我对Regex非常陌生,所以我确信我遗漏了一些显而易见的东西,但需要解决以下问题。
我想从一个特定的子字符串中提取后面的字符串。我正在处理一个扫描文档列表,并有下面的示例字符串,我希望在"FORENAME“之后提取所有内容
这就是我到目前为止所做的:
regex = r"(?<=(FORE))[A-Z]+"
test_str = 'UNIQUE NUMBER 12345 678910 11 FROM THIS DOCUMENT | . ISSUED ON 2011-04-04 FORENAME GUIDO \\ SURNAME VAN ROSSUM. '
matches = re.finditer(regex, test_str)
for matchNum, match in enumerate(matches, start=1):
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)))
返回以下内容:
Match 1 was found at 78-82: NAME
Group 1 found at 74-78: FORE
我想要的是:
GUIDO \姓VAN ROSSUM。
谢谢!
发布于 2020-04-24 00:04:38
我想要的是: GUIDO \姓VAN ROSSUM。
基于上述,您可以使用:
import re
test_str = 'UNIQUE NUMBER 12345 678910 11 FROM THIS DOCUMENT | . ISSUED ON 2011-04-04 FORENAME GUIDO \\ SURNAME VAN ROSSUM.'
result = re.sub(r"^.*FORENAME(.*?)$", r"\1", test_str)
print(result)
# GUIDO \ SURNAME VAN ROSSUM.
发布于 2020-04-24 02:23:09
对于这么简单的问题,你不需要正则表达式
test_str = 'UNIQUE NUMBER 12345 678910 11 FROM THIS DOCUMENT | . ISSUED ON 2011-04-04 FORENAME GUIDO \\ SURNAME VAN ROSSUM. '
pos = test_str.find("FORENAME") + len("FORENAME")
print(test_str[pos:])
https://stackoverflow.com/questions/61398648
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