- startswith and in, return a Boolean
- The
in operator is a test of membership.
- This can be performed with a
list-comprehension or filter
- Using a
list-comprehension, with in, is the fastest implementation tested.
- If case is not an issue, consider mapping all the words to lowercase.
l = list(map(str.lower, l)).
filter:
- Using filter creates a
filter object, so list() is used to show all the matching values in a list.
l = ['ones', 'twos', 'threes']
wanted = 'three'
# using startswith
result = list(filter(lambda x: x.startswith(wanted), l))
# using in
result = list(filter(lambda x: wanted in x, l))
print(result)
[out]:
['threes']
list-comprehension
l = ['ones', 'twos', 'threes']
wanted = 'three'
# using startswith
result = [v for v in l if v.startswith(wanted)]
# using in
result = [v for v in l if wanted in v]
print(result)
[out]:
['threes']
Which implementation is faster?
- Using the
words corpus from nltk
- Words with
'three'
['three', 'threefold', 'threefolded', 'threefoldedness', 'threefoldly', 'threefoldness', 'threeling', 'threeness', 'threepence', 'threepenny', 'threepennyworth', 'threescore', 'threesome']
from nltk.corpus import words
%timeit list(filter(lambda x: x.startswith(wanted), words.words()))
[out]:
47.4 ms ± 1.9 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit list(filter(lambda x: wanted in x, words.words()))
[out]:
27 ms ± 1.78 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit [v for v in words.words() if v.startswith(wanted)]
[out]:
34.1 ms ± 768 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit [v for v in words.words() if wanted in v]
[out]:
14.5 ms ± 63.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
