python正则表达式向前看正负 [英] python regex look ahead positive + negative

问题描述

此正则表达式将得到456.我的问题是为什么它不能从1-234-56变为234? 56是否限定(?！\ d))模式，因为它不是一位数字. (?！\ d))寻找的起点在哪里?

This regex will get 456. My question is why it CANNOT be 234 from 1-234-56 ? Does 56 qualify the (?!\d)) pattern since it is NOT a single digit. Where is the beginning point that (?!\d)) will look for?

import re
pattern = re.compile(r'\d{1,3}(?=(\d{3})+(?!\d))')
a = pattern.findall("The number is: 123456") ; print(a)

在第一阶段添加逗号分隔符，例如123,456.

It is in the first stage to add the comma separator like 123,456.

a = pattern.findall("The number is: 123456") ; print(a)
results = pattern.finditer('123456')
  for result in results:
    print ( result.start(), result.end(), result)

推荐答案

我的问题是为什么它不能从1-234-56变为234?

这是不可能的，因为(?=(\d{3})+(?!\d))要求1位数至3位数的序列后出现3位数的序列. 56(您想象中的场景中的最后一个数字组)是一个2位数的组.由于量词可以是懒惰的，也可以是贪婪的，因此您不能同时将一个，两个和三个数字组与\d{1,3}进行匹配.要从123456获得234，您需要为其专门定制的正则表达式: \B\d{3} 或 (?<=1)\d{3} 甚至是

It is not possible as (?=(\d{3})+(?!\d)) requires 3-digit sequences appear after a 1-3-digit sequence. 56 (the last digit group in your imagined scenario) is a 2-digit group. Since a quantifier can be either lazy or greedy, you cannot match both one, two and three digit groups with \d{1,3}. To get 234 from 123456, you'd need a specifically tailored regex for it: \B\d{3}, or (?<=1)\d{3} or even \d{3}(?=\d{2}(?!\d)).

56是否匹配(?!\d))模式? (?！\ d))寻找的起点在哪里?

Does 56 match the (?!\d)) pattern? Where is the beginning point that (?!\d)) will look for?

否，这是一个否定的超前查询，它不匹配，它仅检查输入字符串中当前位置之后是否没有数字.如果有数字，则匹配失败(找不到并返回结果).

No, this is a negative lookahead, it does not match, it only checks if there is no digit right after the current position in the input string. If there is a digit, the match is failed (not result found and returned).

关于前瞻的更多说明:它位于(\d{3})+子模式之后，因此正则表达式引擎会在最后一个3位数字组之后立即开始搜索一个数字，如果找到该数字，则匹配失败(因为它是负面的前瞻).用简单的话来说， (?!\d)是此正则表达式中的数字闭合/跟踪边界.

More clarification on the look-ahead: it is located after (\d{3})+ subpattern, thus the regex engine starts searching for a digit right after the last 3-digit group, and fails a match if the digit is found (as it is a negative lookahead). In plain words, the (?!\d) is a number closing/trailing boundary in this regex.

更详细的细分:

• \d{1,3}-1至3位数字，尽可能多(使用贪婪量词)
• (?=(\d{3})+(?!\d))-正前瞻((?=...))，用于检查前面匹配的1-3位数字序列是否跟随
  • (\d{3})+-1个或更多(+)正好3位数字的序列...
  • (?!\d)-后面没有数字.
  • \d{1,3} - 1 to 3 digit sequence, as many as possible (greedy quantifier is used)
  • (?=(\d{3})+(?!\d)) - a positive look-ahead ((?=...)) that checks if the 1-3 digit sequence matched before are followed by
    • (\d{3})+ - 1 or more (+) sequences of exactly 3 digits...
    • (?!\d) - not followed by a digit.

    黑头字符不匹配，不消耗字符，但是您仍然可以在其中捕获.当执行前瞻时，正则表达式索引与以前的字符相同. 使用您的正则表达式和输入，将123与\d{1,3}匹配，因为您拥有3位数的序列(456).但是456在先行之内具有防护能力，并且 re.findall 仅在设置了捕获组的情况下返回捕获的文本.

    Lookaheads do not match, do not consume characters, but you still can capture inside them. When a lookahead is executed, the regex index is at the same character as before. With your regex and input, you match 123 with \d{1,3} as then you have 3-digit sequence (456). But 456 is capured within a lookahead, and re.findall returns only captured texts if capturing groups are set.

    要仅将逗号添加为数字分组符号，请使用

    To just add comma as digit grouping symbol, use

    rx = r'\d(?=(?:\d{3})+(?!\d))'
    

    请参见 IDEONE演示

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