子字符串在字符串中的位置

Question

我需要知道一个单词在文本中的所有位置 - 字符串中的子字符串。到目前为止的解决方案是使用正则表达式，但我不确定是否有更好的内置标准库策略。有什么想法吗？

import re

text = "The quick brown fox jumps over the lazy dog. fox. Redfox."
links = {'fox': [], 'dog': []}
re_capture = u"(^|[^\w\-/])(%s)([^\w\-/]|$)" % "|".join(links.keys())

iterator = re.finditer(re_capture, text)

if iterator:
    for match in iterator:

        # fix position by context 
        # (' ', 'fox', ' ')
        m_groups = match.groups()
        start, end = match.span()
        start = start + len(m_groups[0])
        end = end - len(m_groups[2])

        key = m_groups[1]
        links[key].append((start, end))

print links

{'fox': [(16, 19), (45, 48)], 'dog': [(40, 43)]}

Edit: Partial words are not allowed to match - see fox of Redfox is not in links.

谢谢。

Answer 1

不像 pythonic 且没有正则表达式：

text = "The quick brown fox jumps over the lazy dog. fox."
links = {'fox': [], 'dog': []}

for key in links:
    pos = 0
    while(True):
        pos = text.find(key, pos)
        if pos < 0:
            break
        links[key].append((pos, pos + len(key)))
        pos = pos + 1
print(links)

Answer 2

如果你想匹配实际的单词并且你的字符串包含 ascii:

text = "fox The quick brown fox jumps over the fox! lazy dog. fox!."
links = {'fox': [], 'dog': []}

from string import punctuation
def yield_words(s,d):
    i = 0
    for ele in s.split(" "):
        tot = len(ele) + 1
        ele = ele.rstrip(punctuation)
        ln = len(ele)
        if ele in d:
            d[ele].append((i, ln + i))
        i += tot
    return d

这与查找解决方案不同，它不会匹配部分单词并在 O(n) 时间内完成：

In [2]: text = "The quick brown fox jumps over the lazy dog. fox. Redfox."

In [3]: links = {'fox': [], 'dog': []}

In [4]: yield_words(text,links)
Out[4]: {'dog': [(40, 43)], 'fox': [(16, 19), (45, 48)]}

这可能是正则表达式是一种好方法的一种情况，它可以简单得多：

def reg_iter(s,d):
    r = re.compile("|".join([r"\b{}\b".format(w) for w in d]))
    for match in r.finditer(s):
        links[match.group()].append((match.start(),match.end()))
    return d

输出：

In [6]: links = {'fox': [], 'dog': []}

In [7]: text = "The quick brown fox jumps over the lazy dog. fox. Redfox."

In [8]: reg_iter(text, links)
Out[8]: {'dog': [(40, 43)], 'fox': [(16, 19), (45, 48)]}

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