找到第一个子字符串拆分字符串

Question

我希望在第一次出现这些词时，用某些词语来分句。 让我说明一下：

message = 'I wish to check my python code for errors to run the program properly with fluency'

我希望在第一次出现for/to/with拆分上面的消息，因此上面消息的结果将check my python code for errors to run the program properly with fluency

我还希望包含我将句子拆分的单词，因此我的最终结果将是： to check my python code for errors to run the program properly with fluency

我的代码不起作用：

import re
message = 'I wish to check my python code for errors to run the program properly with fluency'
result = message.split(r"for|to|with",1)[1]
print(result)

我能做什么？

Answer 1

message = 'I wish to check my python code for errors to run the program properly with fluency'
array = message.split(' ')
number = 0
message_new = ''
for i in range(len(array)):
    if array[i] == 'to' or array[i] == 'for':
        number=i
        break
for j in range(number,len(array)):
    message_new += array[j] + ' '
print(message_new) 

输出：

to check my python code for errors to run the program properly with fluency 

Answer 2

split不会将正则表达式作为参数（也许你正在考虑Perl）。

以下是您想要的：

import re
message = 'I wish to check my python code for errors to run the program properly with fluency'
result = re.search(r'\b(for|to|with)\b', message)
print message[result.start(1):]

这不使用替换，重新加入或循环，而只是简单搜索所需的字符串并使用其位置结果。

Answer 3

这个问题已经回答： 如何删除python中特定字符之前的所有字符，但它只适用于一个特定的分隔符，对于多个分隔符，你首先要找出哪个首先出现，可以在这里找到： 怎么能我发现python字符串中第一次出现一个子字符串，你从第一个猜测开始，我没有太多的想象力所以让我们称之为bestDelimiter = firstDelimiter，找出它第一次出现的位置，将位置保存到bestPosition =第一次出现的位置，继续找出其余分隔符的位置，每次你找到一个在当前bestPosition之前出现的分隔符你更新两个变量bestDelimiter和bestPosition，最后出现的那个是最好的分辨符，然后使用bestDelimiter继续应用您需要的操作

Answer 4

我的猜测是，这个简单的表达可能就是这么做的

.*?(\b(?:to|for|with)\b.*)

和re.match可能是这五种方法中最快的一种：

用re.findall测试

import re

regex = r".*?(\b(?:to|for|with)\b.*)"
test_str = "I wish to check my python code for errors to run the program properly with fluency"
print(re.findall(regex, test_str))

用re.sub测试

import re

regex = r".*?(\b(?:to|for|with)\b.*)"
test_str = "I wish to check my python code for errors to run the program properly with fluency"
subst = "\\1"

result = re.sub(regex, subst, test_str)

if result:
    print (result)

用re.finditer测试

import re

regex = r".*?(\b(?:to|for|with)\b.*)"

test_str = "I wish to check my python code for errors to run the program properly with fluency"

matches = re.finditer(regex, test_str, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):

    # FULL MATCH
    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

用re.match测试

import re

regex = r".*?(\b(?:to|for|with)\b.*)"
test_str = "I wish to check my python code for errors to run the program properly with fluency"

print(re.match(regex, test_str).group(1))

用re.search测试

import re

regex = r".*?(\b(?:to|for|with)\b.*)"
test_str = "I wish to check my python code for errors to run the program properly with fluency"

print(re.search(regex, test_str).group(1))

如果您希望进一步探索或修改它，可以在本演示的右上方面板中解释该表达式，如果您愿意，可以在此链接中查看它与某些示例输入的匹配情况。

Answer 5

您可以先找到for ， to和with所有实例，拆分所需的值，然后拼接并重新加入：

import re
message = 'I wish to check my python code for errors to run the program properly with fluency'
vals, [_, *s] = re.findall(r"\bfor\b|\bto\b|\bwith\b", message), re.split(r"\bfor\b|\bto\b|\bwith\b", message)
result = ''.join('{} {}'.format(a, re.sub("^\s+", "", b)) for a, b in zip(vals, s))

输出：

'to check my python code for errors to run the program properly with fluency'