For approaches to retrieving partial matches in a numeric list, go to:

• How to return a subset of a list that matches a condition?

• Python: Find in list

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?

• 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)).
• Tested with python 3.10.0

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?

• Tested in Jupyter Lab using the words corpus from nltk v3.6.5, 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()))
[out]:
64.8 ms ± 856 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit list(filter(lambda x: wanted in x, words.words()))
[out]:
54.8 ms ± 528 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit [v for v in words.words() if v.startswith(wanted)]
[out]:
57.5 ms ± 634 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit [v for v in words.words() if wanted in v]
[out]:
50.2 ms ± 791 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)