python - 在 Python 中获取 2 个子字符串之间的所有可能字符串

Question

获取中间字符串的问题has been asked之前但不涵盖某些条件。在我的例子中，我可能有如下字符串:

subject = '"lorem ipsum", "foo", "baz", "bar", "lorem ipsum", "bar", "ipsum", "foo", "baz", "bar"'

我想提取 foo"、" 和 "、"bar" 之间的所有文本实例。传统的“Get between”答案是:

import re
result = re.findall('foo", "(.*)", "bar', subject)
print(result)

只返回 1 个结果字符串:

'baz", "bar", "lorem ipsum", "bar", "ipsum", "foo", "baz'

但我想返回的是所有可能的“Between”字符串的列表，例如:

[
    'baz',
    'baz", "bar", "lorem ipsum',
    'baz", "bar", "lorem ipsum", "bar", "ipsum", "foo", "baz',
    'baz'
]

所以给定两个子字符串(开始和结束)和一个主题字符串，如何在 subject 中获取 start 和 end 之间的所有可能子字符串>?计算效率最高的解决方案当然是可取的。

Answer 1

这是一种可行的方法，我认为相当有效。

• 使用 re.finditer，找到起始和结束模式的索引/跨度
• 创建索引/跨度的所有合理组合
• 切出结果

subject = '"lorem ipsum", "foo", "baz", "bar", "lorem ipsum", "bar", "ipsum", "foo", "baz", "bar"'

import re
# spans of "start" pattern
sSpans = [match.span() for match in re.finditer('foo", "' ,subject)]
# spans of "end"   pattern
eSpans = [match.span() for match in re.finditer('", "bar"',subject)]
# all possible combination of "between" spans
spans = [(s[1],e[0]) for s in sSpans for e in eSpans] 
# filter only reasonable spand where end > start
spans = [(s,e) for s,e in spans if e > s]
# slice out the "between strings"
result = [subject[s:e] for s,e in spans]
for r in result: print(r)

也可以压缩成一行:

result = [subject[s.span()[1]:e.span()[0]] for s in re.finditer('foo", "' ,subject) for e in re.finditer('", "bar"',subject) if e.span()[0] > s.span()[1]]
for r in result: print(r)