Python索引不止一次

Question

我知道.index()將返回子串在python中的位置。 但是，我想要的是找到子串在第n次的位置，這將是這樣的：

>> s = 'abcdefacbdea'
>> s.index('a')
0
>> s.nindex('a', 1)
6
>>s.nindex('a', 2)
11

有沒有辦法在python中執行此操作？

Answer 1

怎么樣...

def nindex(mystr, substr, n=0, index=0):
    for _ in xrange(n+1):
        index = mystr.index(substr, index) + 1
    return index - 1

Obs：正如str.index()所做的那樣，當找不到substr時， nindex()會引發ValueError 。

Answer 2

是。 使用s.index('yourstring', start)寫一個循環s.index('yourstring', start)

找到一個大胖子后更新-1 ...我沒有寫一些代碼???

這是我的兌換嘗試，如果需要允許不重疊，並按照顯示的程度進行測試：

>>> 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
>>>

Answer 3

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

我有興趣看到其他解決方案，我覺得這不是特別pythonic。

Answer 4

我可能會用

[index for index, value in enumerate(s) if s == 'a'][n]

要么

from itertools import islice
next(islice((index for index, value in enumerate(s) if s == 'a'), n, None))

或者完全避免處理指數。

Answer 5

這是一個記憶版本，盡可能地避免浪費工作，同時保持與您的規格接近[1]（而不是做一些更健全的事情，例如循環瀏覽所有命中; - ）......：

[1]：只是關閉 - 當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)

按要求打印0，然后是6，然后是11。

Answer 6

>>> 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

Answer 7

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'))              # []

Answer 8

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

Answer 9

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)))

沒有索引：

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', '')
"""

Answer 10

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)

Answer 11

只需使用最后一次調用（+ 1）的結果作為起始位置，重復調用'index'：

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：我還沒有測試過。

Answer 12

怎么樣...

# 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)

Answer 13

這個在正則表達式中工作..如果你修改它以緩存已編譯的正則表達式（或記住它），那么它（測試后）可能會更快。

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 

輸出是：

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

這是備忘版本：

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