需要部分正则表达式代码的具体说明

Question

I'm developing a calculator program in Python, and need to remove leading zeros from numbers so that calculations work as expected. For example, if the user enters "02+03" into the calculator, the result should return 5. In order to remove these leading zeroes in-front of digits, I asked a question on here and got the following answer.

self.answer = eval(re.sub(r"((?<=^)|(?<=[^\.\d]))0+(\d+)", r"\1\2", self.equation.get()))

I fully understand how the positive lookbehind to the beginning of the string and lookbehind to the non digit, non period character works. What I'm confused about is where in this regex code can I find the replacement for the matched patterns?

I found this online when researching regex expressions.

result = re.sub(pattern, repl, string, count=0, flags=0)

Where is the "repl" in the regex code above? If possible, could somebody please help to explain what the r"\\1\\2" is used for in this regex also?

Thanks for your help! :)

Answer 1

The "repl" part of the regex is this component:

r"\1\2"

In the "find" part of the regex, group capturing is taking place (ordinarily indicated by "()" characters around content, although this can be overridden by specific arguments).

In python regex, the syntax used to indicate a reference to a positional captured group (sometimes called a "backreference") is "\\n" (where "n" is a digit refering to the position of the group in the "find" part of the regex).

So, this regex is returning a string in which the overall content is being replaced specifically by parts of the input string matched by numbered groups.

Note: I don't believe the "\\1" part of the "repl" is actually required. I think:

r"\2"

...would work just as well.

Further reading: https://www.regular-expressions.info/brackets.html

Answer 2

See example:

>>> import re
>>> s='awd232frr2cr23'
>>> re.sub('\d',' ',s)
'awd   frr cr  '
>>> 

Explanation:

• As it is, '\\d' is for integer so removes them and replaces with repl (in this case ' ' ).

Answer 3

Firstly, repl includes what you are about to replace.

To understand \\1\\2 you need to know what capture grouping is.
Check this video out for basics of Group capturing .
Here , since your regex splits every match it finds into groups which are 1,2... so on. This is so because of the parenthesis () you have placed in the regex. $1 , $2 or \\1,\\2 can be used to refer to them.

In this case:
The regex is replacing all numbers after the leading 0 (which is caught by group 2) with itself.
Note: \\1 is not necessary. works fine without it.