Python dataframe separate cell values containing lists

Question

I have a dataframe df:

        0               1               2   
Mon ['x','y','z']   ['a','b','c']   ['a','b','c']
Tue ['a','b','c']   ['a','b','c']   ['x','y','z']
Wed ['a','b','c']   ['a','b','c']   ['a','b','c']

Lists are all of diff from each other (Maybe similar too) and I wish to convert it to the form:

    0 1 2
Mon x a a
Mon y b b
Mon z c c
Tue a a x
Tue b b y
Tue c c z
Wed a a a
Wed b b b
Wed c c c

Referring to some previous SO questions, Explode lists with different lengths in Pandas,
Split (explode) pandas dataframe string entry to separate rows

I have tried to use their solutions but I am unable to get the desired output. How can I achieve this?

s1 = df[0]
s2 = df[1]
s3 = df[2]
i1 = np.arange(len(df)).repeat(s1.str.len())
i2 = np.arange(len(df)).repeat(s2.str.len())
i3 = np.arange(len(df)).repeat(s3.str.len())
df.iloc[i1, :-1].assign(**{'Shared Codes': np.concatenate(s1.values)})
df.iloc[i2, :-1].assign(**{'Shared Codes': np.concatenate(s2.values)})
df.iloc[i3, :-1].assign(**{'Shared Codes': np.concatenate(s3.values)})

Also, this doesn’t seem like a very reasonable way to do it, provided I have even more columns. Using python 2.7.

Asked By: inquisitiveProgrammer

||

Source

Answer 1

This is one way using itertools.chain and numpy.repeat:

import pandas as pd, numpy as np
from itertools import chain

df = pd.DataFrame({0: [['x', 'y', 'z'], ['a', 'b', 'c'], ['a', 'b', 'c']],
                   1: [['a', 'b', 'c'], ['a', 'b', 'c'], ['a', 'b', 'c']],
                   2: [['a', 'b', 'c'], ['x', 'y', 'z'], ['a', 'b', 'c']]},
                  index=['Mon', 'Tue', 'Wed'])

res = pd.DataFrame({k: list(chain.from_iterable(df[k])) for k in df},
                   index=np.repeat(df.index, list(map(len, df[0]))))

print(res)

#      0  1  2
# Mon  x  a  a
# Mon  y  b  b
# Mon  z  c  c
# Tue  a  a  x
# Tue  b  b  y
# Tue  c  c  z
# Wed  a  a  a
# Wed  b  b  b
# Wed  c  c  c

Answered By: jpp

Answer 2

A simple iteration might help if the columns contain list made up of 3 elements each i.e :

ndf = pd.concat([df.apply(lambda x : [i[j] for i in x],1) for j in range(3)]).sort_index()

     0  1  2
Mon  x  a  a
Mon  y  b  b
Mon  z  c  c
Tue  a  a  x
Tue  b  b  y
Tue  c  c  z
Wed  a  a  a
Wed  b  b  b
Wed  c  c  c

Answered By: Bharath M Shetty

Answer 3

I’d do it this way:

dfs = []
for day in df.index:
    part = pd.DataFrame(df.loc[day].tolist()).T
    part.index = np.repeat(day, len(df.columns))
    dfs.append(part)
result = pd.concat(dfs)

Answered By: John Zwinck

Answer 4

from pandas import DataFrame
import numpy as np
import numpy as np
import pandas as pd
import pandas as pd




df = pd.DataFrame({0: [['x', 'y', 'z'], ['a', 'b', 'c'], ['a', 'b', 'c']],
                   1: [['a', 'b', 'c'], ['a', 'b', 'c'], ['a', 'b', 'c']],
                   2: [['a', 'b', 'c'], ['x', 'y', 'z'], ['a', 'b', 'c']]},
                  index=['Mon', 'Tue', 'Wed'])

print(df)
"""
             0          1          2
Mon  [x, y, z]  [a, b, c]  [a, b, c]
Tue  [a, b, c]  [a, b, c]  [x, y, z]
Wed  [a, b, c]  [a, b, c]  [a, b, c]

"""

idx = df.index.repeat(df.apply(len))
print(idx)
"""
Index(['Mon', 'Mon', 'Mon', 'Tue', 'Tue', 'Tue', 'Wed', 'Wed', 'Wed'], dtype='object')
"""

res = pd.DataFrame(df.explode([0,1,2]) , index = df.index.repeat(df.apply(len)))
print(res)

"""
     0  1  2
Mon  x  a  a
Mon  y  b  b
Mon  z  c  c
Tue  a  a  x
Tue  b  b  y
Tue  c  c  z
Wed  a  a  a
Wed  b  b  b
Wed  c  c  c
"""

Answered By: Soudipta Dutta

Python dataframe separate cell values containing lists

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