CODEWITHSUNDEEP
pythonlistpandasunpack
Alexey Trofimov
Alexey Trofimov

Reputation: 5007

Unpack the list element of DataFrame

I have this df:

l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']
df = pd.DataFrame(list(zip(l1, l2)), columns = ['l1', 'l2'])

result:

  l1            l2
0  a             x
1  b  [y1, y2, y3]
2  c             z

What i need is to unpack the inner list in l2 and spread the corresponding value in l1 like this:

  l1  l2
0  a   x
1  b  y1
2  b  y2
3  b  y3
4  c   z

What is the proper way to do this? Thanks.

Upvotes: 2

Views: 3154

Answers (4)

Alexey Trofimov
Alexey Trofimov

Reputation: 5007

For DataFrames with not constatnt number of columns i now do something like this:

l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']
df = pd.DataFrame(list(zip(l1, l2)), columns = ['l1', 'l2'])

Since pandas 0.25.0 there is a built in explode method, which does exactly what this, preserving index:

df.explode('l2')

result:

  l1  l2
0  a   x
1  b  y1
1  b  y2
1  b  y3
2  c   z

If you need to refresh the index:

df.explode('l2').reset_index(drop=True)

result:

  l1  l2
0  a   x
1  b  y1
2  b  y2
3  b  y3
4  c   z

OLD ANSWER:

df2 = pd.DataFrame(columns=df.columns,index=df.index)[0:0]

for idx in df.index:
    new_row = df.loc[idx, :].copy()
    for res in df.ix[idx, 'l2']:
        new_row.set_value('l2', res)
        df2.loc[len(df2)] = new_row

It works, but this looks pretty much like bruteforce.

Upvotes: 2

jezrael
jezrael

Reputation: 862581

I think you can use numpy.repeat for repeat values by legths by str.len and flat values of nested lists by chain:

from  itertools import chain

df1 = pd.DataFrame({
        "l1": np.repeat(df.l1.values, df.l2.str.len()),
        "l2": list(chain.from_iterable(df.l2))})
print (df1)
  l1  l2
0  a   x
1  b  y1
2  b  y2
3  b  y3
4  c   z

Timings:

#[100000 rows x 2 columns]
np.random.seed(10)
N = 100000
l1 = ['a', 'b', 'c']
l1 = np.random.choice(l1, N)
l2 = [list(tuple(string.ascii_letters[:np.random.randint(1, 10)])) for _ in np.arange(N)]
df = pd.DataFrame({"l1":l1, "l2":l2})
df.l2 = df.l2.apply(lambda x: x if len(x) !=1 else x[0])
#print (df)


In [91]: %timeit (pd.DataFrame([(left, right) for outer in zip(l1, l2) for left, right in zip_longest(*outer, fillvalue=outer[0])]))
1 loop, best of 3: 242 ms per loop

In [92]: %timeit (pd.DataFrame({ "l1": np.repeat(df.l1.values, df.l2.str.len()), "l2": list(chain.from_iterable(df.l2))}))
10 loops, best of 3: 84.6 ms per loop

Conclusion:

numpy.repeat is 3 times faster as zip_longest solution in larger df.

EDIT:

For compare with loop version is necessery smaller df, because very slow:

#[1000 rows x 2 columns]
np.random.seed(10)
N = 1000
l1 = ['a', 'b', 'c']
l1 = np.random.choice(l1, N)
l2 = [list(tuple(string.ascii_letters[:np.random.randint(1, 10)])) for _ in np.arange(N)]
df = pd.DataFrame({"l1":l1, "l2":l2})
df.l2 = df.l2.apply(lambda x: x if len(x) !=1 else x[0])
#print (df)

def alexey(df):
    df2 = pd.DataFrame(columns=df.columns,index=df.index)[0:0]

    for idx in df.index:
        new_row = df.loc[idx, :].copy()
        for res in df.ix[idx, 'l2']:
            new_row.set_value('l2', res)
            df2.loc[len(df2)] = new_row
    return df2

print (alexey(df))

In [20]: %timeit (alexey(df))
1 loop, best of 3: 11.4 s per loop

In [21]: %timeit pd.DataFrame([(left, right) for outer in zip(l1, l2) for left, right in zip_longest(*outer, fillvalue=outer[0])])
100 loops, best of 3: 2.57 ms per loop

In [22]: %timeit pd.DataFrame({ "l1": np.repeat(df.l1.values, df.l2.str.len()), "l2": list(chain.from_iterable(df.l2))})
The slowest run took 4.42 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.41 ms per loop

Upvotes: 1

Bruce Pucci
Bruce Pucci

Reputation: 1901

You could use a nested list comprehension with itertools.zip_longest.

import pandas as pd

from itertools import zip_longest

l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']

expanded = [(left, right) for outer in zip(l1, l2) 
                          for left, right in zip_longest(*outer, fillvalue=outer[0])]

pd.DataFrame(expanded)

Result is... 

   0   1
0  a   x
1  b  y1
2  b  y2
3  b  y3
4  c   z

To me this is on the border of being too long of a list comp. Also assumes that l1 has no lists in it and will be doing the filling.

Upvotes: 2

Sosel
Sosel

Reputation: 1718

Brute force, looping over the dataframe:

for idx in df.index:
    # This transforms the item in "l2" into an iterable list
    item = df.loc[idx, "l2"] if isinstance(df.loc[idx, "l2"], (list, tuple)) else [df.loc[idx, "l2"]]
    for element in item:
        print(df.loc[idx, "l1"], element)

returns 

a x
b y1
b y2
b y3
c z

Upvotes: 1

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