[英]python: find first string in string
给定一个字符串和一个子字符串列表,我想要第一个位置,任何子字符串都出现在字符串中。 如果没有子字符串出现,则返回0.我想忽略大小写。
有什么比pythonic更pythonic:
given = 'Iamfoothegreat'
targets = ['foo', 'bar', 'grea', 'other']
res = len(given)
for t in targets:
i = given.lower().find(t)
if i > -1 and i < res:
res = i
if res == len(given):
result = 0
else:
result = res
该代码有效,但似乎效率低下。
我不会返回0因为可能是起始索引,要么使用-1,无或其他一些不可能的值,您可以简单地使用try / except并返回索引:
def get_ind(s, targ):
s = s.lower()
for t in targets:
try:
return s.index(t.lower())
except ValueError:
pass
return None # -1, False ...
如果你想忽略输入字符串的大小写,那么在循环之前设置s = s.lower()
。
你也可以这样做:
def get_ind_next(s, targ):
s = s.lower()
return next((s.index(t) for t in map(str.lower,targ) if t in s), None)
但是,对于每个子字符串而言,最糟糕的是两次查找,而不是使用try / except。 它至少也会在第一场比赛中短路。
如果你真的想要所有的min,那么改为:
def get_ind(s, targ):
s = s.lower()
mn = float("inf")
for t in targ:
try:
i = s.index(t.lower())
if i < mn:
mn = i
except ValueError:
pass
return mn
def get_ind_next(s, targ):
s = s.lower()
return min((s.index(t) for t in map(str.lower, targ) if t in s), default=None)
default=None
仅适用于python> = 3.4,因此如果您使用的是python2,那么您将不得不稍微更改逻辑。
Timings python3:
In [29]: s = "hello world" * 5000
In [30]: s += "grea" + s
In [25]: %%timeit
....: targ = [re.escape(x) for x in targets]
....: pattern = r"%(pattern)s" % {'pattern' : "|".join(targ)}
....: firstMatch = next(re.finditer(pattern, s, re.IGNORECASE),None)
....: if firstMatch:
....: pass
....:
100 loops, best of 3: 5.11 ms per loop
In [18]: timeit get_ind_next(s, targets)
1000 loops, best of 3: 691 µs per loop
In [19]: timeit get_ind(s, targets)
1000 loops, best of 3: 627 µs per loop
In [20]: timeit min([s.lower().find(x.lower()) for x in targets if x.lower() in s.lower()] or [0])
1000 loops, best of 3: 1.03 ms per loop
In [21]: s = 'Iamfoothegreat'
In [22]: targets = ['bar', 'grea', 'other','foo']
In [23]: get_ind_next(s, targets) == get_ind(s, targets) == min([s.lower().find(x.lower()) for x in targets if x.lower() in s.lower()] or [0])
Out[24]: True
Python2:
In [13]: s = "hello world" * 5000
In [14]: s += "grea" + s
In [15]: targets = ['foo', 'bar', 'grea', 'other']
In [16]: timeit get_ind(s, targets)1000 loops,
best of 3: 322 µs per loop
In [17]: timeit min([s.lower().find(x.lower()) for x in targets if x.lower() in s.lower()] or [0])
1000 loops, best of 3: 710 µs per loop
In [18]: get_ind(s, targets) == min([s.lower().find(x.lower()) for x in targets if x.lower() in s.lower()] or [0])
Out[18]: True
你也可以将第一个与min结合起来:
def get_ind(s, targ):
s,mn = s.lower(), None
for t in targ:
try:
mn = s.index(t.lower())
yield mn
except ValueError:
pass
yield mn
哪个做同样的工作,它只是更好一点,也许稍快一点:
In [45]: min(get_ind(s, targets))
Out[45]: 55000
In [46]: timeit min(get_ind(s, targets))
1000 loops, best of 3: 317 µs per loop
另一个例子就是使用正则表达式,因为认为python正则表达式实现速度非常快。 不是我的正则表达式功能
import re
given = 'IamFoothegreat'
targets = ['foo', 'bar', 'grea', 'other']
targets = [re.escape(x) for x in targets]
pattern = r"%(pattern)s" % {'pattern' : "|".join(targets)}
firstMatch = next(re.finditer(pattern, given, re.IGNORECASE),None)
if firstMatch:
print firstMatch.start()
print firstMatch.group()
输出是
3
foo
如果什么也没发现输出什么都没有。 应该自我解释,以检查是否找不到任何东西。
也给你匹配的字符串
given = 'Iamfoothegreat'.lower()
targets = ['foo', 'bar', 'grea', 'other']
dct = {'pos' : - 1, 'string' : None};
given = given.lower()
for t in targets:
i = given.find(t)
if i > -1 and (i < list['pos'] or list['pos'] == -1):
dct['pos'] = i;
dct['string'] = t;
print dct
输出是:
{'pos': 3, 'string': 'foo'}
如果未找到元素:
{'pos': -1, 'string': None}
用这个字符串和模式
given = "hello world" * 5000
given += "grea" + given
targets = ['foo', 'bar', 'grea', 'other']
带有timeit的1000个循环:
regex approach: 4.08629107475 sec for 1000
normal approach: 1.80048894882 sec for 1000
10个循环。 现在有更大的目标(目标* 1000):
normal approach: 4.06895017624 for 10
regex approach: 34.8153910637 for 10
您可以使用以下内容:
answer = min([given.lower().find(x.lower()) for x in targets
if x.lower() in given.lower()] or [0])
演示1
given = 'Iamfoothegreat'
targets = ['foo', 'bar', 'grea', 'other']
answer = min([given.lower().find(x.lower()) for x in targets
if x.lower() in given.lower()] or [0])
print(answer)
产量
3
演示2
given = 'this is a different string'
targets = ['foo', 'bar', 'grea', 'other']
answer = min([given.lower().find(x.lower()) for x in targets
if x.lower() in given.lower()] or [0])
print(answer)
产量
0
我也认为以下解决方案非常易读:
given = 'the string'
targets = ('foo', 'bar', 'grea', 'other')
given = given.lower()
for i in range(len(given)):
if given.startswith(targets, i):
print i
break
else:
print -1
您的代码相当不错,但是您可以通过将.lower
转换移出循环来使其更有效:不需要为每个目标子字符串重复它。 使用列表推导可以稍微压缩代码,但这并不一定会使代码更快。 我使用嵌套列表comp来避免为每个t
调用given.find(t)
两次。
我已将代码包装在函数中以便于测试。
def min_match(given, targets):
given = given.lower()
a = [i for i in [given.find(t) for t in targets] if i > -1]
return min(a) if a else None
targets = ['foo', 'bar', 'grea', 'othe']
data = (
'Iamfoothegreat',
'IAMFOOTHEGREAT',
'Iamfothgrease',
'Iamfothgret',
)
for given in data:
print(given, min_match(given, targets))
产量
Iamfoothegreat 3
IAMFOOTHEGREAT 3
Iamfothgrease 7
Iamfothgret None
尝试这个:
def getFirst(given,targets):
try:
return min([i for x in targets for i in [given.find(x)] if not i == -1])
except ValueError:
return 0
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