Replace various regex
Jean-Michel Pichavant
jeanmichel at sequans.com
Mon Feb 15 09:27:40 EST 2010
Martin wrote:
> On Feb 15, 2:03 pm, Jean-Michel Pichavant <jeanmic... at sequans.com>
> wrote:
>
>> Martin wrote:
>>
>>> Hi,
>>>
>>> I am trying to come up with a more generic scheme to match and replace
>>> a series of regex, which look something like this...
>>>
>>> 19.01,16.38,0.79,1.26,1.00 ! canht_ft(1:npft)
>>> 5.0, 4.0, 2.0, 4.0, 1.0 ! lai(1:npft)
>>>
>>> Ideally match the pattern to the right of the "!" sign (e.g. lai), I
>>> would then like to be able to replace one or all of the corresponding
>>> numbers on the line. So far I have a rather unsatisfactory solution,
>>> any suggestions would be appreciated...
>>>
>>> The file read in is an ascii file.
>>>
>>> f = open(fname, 'r')
>>> s = f.read()
>>>
>>> if CANHT:
>>> s = re.sub(r"\d+.\d+,\d+.\d+,\d+.\d+,\d+.\d+,\d+.\d+ !
>>> canht_ft", CANHT, s)
>>>
>>> where CANHT might be
>>>
>>> CANHT = '115.01,16.38,0.79,1.26,1.00 ! canht_ft'
>>>
>>> But this involves me passing the entire string.
>>>
>>> Thanks.
>>>
>>> Martin
>>>
>> I remove all lines containing things like 9*0.0 in your file, cause I
>> don't know what they mean and how to handle them. These are not numbers.
>>
>> import re
>>
>> replace = {
>> 'snow_grnd' : (1, '99.99,'), # replace the 1st number by 99.99
>> 't_soil' : (2, '88.8,'), # replace the 2nd number by 88.88
>> }
>>
>> testBuffer = """
>> 0.749, 0.743, 0.754, 0.759 ! stheta(1:sm_levels)(top to bottom)
>> 0.46 ! snow_grnd
>> 276.78,277.46,278.99,282.48 ! t_soil(1:sm_levels)(top to bottom)
>> 19.01,16.38,0.79,1.26,1.00 ! canht_ft(1:npft)
>> 200.0, 4.0, 2.0, 4.0, 1.0 ! lai(1:npft)
>> """
>>
>> outputBuffer = ''
>> for line in testBuffer.split('\n'):
>> for key, (index, repl) in replace.items():
>> if key in line:
>> parameters = {
>> 'n' : '[\d\.]+', # given you example you have to change
>> this one, I don't know what means 9*0.0 in your file
>> 'index' : index - 1,
>> }
>> # the following pattern will silently match any digit before
>> the <index>th digit is found, and use a capturing parenthesis for the last
>> pattern =
>> '(\s*(?:(?:%(n)s)[,\s]+){0,%(index)s})(?:(%(n)s)[,\s]+)(.*!.*)' %
>> parameters # regexp are sometimes a nightmare to read
>> line = re.sub(pattern, r'\1 '+repl+r'\3' , line)
>> break
>> outputBuffer += line +'\n'
>>
>> print outputBuffer
>>
>
> Thanks I will take a look. I think perhaps I was having a very slow
> day when I posted and realised I could solve the original problem more
> efficiently and the problem wasn't perhaps as I first perceived. It is
> enough to match the tag to the right of the "!" sign and use this to
> adjust what lies on the left of the "!" sign. Currently I have
> this...if anyone thinks there is a neater solution I am happy to hear
> it. Many thanks.
>
> variable_tag = 'lai'
> variable = [200.0, 60.030, 0.060, 0.030, 0.030]
>
> # generate adjustment string
> variable = ",".join(["%s" % i for i in variable]) + ' ! ' +
> variable_tag
>
> # call func to adjust input file
> adjustStandardPftParams(variable, variable_tag, in_param_fname,
> out_param_fname)
>
> and the inside of this func looks like this
>
> def adjustStandardPftParams(self, variable, variable_tag, in_fname,
> out_fname):
>
> f = open(in_fname, 'r')
> of = open(out_fname, 'w')
> pattern_found = False
>
> while True:
> line = f.readline()
> if not line:
> break
> pattern = re.findall(r"!\s+"+variable_tag, line)
> if pattern:
> print 'yes'
> print >> of, "%s" % variable
> pattern_found = True
>
> if pattern_found:
> pattern_found = False
> else:
> of.write(line)
>
> f.close()
> of.close()
>
> return
>
Are you sure a simple
if variable_tag in line:
# do some stuff
is not enough ?
People will usually prefer to write
for line in open(in_fname, 'r') :
instead of your ugly while loop ;-)
JM
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