Python 正则表达式 - re.search() 与 re.findall() [英] Python regular expressions - re.search() vs re.findall()

问题描述

对于学校，我应该编写一个提取 IP 地址的 Python RE 脚本.我使用的正则表达式似乎适用于 re.search() 但不适用于 re.findall().

For school I'm supposed to write a Python RE script that extracts IP addresses. The regular expression I'm using seems to work with re.search() but not with re.findall().

exp = "(\d{1,3}\.){3}\d{1,3}"
ip = "blah blah 192.168.0.185 blah blah"
match = re.search(exp, ip)
print match.group()

匹配总是 192.168.0.185，但是当我做 re.findall()

The match for that is always 192.168.0.185, but its different when I do re.findall()

exp = "(\d{1,3}\.){3}\d{1,3}"
ip = "blah blah 192.168.0.185 blah blah"
matches = re.findall(exp, ip)
print matches[0]

0.

我想知道为什么 re.findall() 产生 0. 当 re.search() 产生 192.168.0.185 时，因为我使用相同的表达式两个功能.

I'm wondering why re.findall() yields 0. when re.search() yields 192.168.0.185, since I'm using the same expression for both functions.

我该怎么做才能使 re.findall() 真正正确地遵循表达式?还是我犯了某种错误?

And what can I do to make it so re.findall() will actually follow the expression correctly? Or am I making some kind of mistake?

推荐答案

findall 返回匹配列表，并从文档中:

findall returns a list of matches, and from the documentation:

如果模式中存在一个或多个组，则返回一个团体名单；这将是一个元组列表，如果模式有不止一组.

If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.

因此，您之前的表达式有一组在最后一次匹配为 0 的字符串中匹配了 3 次.

So, your previous expression had one group that matched 3 times in the string where the last match was 0.

要解决您的问题，请使用:exp = "(?:\d{1,3}\.){3}\d{1,3}";通过使用非分组版本，没有返回组，因此在两种情况下都返回匹配.

To fix your problem use: exp = "(?:\d{1,3}\.){3}\d{1,3}"; by using the non-grouping version, there is no returned groups so the match is returned in both cases.

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