[英]Match two lists of letters in Python
How can I match two lists of letters without considering the order of letters appearance in that lists in Python 如何在不考虑Python中该列表中字母出现顺序的情况下匹配两个字母列表
Eg: Think my first list is ['a','b','c','d']
and I want to match this list with another list ['b','c','a','d']
then to get a out put as true. 例如:认为我的第一个列表是
['a','b','c','d']
,我想将此列表与另一个列表['b','c','a','d']
然后得出真实的结果。 How to do this? 这个怎么做? I'm new to python and want your help!
我是python新手,需要您的帮助!
Thanks in advance 提前致谢
How about: 怎么样:
# if you don't want to consider duplicates either
output = set(your_first_list) == set(your_second_list)
# if duplicates matter
output = sorted(your_first_list) == sorted(your_second_list)
You could sort them: 您可以对它们进行排序:
In [1]: a = list('abcd')
In [2]: b = list('bcad')
In [3]: sorted(a) == sorted(b)
Out[3]: True
In [4]: a == b
Out[4]: False
I had something different in mind, that is, like this: 我的想法有所不同,就是这样:
all(x in a for x in b) and all(x in b for x in a)
This checks if all letters in a
occur in b
, and all letters of b
occur in a
. 此检查,如果在所有字母
a
发生在b
,和所有字母b
出现a
。 This means that they 'match' if a
and b
are sets. 这意味着, 如果设置
a
和b
,则它们“匹配”。
But since there was already a good answer, I decided to do a speed comparison, and it turns out my solution is considerably faster than the solution Daren and Lev suggested based on sorted()
. 但是由于已经有了一个很好的答案,所以我决定进行速度比较,结果证明我的解决方案比Daren和Lev基于
sorted()
提出的解决方案要快得多 。 For strings with a length under 100 characters, it also outperformed Daren's set(a) == set(b)
. 对于长度少于100个字符的字符串,它的性能也优于Daren的
set(a) == set(b)
。
import timeit, random, string
def randstring(length):
return ''.join(random.choice(string.ascii_lowercase) \
for i in xrange(length))
def sortmatch(a,b):
return sorted(a) == sorted(b)
def bothways(a,b):
return all(x in a for x in b) and all(x in b for x in a)
def setmatch(a,b):
return set(a) == set(b)
c1 = "sortmatch(a,b)"
c2 = "setmatch(a,b)"
c3 = "bothways(a,b)"
init = """
from __main__ import randstring, sortmatch, bothways, setmatch
a = randstring(%i)
b = randstring(%i)
"""
lengths = [5,20,100,1000,5000]
times = 10000
for n in lengths:
t1 = timeit.Timer(stmt=c1, setup=init % (n,n))
t2 = timeit.Timer(stmt=c2, setup=init % (n,n))
t3 = timeit.Timer(stmt=c3, setup=init % (n,n))
print("String length: %i" % n)
print("Sort and match: %.2f" % (t1.timeit(times)))
print("Set and match: %.2f" % (t2.timeit(times)))
print("Check both ways: %.2f\n" % (t3.timeit(times)))
Results: 结果:
String length: 5
字符串长度:5
Sort and match: 0.04排序和匹配:0.04
Set and match: 0.03设置并匹配:0.03
Check both ways: 0.02双向检查:0.02
String length: 20
字符串长度:20
Sort and match: 0.11排序和匹配:0.11
Set and match: 0.06设置并匹配:0.06
Check both ways: 0.02双向检查:0.02
String length: 100
字符串长度:100
Sort and match: 0.53排序和匹配:0.53
Set and match: 0.16设置并匹配:0.16
Check both ways: 0.25双向检查:0.25
String length: 1000
字串长度:1000
Sort and match: 6.86排序和匹配:6.86
Set and match: 0.89设置并匹配:0.89
Check both ways: 3.82两种方式都检查:3.82
String length: 5000
字符串长度:5000
Sort and match: 36.67排序和匹配:36.67
Set and match: 4.28设置并匹配:4.28
Check both ways: 19.49两种方式都检查:19.49
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