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Re: Help with regular expression in python

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2011/8/18 Matt Funk <matze999@gmail.com>:
> Hi,
> i am sorry if this doesn't quite match the subject of the list. If someone
> takes offense please point me to where this question should go. Anyway, i have
> a problem using regular expressions. I would like to match the line:
>
> 1.002000e+01 2.037000e+01 2.128000e+01 1.908000e+01 1.871000e+01 1.914000e+01
> 2.007000e+01 1.664000e+01 2.204000e+01 2.109000e+01 2.209000e+01 2.376000e+01
> 2.158000e+01 2.177000e+01 2.152000e+01 2.267000e+01 1.084000e+01 1.671000e+01
> 1.888000e+01 1.854000e+01 2.064000e+01 2.000000e+01 2.200000e+01 2.139000e+01
> 2.137000e+01 2.178000e+01 2.179000e+01 2.123000e+01 2.201000e+01 2.150000e+01
> 2.150000e+01 2.199000e+01 : (instance: 0)       :       some description
>
> The number of floats can vary (in this example there are 32). So what i thought
> i'd do is the following:
> instance_linetype_pattern_str = '([-+]?(\d+(\.\d*)?|\.\d+)([eE][-+]?\d+)?)
> {32}'
> instance_linetype_pattern = re.compile(instance_linetype_pattern_str)
> Basically the expression in the first major set of paranthesis matches a
> scientific number format. The '{32}' is supposed to match the previous 32
> times. However, it doesn't. I  can't figure out why this does not work. I'd
> really like to understand it if someone can shed light on it.
>
> thanks
> matt
> --
> http://mail.python.org/mailman/listinfo/python-list
>
Hi,
the already suggested handling of whitespace with \s+ etc. at the end
of the parenthesised patern should help;
furhtermore, if you are using this pattern in the python source, you
should either double all backslashes or use a raw string for the
pattern - with r prepended before the opening quotation mark:
pattern_str = r"..."
in order to handle backslashes literally and not as escape character.
hth,
vbr

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Re: Help with regular expression in python Vlastimil Brom <vlastimil.brom@gmail.com> - 2011-08-18 23:27 +0200

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