@H_301_3@
我的文件包含45个十六进制数字,由空格分隔或48个十六进制数字,由空格分隔.我个人而不是整体需要所有这些数字.我目前正在使用蛮力方法获得45个数字.
pattern = re.compile("([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s([0-9a-f]{2})\s")
然而,即便如此,我仍然无法弄清楚如何在48十六进制数字实例中提取剩余的三个数字.你能帮我解决这个问题吗？
我会避免像下面的解决方案(如果它工作,没有尝试过),因为我将不得不为每个实例拆分字符串,即考虑它给出正确的输出！
(((?:[0-9a-f]{2})\s){48})|(((?:[0-9a-f]{2})\s){45})
谢谢！
解决方法
在编写长RE时,请考虑使用
re.VERBOSE使其更具可读性.
pattern = re.compile(r"""
^( [0-9a-fA-F]{2} (?: \s [0-9a-fA-F]{2} ){44}
(?:(?: \s [0-9a-fA-F]{2} ){3} )? )$
""",re.VERBOSE)
读为：两个十六进制数字,后跟44次(空格后跟两个十六进制数字),可选地后跟3次(空格后跟两个十六进制数字).
测试：
>>> pattern.match(" ".join(["0f"] * 44))
>>> pattern.match(" ".join(["0f"] * 45))
<_sre.SRE_Match object at 0x7fd8f27e0738>
>>> pattern.match(" ".join(["0f"] * 46))
>>> pattern.match(" ".join(["0f"] * 47))
>>> pattern.match(" ".join(["0f"] * 48))
<_sre.SRE_Match object at 0x7fd8f27e0990>
>>> pattern.match(" ".join(["0f"] * 49))
最后,要检索各个数字,请对匹配结果执行.group(0).split().这比编写将所有数字放入不同组的RE要容易得多.
编辑：好的,这是如何解决原始问题.只需动态构建RE.
chunk = r"""([0-9a-fA-F]{2}\s)"""
pattern = re.compile(chunk * 45 + "(?:" + chunk * 3 + ")?")
@H_301_3@
