Check whether tuple column in pandas contains some value from a list

Question

I have a pandas DataFrame with a tuple column. I would like a mask identifying for each row whether any of the values in the tuple column matches any value in a predetermined tuple. My attempt is below:

import pandas as pd

df = pd.DataFrame([{'a': 1, 'b': (2, 3, 4)}, {'a': 5, 'b': (6, 7, 8)}])
print(df)

codes = (3, 4, 20, 22)
mask = df.b.str.contains_any(codes)  # This line is incorrect

Desired output:

0     True
1    False

I was hopeful based on https://stackoverflow.com/a/51689894/10499953 that str functions would work for tuples, but I couldn't get that to work for even a single value from codes:

a = df['has_code'] = df['b'].str.contains(4)

gives

TypeError: first argument must be string or compiled pattern.

Answer 1

Try this:

res = df['b'].apply(lambda x: any(val in x for val in codes))
print(res)

Output:

0     True
1    False

Answer 2

Another option

df['b'].apply(lambda x: any(set(x).intersection(codes)))

Answer 3

You can use set.intersection and use astype(bool)

code = set(codes)
df.b.map(code.intersection).astype(bool)

0     True
1    False
Name: b, dtype: bool

Timeit analysis

#setup
o = [np.random.randint(0,10,(3,)) for _ in range(10_000)]
len(o)
# 10000

s = pd.Series(o)
s
0       [6, 2, 5]
1       [7, 4, 0]
2       [1, 8, 2]
3       [4, 8, 9]
4       [7, 3, 4]
          ...
9995    [3, 9, 4]
9996    [6, 2, 9]
9997    [2, 0, 5]
9998    [5, 0, 7]
9999    [7, 4, 2]
Length: 10000, dtype: object

---

# Adam's answer
In [38]: %timeit s.apply(lambda x: any(set(x).intersection(codes)))
19.1 ms ± 193 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#komatiraju's answer
In [39]: %timeit s.apply(lambda x: any(val in x for val in codes))
83.8 ms ± 974 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

#My answer
In [42]: %%timeit
    ...: code = set(codes)
    ...: s.map(code.intersection).astype(bool)
    ...:
    ...:
15.5 ms ± 300 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#wwnde's answer
In [74]: %timeit s.apply(lambda x:len([*{*x}&{*codes}])>0)
19.5 ms ± 372 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

For Series of size 1 million

bigger_o = np.repeat(o,100,axis=0)
bigger_o.shape
# (1000000, 3)
s = pd.Series((list(bigger_o)))
s
0         [6, 2, 5]
1         [6, 2, 5]
2         [6, 2, 5]
3         [6, 2, 5]
4         [6, 2, 5]
            ...
999995    [7, 4, 2]
999996    [7, 4, 2]
999997    [7, 4, 2]
999998    [7, 4, 2]
999999    [7, 4, 2]
Length: 1000000, dtype: object

---

In [54]: %timeit s.apply(lambda x: any(set(x).intersection(codes)))
1.89 s ± 28.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [55]: %timeit s.apply(lambda x: any(val in x for val in codes))
8.9 s ± 652 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [56]: %%timeit
    ...: code = set(codes)
    ...: s.map(code.intersection).astype(bool)
    ...:
    ...:
1.54 s ± 4.97 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [79]: %timeit s.apply(lambda x:len([*{*x}&{*codes}])>0)
1.95 s ± 88.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answer 4

df.b.apply(lambda x:len([*{*x}&{*codes}])>0)#my preferred speed wise
#df.b.apply(lambda x:[*{*x}&{*codes}]).str.len()>0 #Works as well

0     True
1    False
Name: b, dtype: bool

%timeit df.b.apply(lambda x:len([*{*x}&{*codes}])>0)
220 µs ± 2.54 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%%timeit 

code = set(codes)
df.b.map(code.intersection).astype(bool)
364 µs ± 1.91 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df['b'].apply(lambda x: any(val in x for val in codes))
210 µs ± 1.95 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df['b'].apply(lambda x: any(set(x).intersection(codes)))
211 µs ± 1.77 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)