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Question

我无法将RegEx表达式转换为python。我知道'(\\d+)'是单个整数的表达式，但我无法弄清楚如何得到一个[2-9]的整数。

RegEx表达式如下：

[2-9][p-z][a-h][2-9][a-z]*[p-z][2-9][p-z][2-9][p-z]

这是我目前的表达，但它产生了许多误报，因为它不够具体：

          re1='(\\d+)'    # Integer Number 1
            re2='([a-z])'   # Any Single Word Character (Not Whitespace) 1
            re3='([a-z])'   # Any Single Word Character (Not Whitespace) 2
            re4='(\\d+)'    # Integer Number 2
            re5='((?:[a-z][a-z]+))' # Word 1
            re6='(\\d+)'    # Integer Number 3
            re7='([a-z])'   # Any Single Word Character (Not Whitespace) 3
            re8='(.)'   # Any Single Character 1
            re9='([a-z])'   # Any Single Word Character (Not Whitespace) 4

            ## Regex search for passcodes ## Thanks to Pierluigi Failla
            rg = re.compile(re1+re2+re3+re4+re5+re6+re7+re8+re9,re.IGNORECASE|re.DOTALL)
            m = rg.search(txt)
            if m:
                int1=m.group(1)
                w1=m.group(2)
                w2=m.group(3)
                int2=m.group(4)
                word1=m.group(5)
                int3=m.group(6)
                w3=m.group(7)
                c1=m.group(8)
                w4=m.group(9)
                txt2='"'+int1+w1+w2+int2+word1+int3+w3+c1+w4+'"'
                return [txt2]

Answer 1

您应该能够在Python中使用2-9范围，如下所示：re1 = re.compile(r'[2-9]')。然后我的控制台中的测试显示re1.match('7')会根据需要返回MatchObject，而re1.match('0')也会根据需要返回None。

您似乎也使用了[a-z]中的范围re2，您说您想要[p-z] - 其他字符范围内的类似问题。

Answer 2

我根据您在问题中看到的内容提出此代码：

import re

pat = ('([2-9])'        # Integer Number 1
       '([p-z])'        # Any Single Word Character (Not Whitespace) 1
       '([a-h])'        # Any Single Word Character (Not Whitespace) 2
       '([2-9])'        # Integer Number 2
       '([a-z]*[p-z]+)' # Word 1
       '([2-9])'        # Integer Number 3
       '([p-z])'        # Any Single Word Character (Not Whitespace) 3
       '(.)'            # Any Single Character 1
       '([p-z])'        # Any Single Word Character (Not Whitespace) 4
       )
rg = re.compile(pat)

txt = 'jiji4pa6fmlgkfmoaz8p#q,,,,,,,,,,'
m = rg.search(txt)
if m:
    text2 = "%s%s%s%s%s%s%s%s%s" % m.groups()
    print text2

# prints 4pa6fmlgkfmoaz8p#q

修改

text2 = ''.join(m.groups())  # is better

Python正则表达式问题

2 个答案:

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