CODEWITHSUNDEEP
pythonpandas
blue-sky
blue-sky

Reputation: 53896

Access index of of filtered value(s)

Here I access the value of a series depending on a predicate : 

import numpy as np

s = pd.Series(np.array([1,2,3]))
print(type(s))

print([i for i in s if i > 2])

returns : 

<class 'pandas.core.series.Series'>
[3]

How to access the index of the value(s) that match the predicate so in this case the index is 2.

Upvotes: 2

Views: 40

Answers (1)

jezrael
jezrael

Reputation: 863791

Use boolean indexing with index:

print (s.index[s > 2])
#alternative
#print (s[s > 2].index)
Int64Index([2], dtype='int64')

Your solution should be changed with Series.items:

print([k for k, v in s.items() if v > 2])
[2]

Performance is different in larger data:

np.random.seed(45)
s = pd.Series(np.random.randint(5, size=100000))

#print (s)

In [93]: %timeit ([k for k, v in s.items() if v > 2])
13.3 ms ± 768 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [94]: %timeit (s.index[s > 2])
930 µs ± 12.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [95]: %timeit (s[s > 2].index)
1.74 ms ± 26.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [96]: %timeit (s.index[s.values > 2])
742 µs ± 79.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [97]: %timeit (s.index.values[s.values > 2])
647 µs ± 12.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [98]: %timeit (s[s.values > 2].index.values)
1.51 ms ± 12.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Upvotes: 1

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