Example list: mylist = ['abc123', 'def456', 'ghi789']
I want to retrieve an element if there's a match for a substring, like abc
sub = 'abc'
print any(sub in mystring for mystring in mylist)
above prints True
if any of the elements in the list contain the pattern.
I would like to print the element which matches the substring. So if I'm checking 'abc'
I only want to print 'abc123'
from list.
This prints all elements that contain sub:
for s in filter (lambda x: sub in x, list): print (s)
All the answers work but they always traverse the whole list. If I understand your question, you only need the first match. So you don't have to consider the rest of the list if you found your first match:
mylist = ['abc123', 'def456', 'ghi789']
sub = 'abc'
next((s for s in mylist if sub in s), None) # returns 'abc123'
If the match is at the end of the list or for very small lists, it doesn't make a difference, but consider this example:
import timeit
mylist = ['abc123'] + ['xyz123']*1000
sub = 'abc'
timeit.timeit('[s for s in mylist if sub in s]', setup='from __main__ import mylist, sub', number=100000)
# for me 7.949463844299316 with Python 2.7, 8.568840944994008 with Python 3.4
timeit.timeit('next((s for s in mylist if sub in s), None)', setup='from __main__ import mylist, sub', number=100000)
# for me 0.12696599960327148 with Python 2.7, 0.09955992100003641 with Python 3.4
I'd just use a simple regex, you can do something like this
import re
old_list = ['abc123', 'def456', 'ghi789']
new_list = [x for x in old_list if re.search('abc', x)]
for item in new_list:
print item
Use a simple for
loop:
seq = ['abc123', 'def456', 'ghi789']
sub = 'abc'
for text in seq:
if sub in text:
print(text)
yields
abc123
Source: Stackoverflow.com