我知道.index()将返回子串位于python中的位置。
但是，我想要的是找到子串在第n次的位置，这将是这样的：
>> s = 'abcdefacbdea'
>> s.index('a')
0
>> s.nindex('a', 1)
6
>>s.nindex('a', 2)
11
有没有办法在python中执行此操作？
答案 0 :(得分：7)
怎么样......
def nindex(mystr, substr, n=0, index=0):
for _ in xrange(n+1):
index = mystr.index(substr, index) + 1
return index - 1
视频：如str.index()那样，nindex()在找不到子视频时会引发ValueError。
答案 1 :(得分：1)
是。使用s.index('yourstring', start)
找到一个大胖子后更新-1 ...我没有写一些代码???
这是我的兑换尝试，如果需要允许不重叠，并按照显示的程度进行测试：
>>> def nindex(haystack, needle, n, overlapping=True):
... delta = 1 if overlapping else max(1, len(needle))
... start = -delta
... for _unused in xrange(n):
... start = haystack.index(needle, start+delta)
... return start
...
>>> for n in xrange(1, 11):
... print n, nindex('abcdefacbdea', 'a', n)
...
1 0
2 6
3 11
4
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "<stdin>", line 5, in nindex
ValueError: substring not found
>>> for olap in (True, False):
... for n in (1, 2):
... print str(olap)[0], n, nindex('abababab', 'abab', n, olap)
...
T 1 0
T 2 2
F 1 0
F 2 4
>>> for n in xrange(1, 8):
... print n, nindex('abcde', '', n)
...
1 0
2 1
3 2
4 3
5 4
6 5
7
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "<stdin>", line 5, in nindex
ValueError: substring not found
>>>
答案 2 :(得分：1)
def nindex(needle, haystack, index=1):
parts = haystack.split(needle)
position = 0
length = len(needle)
for i in range(index - 1):
position += len(parts[i]) + length
return position
我有兴趣看到其他解决方案，我觉得这不是特别pythonic。
答案 3 :(得分：1)
我可能会用
[index for index, value in enumerate(s) if s == 'a'][n]
或
from itertools import islice
next(islice((index for index, value in enumerate(s) if s == 'a'), n, None))
或完全避免处理指数。
答案 4 :(得分：1)
这是一个 memoized 版本，可以尽可能地避免浪费的工作，同时保持与规范的接近[1]（而不是做一些更好的事情，比如循环 所有点击; - ）...：
[1]：只是关闭 - 当然不能在字符串中使用新的.nindex方法！ - ）
def nindex(haystack, needle, nrep=1, _memo={}):
if nrep < 1:
raise ValueError('%r < 1' % (nrep,))
k = needle, haystack
if k in _memo:
where = _memo[k]
else:
where = _memo[k] = [-1]
while len(where) <= nrep:
if where[-1] is None:
return -1
w = haystack.find(needle, where[-1] + 1)
if w < 0:
where.append(None)
return -1
where.append(w)
return where[nrep]
s = 'abcdefacbdea'
print nindex(s, 'a')
print nindex(s, 'a', 2)
print nindex(s, 'a', 3)
根据要求打印0，然后是6，然后是11。
答案 5 :(得分：1)
>>> from re import finditer, escape
>>> from itertools import count, izip
>>> def nfind(s1, s2, n=1):
... """return the index of the nth nonoverlapping occurance of s2 in s1"""
... return next(j.start() for i,j in izip(count(1), finditer(escape(s2),s1)) if i==n)
...
>>> nfind(s,'a')
0
>>> nfind(s,'a',2)
6
>>> nfind(s,'a',3)
11
答案 6 :(得分：0)
def ifind( s, word, start=0 ):
pos = s.find(word,start)
while -1 < pos:
yield pos
pos = s.find(word,pos+1)
print list(ifind('abcdefacbdea', 'a')) # [0, 6, 11]
print list(ifind('eee', 'a')) # []
答案 7 :(得分：0)
def nindex(str, substr, index):
slice = str
n = 0
while index:
n += slice.index(substr) + len(substr)
slice = str[n:]
index -= 1
return slice.index(substr) + n
答案 8 :(得分：0)
import itertools
def multis(search,text,start=0):
while start>-1:
f=text.find(search,start)
start=f
if start>-1:
yield f
start+=1
# one based function for nth result only
def nindex(text,search,n):
return itertools.islice(multis(search,text),n-1,n).next()
text = 'abcdefacbdea'
search = 'a'
print("Hit %i: %i" % (3, nindex(text,search,3)))
print ('All hits: %s' % list(multis(search,text)))
没有索引：
def nthpartition(search,text,n=None):
## nth partition before and after or all if not n
if not n:
n=len(text) # bigger always than maximum number of n
for i in range(n):
before,search,text = text.partition(search)
if not search:
return
yield before,text
text = 'abcdefacbdea'
search = 'a'
print("Searching %r in %r" % (search,text))
for parts in nthpartition(search,text): print(parts)
"""Output:
Searching 'a' in 'abcdefacbdea'
('', 'bcdefacbdea')
('bcdef', 'cbdea')
('cbde', '')
"""
答案 9 :(得分：0)
import re
def nindex(text, n=1, default=-1):
return next(
itertools.islice((m.start() for m in re.finditer('a', text)), n - 1, None),
default
)
print nindex(s)
print nindex(s, 1)
print nindex(s, 2)
print nindex(s, 3)
print nindex(s, 4)
答案 10 :(得分：0)
重复调用'index'，使用最后一次调用的结果（+ 1）作为起始位置：
def nindex(needle, haystack, n):
"find the nth occurrence of needle in haystack"
pos = -1
for dummy in range(n):
pos = haystack.index(needle, pos + 1)
return pos
注意：我没有测试过。
答案 11 :(得分：0)
怎么样......
# index is 0-based
def nindex(needle, haystack, index=0):
parts = haystack.split(needle)
if index >= len(parts)-1:
return -1
return sum(len(x) for x in parts[:index+1])+index*len(needle)
答案 12 :(得分：0)
这个在正则表达式中工作..如果你修改它以缓存已编译的正则表达式（或记住它），那么 NOT（测试后）可能会更快。
import re
def nindex(s, substr, n = 1):
"""Find the nth occurrence of substr in s."""
safe_substr = re.escape(substr)
regex_str = ".*?(?:%s.*?){%i}(%s).*?" % (safe_substr, n - 1, safe_substr)
regex = re.compile(regex_str)
match = regex.search(s)
if match is None:
index = None
else:
index = match.start(1)
return index
# The rest of this code is just test cases...
for search_str in ("a", "bc"):
print "Looking for %s" % search_str
for test_str in ('abcdefacbdea',
'abcdefacbdeaxxx',
'xxxabcdefacbdeaxxx'):
for i in (0, 1, 2, 3, 4):
print("%s %i index: %s" %
(test_str, i, nindex(test_str, search_str, i)))
print
输出是：
Looking for a
abcdefacbdea 0 index: None
abcdefacbdea 1 index: 0
abcdefacbdea 2 index: 6
abcdefacbdea 3 index: 11
abcdefacbdea 4 index: None
abcdefacbdeaxxx 0 index: None
abcdefacbdeaxxx 1 index: 0
abcdefacbdeaxxx 2 index: 6
abcdefacbdeaxxx 3 index: 11
abcdefacbdeaxxx 4 index: None
xxxabcdefacbdeaxxx 0 index: None
xxxabcdefacbdeaxxx 1 index: 3
xxxabcdefacbdeaxxx 2 index: 9
xxxabcdefacbdeaxxx 3 index: 14
xxxabcdefacbdeaxxx 4 index: None
Looking for bc
abcdefacbdea 0 index: None
abcdefacbdea 1 index: 1
abcdefacbdea 2 index: None
abcdefacbdea 3 index: None
abcdefacbdea 4 index: None
abcdefacbdeaxxx 0 index: None
abcdefacbdeaxxx 1 index: 1
abcdefacbdeaxxx 2 index: None
abcdefacbdeaxxx 3 index: None
abcdefacbdeaxxx 4 index: None
xxxabcdefacbdeaxxx 0 index: None
xxxabcdefacbdeaxxx 1 index: 4
xxxabcdefacbdeaxxx 2 index: None
xxxabcdefacbdeaxxx 3 index: None
xxxabcdefacbdeaxxx 4 index: None
以下是备忘版本：
def memoized_hedgehog_nindex(s, substr, n = 1, _memoized_regexes = {}):
safe_substr = re.escape(substr)
regex_str = ".*?(?:%s.*?){%i}(%s).*?" % (safe_substr, n - 1, safe_substr)
# memoize
key = (n, safe_substr)
if key in _memoized_regexes:
regex = _memoized_regexes[key]
else:
regex = re.compile(regex_str)
_memoized_regexes[key] = regex
match = regex.search(s)
if match is None:
index = None
else:
index = match.start(1)
return index