For approaches to retrieving partial matches in a numeric list, go to:
But if you’re looking for how to retrieve partial matches for a list of strings, you’ll find the best approaches concisely explained in the answer below.
SO: Python list lookup with partial match shows how to return a bool
, if a list
contains an element that partially matches (e.g. begins
, ends
, or contains
) a certain string. But how can you return the element itself, instead of True
or False
Example:
l = ['ones', 'twos', 'threes'] wanted = 'three'
Here, the approach in the linked question will return True
using:
any(s.startswith(wanted) for s in l)
So how can you return the element 'threes'
instead?
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Answer
startswith
andin
, return a Boolean.- The
in
operator is a test of membership. - This can be performed with a
list-comprehension
orfilter
. - Using a
list-comprehension
, within
, is the fastest implementation tested. - If case is not an issue, consider mapping all the words to lowercase.
l = list(map(str.lower, l))
.
- Tested with python 3.11.0
filter
:
- Using
filter
creates afilter
object, solist()
is used to show all the matching values in alist
.
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?
- Tested in Jupyter Lab using the
words
corpus fromnltk v3.7
, which has 236736 words - 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())) %timeit list(filter(lambda x: wanted in x, words.words())) %timeit [v for v in words.words() if v.startswith(wanted)] %timeit [v for v in words.words() if wanted in v]
%timeit
results
62.8 ms ± 816 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) 53.8 ms ± 982 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) 56.9 ms ± 1.33 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 47.5 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)