CODEWITHSUNDEEP
pythonpandas
AVal
AVal

Reputation: 617

Check if pandas column contains all elements from a list

I have a df like this:

frame = pd.DataFrame({'a' : ['a,b,c', 'a,c,f', 'b,d,f','a,z,c']})

And a list of items:

letters = ['a','c']

My goal is to get all the rows from frame that contain at least the 2 elements in letters

I came up with this solution:

for i in letters:
    subframe = frame[frame['a'].str.contains(i)]

This gives me what I want, but it might not be the best solution in terms of scalability. Is there any 'vectorised' solution? Thanks

Upvotes: 25

Views: 28065

Answers (7)

Hermes Morales
Hermes Morales

Reputation: 637

frame.iloc[[x for x in range(len(frame)) if set(letters).issubset(frame.iloc[x,0])]]

output:

        a
 0  a,b,c
 1  a,c,f
 3  a,z,c

timeit

%%timeit
#hermes
frame.iloc[[x for x in range(len(frame)) if set(letters).issubset(frame.iloc[x,0])]]

output

300 µs ± 32.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Upvotes: 1

ManojK
ManojK

Reputation: 1640

Use set.issubset:

frame = pd.DataFrame({'a' : ['a,b,c', 'a,c,f', 'b,d,f','a,z,c','x,y']})
letters = ['a','c']

frame[frame['a'].apply(lambda x: set(letters).issubset(x))]

Out:

       a
0  a,b,c
1  a,c,f
3  a,z,c

Upvotes: 8

yatu
yatu

Reputation: 88305

One way is to split the column values into lists using str.split, and check if set(letters) is a subset of the obtained lists:

letters_s = set(letters)
frame[frame.a.str.split(',').map(letters_s.issubset)]

     a
0  a,b,c
1  a,c,f
3  a,z,c
​

Benchmark:

def serge(frame):
    contains = [frame['a'].str.contains(i) for i in letters]
    return frame[np.all(contains, axis=0)]

def yatu(frame):
    letters_s = set(letters)
    return frame[frame.a.str.split(',').map(letters_s.issubset)]

def austin(frame):
    mask =  frame.a.apply(lambda x: np.intersect1d(x.split(','), letters).size > 0)
    return frame[mask]

def datanovice(frame):
    s = frame['a'].str.split(',').explode().isin(letters).groupby(level=0).cumsum()
    return frame.loc[s[s.ge(2)].index.unique()]

perfplot.show(
    setup=lambda n: pd.concat([frame]*n, axis=0).reset_index(drop=True), 

    kernels=[
        lambda df: serge(df),
        lambda df: yatu(df),
        lambda df: df[df['a'].apply(lambda x: np.all([*map(lambda l: l in x, letters)]))],
        lambda df: austin(df),
        lambda df: datanovice(df),
    ],

    labels=['serge', 'yatu', 'bruno','austin', 'datanovice'],
    n_range=[2**k for k in range(0, 18)],
    equality_check=lambda x, y: x.equals(y),
    xlabel='N'
)

enter image description here

Upvotes: 19

Bruno Mello
Bruno Mello

Reputation: 4628

This also solves it:

frame[frame['a'].apply(lambda x: np.all([*map(lambda l: l in x, letters)]))]

Upvotes: 10

Austin
Austin

Reputation: 26057

You can use np.intersect1d:

import pandas as pd
import numpy as np

frame = pd.DataFrame({'a' : ['a,b,c', 'a,c,f', 'b,d,f','a,z,c']})
letters = ['a','c']

mask =  frame.a.apply(lambda x: np.intersect1d(x.split(','), letters).size > 0)
print(frame[mask])

    a
0  a,b,c
1  a,c,f
3  a,z,c

Upvotes: 9

Umar.H
Umar.H

Reputation: 23099

IIUC, explode and a boolean filter

the idea is to create a single series then we can groupby the index the count the true occurrences of your list using a cumulative sum

s = frame['a'].str.split(',').explode().isin(letters).groupby(level=0).cumsum()

print(s)

0    1.0
0    1.0
0    2.0
1    1.0
1    2.0
1    2.0
2    0.0
2    0.0
2    0.0
3    1.0
3    1.0
3    2.0

frame.loc[s[s.ge(2)].index.unique()]

out:

       a
0  a,b,c
1  a,c,f
3  a,z,c

Upvotes: 5

Serge Ballesta
Serge Ballesta

Reputation: 149185

I would build a list of Series, and then apply a vectorized np.all:

contains = [frame['a'].str.contains(i) for i in letters]
resul = frame[np.all(contains, axis=0)]

It gives as expected:

       a
0  a,b,c
1  a,c,f
3  a,z,c

Upvotes: 21

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