Tukukkk

As an user with both R and python, I have seen this type of question a couple of times.

In R, they have the built-in function from package tidyr called unnest. But in Python(pandas) there is no built-in function for this type of question.

I know object columns type always make the data hard to convert with a pandas' function. When I received the data like this , the first thing that came to mind was to 'flatten' or unnest the columns .

Method 1
apply + pd.Series (easy to understand but in terms of performance not recommended . )

df.set_index('A').B.apply(pd.Series).stack().reset_index(level=0).rename(columns={0:'B'})

Out[463]: 

   A  B

0  1  1

1  1  2

0  2  1

1  2  2

Method 2 using repeat with DataFrame constructor , re-create your dataframe (good at performance, not good at multiple columns )

df=pd.DataFrame({'A':df.A.repeat(df.B.str.len()),'B':np.concatenate(df.B.values)})

df

Out[465]: 

   A  B

0  1  1

0  1  2

1  2  1

1  2  2

Method 2.1 for example besides A we have A.1 .....A.n. If we still use the method(Method 2) above it is hard for us to re-create the columns one by one .

Solution : join or merge with the index after 'unnest' the single columns

s=pd.DataFrame({'B':np.concatenate(df.B.values)},index=df.index.repeat(df.B.str.len()))

s.join(df.drop('B',1),how='left')

Out[477]: 

   B  A

0  1  1

0  2  1

1  1  2

1  2  2

If you need the column order exactly the same as before, add reindex at the end.

s.join(df.drop('B',1),how='left').reindex(columns=df.columns)

Method 3 recreate the list

pd.DataFrame([[x] + [z] for x, y in df.values for z in y],columns=df.columns)

Out[488]: 

   A  B

0  1  1

1  1  2

2  2  1

3  2  2

If more than two columns

s=pd.DataFrame([[x] + [z] for x, y in zip(df.index,df.B) for z in y])

s.merge(df,left_on=0,right_index=True)

Out[491]: 

   0  1  A       B

0  0  1  1  [1, 2]

1  0  2  1  [1, 2]

2  1  1  2  [1, 2]

3  1  2  2  [1, 2]

Method 4 using reindex or loc

df.reindex(df.index.repeat(df.B.str.len())).assign(B=np.concatenate(df.B.values))

Out[554]: 

   A  B

0  1  1

0  1  2

1  2  1

1  2  2



#df.loc[df.index.repeat(df.B.str.len())].assign(B=np.concatenate(df.B.values))

Method 5 when the list only contains unique values:

df=pd.DataFrame({'A':[1,2],'B':[[1,2],[3,4]]})

from collections import ChainMap

d = dict(ChainMap(*map(dict.fromkeys, df['B'], df['A'])))

pd.DataFrame(list(d.items()),columns=df.columns[::-1])

Out[574]: 

   B  A

0  1  1

1  2  1

2  3  2

3  4  2

Method 6 using numpy for high performance:

newvalues=np.dstack((np.repeat(df.A.values,list(map(len,df.B.values))),np.concatenate(df.B.values)))

pd.DataFrame(data=newvalues[0],columns=df.columns)

   A  B

0  1  1

1  1  2

2  2  1

3  2  2

Method 7 : using base function itertools cycle and chain: Pure python solution just for fun

from itertools import cycle,chain

l=df.values.tolist()

l1=[list(zip([x[0]], cycle(x[1])) if len([x[0]]) > len(x[1]) else list(zip(cycle([x[0]]), x[1]))) for x in l]

pd.DataFrame(list(chain.from_iterable(l1)),columns=df.columns)

   A  B

0  1  1

1  1  2

2  2  1

3  2  2

Special case (two columns type object)

df=pd.DataFrame({'A':[1,2],'B':[[1,2],[3,4]],'C':[[1,2],[3,4]]})

df

Out[592]: 

   A       B       C

0  1  [1, 2]  [1, 2]

1  2  [3, 4]  [3, 4]

Self-def function

def unnesting(df, explode):

    idx=df.index.repeat(df[explode[0]].str.len())

    df1=pd.concat([pd.DataFrame({x:np.concatenate(df[x].values)} )for x in explode],axis=1)

    df1.index=idx

    return df1.join(df.drop(explode,1),how='left')



unnesting(df,['B','C'])

Out[609]: 

   B  C  A

0  1  1  1

0  2  2  1

1  3  3  2

1  4  4  2

Summary :

I am using pandas and python functions for this type of question. If you are worried about the speed of the above solutions, check user3483203's answer , since he is using numpy and most of the time numpy is faster . I recommend Cpython and numba if speed matters in your case.

Option 1

If all of the sublists in the other column are the same length, numpy can be an efficient option here:

vals = np.array(df.B.values.tolist())    

a = np.repeat(df.A, vals.shape[1])



pd.DataFrame(np.column_stack((a, vals.ravel())), columns=df.columns)

   A  B

0  1  1

1  1  2

2  2  1

3  2  2

Option 2

If the sublists have different length, you need an additional step:

vals = df.B.values.tolist()

rs = [len(r) for r in vals]    

a = np.repeat(df.A, rs)



pd.DataFrame(np.column_stack((a, np.concatenate(vals))), columns=df.columns)

   A  B

0  1  1

1  1  2

2  2  1

3  2  2

Option 3

I took a shot at generalizing this to work to flatten N columns and tile M columns, I'll work later on making it more efficient:

df = pd.DataFrame({'A': [1,2,3], 'B': [[1,2], [1,2,3], [1]],

                   'C': [[1,2,3], [1,2], [1,2]], 'D': ['A', 'B', 'C']})

   A          B          C  D

0  1     [1, 2]  [1, 2, 3]  A

1  2  [1, 2, 3]     [1, 2]  B

2  3        [1]     [1, 2]  C

def unnest(df, tile, explode):

    vals = df[explode].sum(1)

    rs = [len(r) for r in vals]

    a = np.repeat(df[tile].values, rs, axis=0)

    b = np.concatenate(vals.values)

    d = np.column_stack((a, b))

    return pd.DataFrame(d, columns = tile +  ['_'.join(explode)])



unnest(df, ['A', 'D'], ['B', 'C'])

    A  D B_C

0   1  A   1

1   1  A   2

2   1  A   1

3   1  A   2

4   1  A   3

5   2  B   1

6   2  B   2

7   2  B   3

8   2  B   1

9   2  B   2

10  3  C   1

11  3  C   1

12  3  C   2

Functions

def wen1(df):

    return df.set_index('A').B.apply(pd.Series).stack().reset_index(level=0).rename(columns={0: 'B'})



def wen2(df):

    return pd.DataFrame({'A':df.A.repeat(df.B.str.len()),'B':np.concatenate(df.B.values)})



def wen3(df):

    s = pd.DataFrame({'B': np.concatenate(df.B.values)}, index=df.index.repeat(df.B.str.len()))

    return s.join(df.drop('B', 1), how='left')



def wen4(df):

    return pd.DataFrame([[x] + [z] for x, y in df.values for z in y],columns=df.columns)



def chris1(df):

    vals = np.array(df.B.values.tolist())

    a = np.repeat(df.A, vals.shape[1])

    return pd.DataFrame(np.column_stack((a, vals.ravel())), columns=df.columns)



def chris2(df):

    vals = df.B.values.tolist()

    rs = [len(r) for r in vals]

    a = np.repeat(df.A.values, rs)

    return pd.DataFrame(np.column_stack((a, np.concatenate(vals))), columns=df.columns)

Timings

import pandas as pd

import matplotlib.pyplot as plt

import numpy as np

from timeit import timeit



res = pd.DataFrame(

       index=['wen1', 'wen2', 'wen3', 'wen4', 'chris1', 'chris2'],

       columns=[10, 50, 100, 500, 1000, 5000, 10000],

       dtype=float

)



for f in res.index:

    for c in res.columns:

        df = pd.DataFrame({'A': [1, 2], 'B': [[1, 2], [1, 2]]})

        df = pd.concat([df]*c)

        stmt = '{}(df)'.format(f)

        setp = 'from __main__ import df, {}'.format(f)

        res.at[f, c] = timeit(stmt, setp, number=50)



ax = res.div(res.min()).T.plot(loglog=True)

ax.set_xlabel("N")

ax.set_ylabel("time (relative)")

Performance

One alternative is to apply the meshgrid recipe over the rows of the columns to unnest:

import numpy as np

import pandas as pd





def unnest(frame, explode):

    def mesh(values):

        return np.array(np.meshgrid(*values)).T.reshape(-1, len(values))



    data = np.vstack(mesh(row) for row in frame[explode].values)

    return pd.DataFrame(data=data, columns=explode)





df = pd.DataFrame({'A': [1, 2], 'B': [[1, 2], [1, 2]]})

print(unnest(df, ['A', 'B']))  # base

print()



df = pd.DataFrame({'A': [1, 2], 'B': [[1, 2], [3, 4]], 'C': [[1, 2], [3, 4]]})

print(unnest(df, ['A', 'B', 'C']))  # multiple columns

print()



df = pd.DataFrame({'A': [1, 2, 3], 'B': [[1, 2], [1, 2, 3], [1]],

                   'C': [[1, 2, 3], [1, 2], [1, 2]], 'D': ['A', 'B', 'C']})



print(unnest(df, ['A', 'B']))  # uneven length lists

print()

print(unnest(df, ['D', 'B']))  # different types

print()

Output

   A  B

0  1  1

1  1  2

2  2  1

3  2  2



   A  B  C

0  1  1  1

1  1  2  1

2  1  1  2

3  1  2  2

4  2  3  3

5  2  4  3

6  2  3  4

7  2  4  4



   A  B

0  1  1

1  1  2

2  2  1

3  2  2

4  2  3

5  3  1



   D  B

0  A  1

1  A  2

2  B  1

3  B  2

4  B  3

5  C  1

My 5 cents:

df[['B', 'B2']] = pd.DataFrame(df['B'].values.tolist())



df[['A', 'B']].append(df[['A', 'B2']].rename(columns={'B2': 'B'}),

                      ignore_index=True)

and another 5

df[['B1', 'B2']] = pd.DataFrame([*df['B']]) # if values.tolist() is too boring



(pd.wide_to_long(df.drop('B', 1), 'B', 'A', '')

 .reset_index(level=1, drop=True)

 .reset_index())

both resulting in the same

   A  B

0  1  1

1  2  1

2  1  2

3  2  2

Something pretty not recommended (at least work in this case):

df=pd.concat([df]*2).sort_index()

it=iter(df['B'].tolist()[0]+df['B'].tolist()[0])

df['B']=df['B'].apply(lambda x:next(it))

concat + sort_index + iter + apply + next.

Now:

print(df)

Is:

   A  B

0  1  1

0  1  2

1  2  1

1  2  2