如何在python 3中打印正则表达式匹配结果? [英] How to print regex match results in python 3?

问题描述

我当时处于空闲状态，因此决定使用正则表达式来对字符串进行排序.但是，当我输入在线教程告诉我的内容时，所要做的只是打印:

I was in IDLE, and decided to use regex to sort out a string. But when I typed in what the online tutorial told me to, all it would do was print:

<_sre.SRE_Match object at 0x00000000031D7E68>

完整程序:

import re
reg = re.compile("[a-z]+8?")
str = "ccc8"
print(reg.match(str))

结果:

<_sre.SRE_Match object at 0x00000000031D7ED0>

有人可以告诉我如何实际打印结果吗?

Could anybody tell me how to actually print the result?

推荐答案

您需要在match函数后添加.group()，以便它打印匹配的字符串，否则仅显示是否发生匹配.要打印由捕获组捕获的字符，需要将相应的组索引传递给.group()函数.

You need to include .group() after to the match function so that it would print the matched string otherwise it shows only whether a match happened or not. To print the chars which are captured by the capturing groups, you need to pass the corresponding group index to the .group() function.

>>> import re
>>> reg = re.compile("[a-z]+8?")
>>> str = "ccc8"
>>> print(reg.match(str).group())
ccc8

带有捕获组的正则表达式.

Regex with capturing group.

>>> reg = re.compile("([a-z]+)8?")
>>> print(reg.match(str).group(1))
ccc

re.match (样式，字符串，标志= 0)

re.match(pattern, string, flags=0)

如果字符串开头的零个或多个字符与正则表达式模式匹配，则返回相应的MatchObject实例.如果字符串与模式不匹配，则返回None；否则返回false.请注意，这与零长度匹配不同.

If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding MatchObject instance. Return None if the string does not match the pattern; note that this is different from a zero-length match.

请注意，即使在MULTILINE模式下，re.match()也只会在字符串的开头而不是每行的开头进行匹配.

Note that even in MULTILINE mode, re.match() will only match at the beginning of the string and not at the beginning of each line.

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