McColgst wrote:
On Feb 12, 2:39 pm, Martin <mdeka...@gmail.com> 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
If I understand correctly, there are a couple ways to do it.
One is to use .split() and split by the '!' sign, given that you wont
have more than one '!' on a line. This will return a list of the words
split by the delimiter, in this case being '!', so you should get back
(19.01,16.38,0.79,1.26,1.00 , canht_ft(1:npft) ) and you can do
whatever replace functions you want using the list.
check out split: http://docs.python.org/library/stdtypes.html#str.split
The .split method is the best way if you process the file a line at a
time. The .split method, incidentally, accepts a maxcount argument so
that you can split a line no more than once.
Another, is in your regular expression, you can match the first part
or second part of the string by specifying where the '!' is,
if you want to match the part after the '!' I would do something like
r"[^! cahnt_ft]", or something similar (i'm not particularly up-to-
date with my regex syntax, but I think you get the idea.)
The regex would be r"(?m)^[^!]*(!.*)" to capture the '!' and the rest of
the line.
I hope I understood correctly, and I hope that helps.
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