将一行中的元组转换为 pandas Dataframe 中的新列

Question

I have column with list of tuples, and would like to convert this tuples into a new columns. Please see the example below

df = pd.DataFrame(dict(a=[1,2,3],
                  b=['a', 'a', 'b'],
                  c=[[('pear', 1), ('apple', 2)], [('pear', 7), ('orange', 1)], [('apple', 9)] ]))
df

    a   b   c
0   1   a   [(pear, 1), (apple, 2)]
1   2   a   [(pear, 7), (orange, 1)]
2   3   b   [(apple, 9)]

and would like to convert it to

    a   b   fruit   value
0   1   a   pear    1
1   1   a   apple   2
2   2   a   pear    7
3   2   a   orange  1
4   3   b   apple   9

I can do it but it is not really efficient, in my case I have more than 500K of rows. Is there a more efficient way of doing it?

Note: I'm using pandas 0.21 and currently cannot upgrade due to my project requirements.

Thanks

UPDATE:

All three solutions proposed below are great for pandas >=0.25 . For earlier versions df.explode is not an option. And for pandas < 0.24 there is no df.to_numpy so only solution for earlier versions is @jezreal's solution

A small benchmark is below (pandas == 0.25) (surprisingly explode is slower):

from itertools import product, chain

def sol_1(df):
    phase1 = (product([a],b,c) for a,b,c in df.to_numpy())
    phase2 = [(a,b,*c) for a, b, c in chain.from_iterable(phase1)]
    return pd.DataFrame(phase2, columns = ["a","b","fruit","value"])


def sol_2(df): 
    df1 = pd.DataFrame([(k, *x) for k, v in df.c.items() for x in v],
                   columns=['i','fruit','value'])
    df = df.merge(df1, left_index=True, right_on='i').drop('i', axis=1)
    return df

def sol_3(df):
    df = df.explode('c')
    df[['fruit', 'value']] = pd.DataFrame(df['c'].tolist(), index=df.index)
    del df['c']
    return df

%timeit sol_1(df)
%timeit sol_2(df)
%timeit sol_3(df)

586 µs ± 6.39 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
2.8 ms ± 206 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
3.14 ms ± 28.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 1

Give this a go and see if it works on your version:

from itertools import product,chain

#create a cartesian for each row in df
phase1 = (product([a],b,c) for a,b,c in df.to_numpy())

#unpack the third entry per row in the flattened iterable
phase2 = [(a,b,*c) for a, b, c in chain.from_iterable(phase1)]

#create dataframe
result = pd.DataFrame(phase2, columns = ["a","b","fruit","value"])


    a   b   fruit   value
0   1   a   pear    1
1   1   a   apple   2
2   2   a   pear    7
3   2   a   orange  1
4   3   b   apple   9

Answer 2

Idea is reshape values in list comprehension to new DataFrame and then use DataFrame.merge :

df1 = pd.DataFrame([(k, *x) for k, v in df.pop('c').items() for x in v],
                   columns=['i','fruit','value'])

print (df1)
   i   fruit  value
0  0    pear      1
1  0   apple      2
2  1    pear      7
3  1  orange      1
4  2   apple      9        

df = df.merge(df1, left_index=True, right_on='i').drop('i', axis=1)
print (df)
   a  b   fruit  value
0  1  a    pear      1
1  1  a   apple      2
2  2  a    pear      7
3  2  a  orange      1
4  3  b   apple      9

Answer 3

Maybe you can try like this:

df = pd.DataFrame(dict(a=[1,2,3],
                  b=['a', 'a', 'b'],
                  c=[[('pear', 1), ('apple', 2)], [('pear', 7), ('orange', 1)], [('apple', 9)] ]))

df = df.explode('c')

df[['fruit', 'value']] = pd.DataFrame(df['c'].tolist(), index=df.index)
del df['c']