Python Fastest way to check if a string contains specific characters in any of the items in a list
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Fastest way to check if a string contains specific characters in any of the items in a list
提问by Muhammad Taha
What is the fastest way to check if a string contains some characters from any items of a list?
What is the fastest way to check if a string contains some characters from any items of a list?
Currently, I'm using this method:
Currently, I'm using this method:
lestring = "Text123"
lelist = ["Text", "foo", "bar"]
for x in lelist:
if lestring.count(x):
print 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)
Is there any way to do it without iteration (which will make it faster I suppose.)?
Is there any way to do it without iteration (which will make it faster I suppose.)?
采纳答案by Abhijit
You can try list comprehension with membership check
You can try list comprehension with membership check
>>> lestring = "Text123"
>>> lelist = ["Text", "foo", "bar"]
>>> [e for e in lelist if e in lestring]
['Text']
Compared to your implementation, though LC has an implicit loop but its faster as there is no explicit function call as in your case with count
Compared to your implementation, though LC has an implicit loop but its faster as there is no explicit function call as in your case with count
Compared to Joe's implementation, yours is way faster, as the filter function would require to call two functions in a loop, lambdaand count
Compared to Joe's implementation, yours is way faster, as the filter function would require to call two functions in a loop, lambdaand count
>>> def joe(lelist, lestring):
return ''.join(random.sample(x + 'b'*len(x), len(x)))
>>> def uz(lelist, lestring):
for x in lelist:
if lestring.count(x):
return 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)
>>> def ab(lelist, lestring):
return [e for e in lelist if e in lestring]
>>> t_ab = timeit.Timer("ab(lelist, lestring)", setup="from __main__ import lelist, lestring, ab")
>>> t_uz = timeit.Timer("uz(lelist, lestring)", setup="from __main__ import lelist, lestring, uz")
>>> t_joe = timeit.Timer("joe(lelist, lestring)", setup="from __main__ import lelist, lestring, joe")
>>> t_ab.timeit(100000)
0.09391469893125759
>>> t_uz.timeit(100000)
0.1528471407273173
>>> t_joe.timeit(100000)
1.4272649857800843
Jamie's commented solution is slower for shorter string's. Here is the test result
Jamie's commented solution is slower for shorter string's. Here is the test result
>>> def jamie(lelist, lestring):
return next(itertools.chain((e for e in lelist if e in lestring), (None,))) is not None
>>> t_jamie = timeit.Timer("jamie(lelist, lestring)", setup="from __main__ import lelist, lestring, jamie")
>>> t_jamie.timeit(100000)
0.22237164127909637
If you need Boolean values, for shorter strings, just modify the above LC expression
If you need Boolean values, for shorter strings, just modify the above LC expression
[e in lestring for e in lelist if e in lestring]
Or for longer strings, you can do the following
Or for longer strings, you can do the following
>>> next(e in lestring for e in lelist if e in lestring)
True
or
or
>>> any(e in lestring for e in lelist)
回答by Joe
filter(lambda x: lestring.count(x), lelist)
That will return all the strings that you're trying to find as a list.
That will return all the strings that you're trying to find as a list.
回答by theodox
if the test is to see if there are any characters in common (not words or segments), create a set out of the letters in the list and then check the letters agains the string:
if the test is to see if there are any characters in common (not words or segments), create a set out of the letters in the list and then check the letters agains the string:
char_list = set(''.join(list_of_words))
test_set = set(string_to_teat)
common_chars = char_list.intersection(test_set)
However I'm assuming you're looking for as little as one character in common...
However I'm assuming you're looking for as little as one character in common...
回答by moomima
The esmrelibrary does the trick. In your case, the simpler, esm(part of esmre) is what you want.
The esmrelibrary does the trick. In your case, the simpler, esm(part of esmre) is what you want.
https://pypi.python.org/pypi/esmre/
https://pypi.python.org/pypi/esmre/
https://code.google.com/p/esmre/
https://code.google.com/p/esmre/
They have good documentation and examples: Taken from their examples:
They have good documentation and examples: Taken from their examples:
>>> import esm
>>> index = esm.Index()
>>> index.enter("he")
>>> index.enter("she")
>>> index.enter("his")
>>> index.enter("hers")
>>> index.fix()
>>> index.query("this here is history")
[((1, 4), 'his'), ((5, 7), 'he'), ((13, 16), 'his')]
>>> index.query("Those are his sheep!")
[((10, 13), 'his'), ((14, 17), 'she'), ((15, 17), 'he')]
>>>
I ran some performance tests:
I ran some performance tests:
import random, timeit, string, esm
def uz(lelist, lestring):
for x in lelist:
if lestring.count(x):
return 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)
def ab(lelist, lestring):
return [e for e in lelist if e in lestring]
def use_esm(index, lestring):
return index.query(lestring)
for TEXT_LEN in [5, 50, 1000]:
for SEARCH_LEN in [5, 20]:
for N in [5, 50, 1000, 10000]:
if TEXT_LEN < SEARCH_LEN:
continue
print 'TEXT_LEN:', TEXT_LEN, 'SEARCH_LEN:', SEARCH_LEN, 'N:', N
lestring = ''.join((random.choice(string.ascii_uppercase + string.digits) for _ in range(TEXT_LEN)))
lelist = [''.join((random.choice(string.ascii_uppercase + string.digits) for _ in range(SEARCH_LEN))) for _
in range(N)]
index = esm.Index()
for i in lelist:
index.enter(i)
index.fix()
t_ab = timeit.Timer("ab(lelist, lestring)", setup="from __main__ import lelist, lestring, ab")
t_uz = timeit.Timer("uz(lelist, lestring)", setup="from __main__ import lelist, lestring, uz")
t_esm = timeit.Timer("use_esm(index, lestring)", setup="from __main__ import index, lestring, use_esm")
ab_time = t_ab.timeit(1000)
uz_time = t_uz.timeit(1000)
esm_time = t_esm.timeit(1000)
min_time = min(ab_time, uz_time, esm_time)
print ' ab%s: %f' % ('*' if ab_time == min_time else '', ab_time)
print ' uz%s: %f' % ('*' if uz_time == min_time else '', uz_time)
print ' esm%s %f:' % ('*' if esm_time == min_time else '', esm_time)
And got that results depends mostly on the number of items that one is looking for (in my case, 'N'):
And got that results depends mostly on the number of items that one is looking for (in my case, 'N'):
TEXT_LEN: 1000 SEARCH_LEN: 20 N: 5
ab*: 0.001733
uz: 0.002512
esm 0.126853:
TEXT_LEN: 1000 SEARCH_LEN: 20 N: 50
ab*: 0.017564
uz: 0.023701
esm 0.079925:
TEXT_LEN: 1000 SEARCH_LEN: 20 N: 1000
ab: 0.370371
uz: 0.489523
esm* 0.133783:
TEXT_LEN: 1000 SEARCH_LEN: 20 N: 10000
ab: 3.678790
uz: 4.883575
esm* 0.259605:
