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Python Pandas Series.isin doesn't work

I have a Series

8 [11820]
9 [11820]
10 [11820]
11 [11820]
12 [11820]
27 [10599]
28 [10599]
29 [10599]
31 [661, 10599]
32 [661, 10599]
33 [7322]
34 [0]
37 [661]
39 [661]
40 [661]
49 [0, 661, 662, 663]


I want to filter this Series with something like points[points.isin([0])] to get

points[points.isin([0])]

34 [0]
49 [0, 661, 662, 663]


but as a result I get 0 Features.

2 Answers
2

Simple way to check, if your value (0) is in the list, is by using apply on your series:

apply

s = s[s.apply(lambda x: 0 in x)]


Some explanation:

For every row it checks, whether 0 is in the list.

Apply returns a "True/False" series, where for every row True means that 0 is in the list inside the row.

After that your first series (s) is being filtered by this "True/False" series via .

Sample code:

# This is your series
s = pd.Series([[0],
[11820],
[11820],
[10599],
[0, 661, 662, 663]])

# This is the solution
s = s[s.apply(lambda x: 0 in x)]

# Print the result
print(s)

0 [0]
4 [0, 661, 662, 663]
Name: A, dtype: object


pd.Series.isin works by hashing and works on the entire element, i.e. it won't consider a partial match. Even for an exact match, since a list cannot be hashed, pd.Series.isin won't work with a series of lists.

pd.Series.isin

pd.Series.isin

You can use a custom function with pd.Series.apply:

pd.Series.apply

df = pd.DataFrame('A': [[1, 2], [0], [0, 2, 3]])

search_list = [0] # list of scalars

mask = df['A'].apply(lambda x: any(i in x for i in search_list))
res = df[mask]

print(res)

A
1 [0]
2 [0, 2, 3]


You can convert your series to tuples, which are hashable, before any comparison. Then compare your series of tuples with a list of tuples.

search_list = [[0]] # list of lists

mask = df['A'].map(tuple).isin(list(map(tuple, search_list)))
res = df[mask]

print(res)

A
1 [0]


Note operations with object dtype series will necessarily be inefficient. If possible, you should split your series of lists into multiple series of integers. Although, in this case, this may be cumbersome given the inconsistent list lengths.

object

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