获取Python中2个子字符串之间的所有可能字符串

Question

之前获取中间字符串has been asked但不涵盖某些条件的问题。就我而言，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“介于两者之间”的答案是：

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'

但我想要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，我怎样才能在subject 中获得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

您可以使用this one 之类的解决方案，您只需稍微修改一下即可满足您的正则表达式要求：

import re

def findall_overlapped(r, s):
  res = []                     # Resulting list
  reg = r'^{}$'.format(r)      # Regex must match full string
  for q in range(len(s)):      # Iterate over all chars in a string
    for w in range(q,len(s)):  # Iterate over the rest of the chars to the right
        cur = s[q:w+1]         # Currently tested slice
        m = re.match(reg, cur) # If there is a full slice match
        if m:
            res.append(m.group(1))    # Append Group 1 value to the resulting list
  return res

rx = r'foo", "(.*?)", "bar"'
text = '"lorem ipsum", "foo", "baz", "bar", "lorem ipsum", "bar", "ipsum", "foo", "baz", "bar"'
for s in findall_overlapped(rx, text):
    print(s)

Python code demo 的输出：

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

Answer 2

这是获得正确答案的一种方法。它搜索第一个词条，并为每个条目搜索列表的其余部分以查找第二个词条。

subject = ["lorem ipsum", "foo", "baz", "bar", "lorem ipsum", "bar", "ipsum", "foo", "baz", "bar"]

sch1 = "foo"
sch2 = "bar"

all = []

for i1 in  range(len(subject)):  # search first term
   if subject[i1] == sch1: # found first term
       for i2 in range(i1, len(subject)):  # search second term
           if subject[i2] == sch2:   # found second term
               all.extend(subject[i1+1:i2])  # add to main list

print(all)

输出（格式化）

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

Answer 3

这是一种可行的方法，我会说相当有效。

• 使用re.finditer，查找开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)

Answer 4

可能不是最快的解决方案：

from timeit import default_timer as timer                                                                                                                                                                

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

key1 = 'foo", "'
key2 = '", "bar"'

t_start = timer()

keyl1 = []
keyl2 = []
result = []

start = 0 
start2 = 0
end = len(subject)

while True:
    keyl1.append(subject.find(key1,start,end))
    start = keyl1[-1]+1
    if keyl1[-1] == -1:
       break
   start2 = start
   while True:
       keyl2.append(subject.find(key2,start2,end))
       if keyl2[-1] == -1:
           keyl2 = []
           break
       if keyl2[-1]>keyl1[-1]:
           result.append(subject[keyl1[-1]+len(key1):keyl2[-1]])
       start2 = keyl2[-1]+1

for r in result:
    print(r)

t_end = timer()
print('Seconds: ', t_end - t_start)