CODEWITHSUNDEEP
pythonpandasdataframe
Brad Solomon
Brad Solomon

Reputation: 40878

Unnest (explode) a Pandas Series

I have:

df = pd.DataFrame({'col1': ['asdf', 'xy', 'q'], 'col2': [1, 2, 3]})

   col1  col2
0  asdf     1
1    xy     2
2     q     3

I'd like to take the "combinatoric product" of each letter from the strings in col1, with each elementwise int in col2. I.e.:

  col1  col2
0    a    1
1    s    1
2    d    1
3    f    1
4    x    2
5    y    2
6    q    3

Current method:

from itertools import product

pieces = []
for _, s in df.iterrows():
    letters = list(s.col1)
    prods = list(product(letters, [s.col2]))
    pieces.append(pd.DataFrame(prods))

pd.concat(pieces)

Any more efficient workarounds?

Upvotes: 15

Views: 4062

Answers (8)

Pygirl
Pygirl

Reputation: 13349

Using Explode (pandas>=0.25)

df = pd.DataFrame({'col1': ['asdf', 'xy', 'q'], 'col2': [1, 2, 3]})

df.col1=df.col1.apply(list)
df = df.explode('col1')

Result:

  col1  col2
0   a   1
0   s   1
0   d   1
0   f   1
1   x   2
1   y   2
2   q   3

Upvotes: 2

Eric
Eric

Reputation: 2729

You can also try to itertools.chain and itertools.repeat functions to achieve similar results.

An example would be

import pandas as pd
from itertools import chain, repeat

d = {'col1': ['asdf', 'xy', 'q'], 'col2': [1, 2, 3]}

expanded_d = {
    "col1": list(chain(*[list(item) for item in d["col1"]])),
    "col2": list(chain(*[list(repeat(d["col2"][idx], len(list(d["col1"][idx])))) for idx in range(len(d["col1"])) ]))
    }

result = pd.DataFrame(data=expanded_d)

  col1  col2
0    a     1
1    s     1
2    d     1
3    f     1
4    x     2
5    y     2
6    q     3

Hope it helps.

Upvotes: 0

cs95
cs95

Reputation: 402413

Using list + str.join and np.repeat -

pd.DataFrame(
{
     'col1' : list(''.join(df.col1)), 
     'col2' : df.col2.values.repeat(df.col1.str.len(), axis=0)
})

  col1  col2
0    a     1
1    s     1
2    d     1
3    f     1
4    x     2
5    y     2
6    q     3

A generalised solution for any number of columns is easily achievable, without much change to the solution -

i = list(''.join(df.col1))
j = df.drop('col1', 1).values.repeat(df.col1.str.len(), axis=0)

df = pd.DataFrame(j, columns=df.columns.difference(['col1']))
df.insert(0, 'col1', i)

df

  col1 col2
0    a    1
1    s    1
2    d    1
3    f    1
4    x    2
5    y    2
6    q    3

Performance

df = pd.concat([df] * 100000, ignore_index=True)


# MaxU's solution

%%timeit
df.col1.str.extractall(r'(.)') \
           .reset_index(level=1, drop=True) \
           .join(df['col2']) \
           .reset_index(drop=True)

1 loop, best of 3: 1.98 s per loop


# piRSquared's solution

%%timeit
pd.DataFrame(
     [[x] + b for a, *b in df.values for x in a],
     columns=df.columns
)

1 loop, best of 3: 1.68 s per loop


# Wen's solution

%%timeit
v = df.col1.apply(list)
pd.DataFrame({'col1':np.concatenate(v.values),'col2':df.col2.repeat(v.apply(len))})

1 loop, best of 3: 835 ms per loop


# Alexander's solution

%%timeit
pd.DataFrame([(letter, i) 
              for letters, i in zip(df['col1'], df['col2']) 
              for letter in letters],
             columns=df.columns)

1 loop, best of 3: 316 ms per loop


%%timeit
pd.DataFrame(
{
     'col1' : list(''.join(df.col1)), 
     'col2' : df.col2.values.repeat(df.col1.str.len(), axis=0)
})

10 loops, best of 3: 124 ms per loop

I tried timing Vaishali's, but it took too long on this dataset.

Upvotes: 16

piRSquared
piRSquared

Reputation: 294248

General solution with a list comprehension and clever unpacking:

pd.DataFrame(
    [[x] + b for a, *b in df.values for x in a],
    columns=df.columns
)

  col1  col2
0    a     1
1    s     1
2    d     1
3    f     1
4    x     2
5    y     2
6    q     3

Upvotes: 4

Vaishali
Vaishali

Reputation: 38415

One more:)

df.set_index('col2').col1.apply(lambda x: pd.Series(list(x))).stack()\
.reset_index(1,drop = True).reset_index(name = 'col1')

    col2    col1
0   1       a
1   1       s
2   1       d
3   1       f
4   2       x
5   2       y
6   3       q

Upvotes: 7

BENY
BENY

Reputation: 323226

Trick from the list :-)

df.col1=df.col1.apply(list)
df
Out[489]: 
           col1  col2
0  [a, s, d, f]     1
1        [x, y]     2
2           [q]     3
pd.DataFrame({'col1':np.concatenate(df.col1.values),'col2':df.col2.repeat(df.col1.apply(len))})
Out[490]: 
  col1  col2
0    a     1
0    s     1
0    d     1
0    f     1
1    x     2
1    y     2
2    q     3

Upvotes: 8

Alexander
Alexander

Reputation: 109536

pd.DataFrame([(letter, i) 
              for letters, i in zip(df['col1'], df['col2']) 
              for letter in letters],
             columns=df.columns)

Upvotes: 8

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210832

In [86]: df.col1.str.extractall(r'(.)') \
           .reset_index(level=1, drop=True) \
           .join(df['col2']) \
           .reset_index(drop=True)
Out[86]:
   0  col2
0  a     1
1  s     1
2  d     1
3  f     1
4  x     2
5  y     2
6  q     3

Upvotes: 7

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