrame.from_records should also work using 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

Answer 5

Try:

import numpy as np, pandas as pd
s = pd.Series([[1, 2, 3], [4, 5, 6]])
pd.DataFrame(np.vstack(s))

Answer 6

You may looking for

s.apply(lambda x:pd.Series(x))
   0  1  2
0  1  2  3
1  4  5  6

Or

 s.apply(lambda x:pd.Series(x)).T

Out[133]: 
   0  1
0  1  4
1  2  5
2  3  6