Say I have a string
假设我有一根绳子
"3434.35353"
and another string
和另一个字符串
"3593"
How do I make a single regular expression that is able to match both without me having to set the pattern to something else if the other fails? I know \d+ would match the 3593, but it would not do anything for the 3434.35353, but (\d+\.\d+) would only match the one with the decimal and return no matches found for the 3593.
如何创建一个既能匹配又能匹配的正则表达式呢?我知道\d+会匹配3593，但是它不会对3434.35353做任何事情，但是(\d+\ d+)只会匹配小数，并且不会返回3593的匹配。
I expect m.group(1) to return:
我期望m.a.(1)返回:
"3434.35353"
or
或
"3593"
4 个解决方案
#1
34
You can put a ? after a group of characters to make it optional.
你可以放一个?在一组字符后使其可选。
You want a dot followed by any number of digits \.\d+, grouped together (\.\d+), optionally (\.\d+)?. Stick that in your pattern:
你想要一个点后面跟着任何数字\。\ d +,组合在一起(\ \ d +),可选(\ \ d +)?。按照你的模式:
import re
print re.match("(\d+(\.\d+)?)", "3434.35353").group(1)
3434.35353
print re.match("(\d+(\.\d+)?)", "3434").group(1)
3434
#2
3
This regex should work:
这个正则表达式应该工作:
\d+(\.\d+)?
It matches one ore more digits (\d+) optionally followed by a dot and one or more digits ((\.\d+)?).
它可选地匹配一个或多个数字(\d+)，后跟一个点和一个或多个数字(\.\d+)?
#3
1
Use the "one or zero" quantifier, ?. Your regex becomes: (\d+(\.\d+)?).
使用“1或0”量词?你的正则表达式变得:(\ d +(\ \ d +)?)。
See Chapter 8 of the TextWrangler manual for more details about the different quantifiers available, and how to use them.
有关可用的不同量词以及如何使用它们的详细信息，请参阅TextWrangler手册的第8章。
#4
1
Read up on the Python RegEx library. The link answers your question and explains why.
阅读Python RegEx库。这个链接回答了你的问题并解释了原因。
However, to match a digit followed by more digits with an optional decimal, you can use
但是，要匹配一个数字，后面跟着更多的数字，并使用一个可选的小数，您可以使用
re.compile("(\d+(\.\d+)?)")
In this example, the ? after the .\d+ capture group specifies that this portion is optional.
在这个例子中，?在。\d+ capture组之后，指定该部分是可选的。
例子
#1
34
You can put a ? after a group of characters to make it optional.
你可以放一个?在一组字符后使其可选。
You want a dot followed by any number of digits \.\d+, grouped together (\.\d+), optionally (\.\d+)?. Stick that in your pattern:
你想要一个点后面跟着任何数字\。\ d +,组合在一起(\ \ d +),可选(\ \ d +)?。按照你的模式:
import re
print re.match("(\d+(\.\d+)?)", "3434.35353").group(1)
3434.35353
print re.match("(\d+(\.\d+)?)", "3434").group(1)
3434
#2
3
This regex should work:
这个正则表达式应该工作:
\d+(\.\d+)?
It matches one ore more digits (\d+) optionally followed by a dot and one or more digits ((\.\d+)?).
它可选地匹配一个或多个数字(\d+)，后跟一个点和一个或多个数字(\.\d+)?
#3
1
Use the "one or zero" quantifier, ?. Your regex becomes: (\d+(\.\d+)?).
使用“1或0”量词?你的正则表达式变得:(\ d +(\ \ d +)?)。
See Chapter 8 of the TextWrangler manual for more details about the different quantifiers available, and how to use them.
有关可用的不同量词以及如何使用它们的详细信息，请参阅TextWrangler手册的第8章。
#4
1
Read up on the Python RegEx library. The link answers your question and explains why.
阅读Python RegEx库。这个链接回答了你的问题并解释了原因。
However, to match a digit followed by more digits with an optional decimal, you can use
但是，要匹配一个数字，后面跟着更多的数字，并使用一个可选的小数，您可以使用
re.compile("(\d+(\.\d+)?)")
In this example, the ? after the .\d+ capture group specifies that this portion is optional.
在这个例子中，?在。\d+ capture组之后，指定该部分是可选的。
例子