Check if pandas column contains all elements from a list
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'
)
This also solves it:
frame[frame['a'].apply(lambda x: np.all([*map(lambda l: l in x, letters)]))]
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