[英]python: find first string in string
給定一個字符串和一個子字符串列表,我想要第一個位置,任何子字符串都出現在字符串中。 如果沒有子字符串出現,則返回0.我想忽略大小寫。
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
該代碼有效,但似乎效率低下。
我不會返回0因為可能是起始索引,要么使用-1,無或其他一些不可能的值,您可以簡單地使用try / except並返回索引:
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 ...
如果你想忽略輸入字符串的大小寫,那么在循環之前設置s = s.lower()
。
你也可以這樣做:
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,對於每個子字符串而言,最糟糕的是兩次查找,而不是使用try / except。 它至少也會在第一場比賽中短路。
如果你真的想要所有的min,那么改為:
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
僅適用於python> = 3.4,因此如果您使用的是python2,那么您將不得不稍微更改邏輯。
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
你也可以將第一個與min結合起來:
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
哪個做同樣的工作,它只是更好一點,也許稍快一點:
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
另一個例子就是使用正則表達式,因為認為python正則表達式實現速度非常快。 不是我的正則表達式功能
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()
輸出是
3
foo
如果什么也沒發現輸出什么都沒有。 應該自我解釋,以檢查是否找不到任何東西。
也給你匹配的字符串
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
輸出是:
{'pos': 3, 'string': 'foo'}
如果未找到元素:
{'pos': -1, 'string': None}
用這個字符串和模式
given = "hello world" * 5000
given += "grea" + given
targets = ['foo', 'bar', 'grea', 'other']
帶有timeit的1000個循環:
regex approach: 4.08629107475 sec for 1000
normal approach: 1.80048894882 sec for 1000
10個循環。 現在有更大的目標(目標* 1000):
normal approach: 4.06895017624 for 10
regex approach: 34.8153910637 for 10
您可以使用以下內容:
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)
產量
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)
產量
0
我也認為以下解決方案非常易讀:
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
您的代碼相當不錯,但是您可以通過將.lower
轉換移出循環來使其更有效:不需要為每個目標子字符串重復它。 使用列表推導可以稍微壓縮代碼,但這並不一定會使代碼更快。 我使用嵌套列表comp來避免為每個t
調用given.find(t)
兩次。
我已將代碼包裝在函數中以便於測試。
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))
產量
Iamfoothegreat 3
IAMFOOTHEGREAT 3
Iamfothgrease 7
Iamfothgret None
嘗試這個:
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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