Replace various regex

[Martin]

On Feb 12, 8:30 pm, MRAB <pyt... at mrabarnett.plus.com> wrote:
> McColgst wrote:
> > On Feb 12, 2:39 pm, Martin <mdeka... at 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.

I guess I could read the file a line at a time and try splitting it,
though I though it would be better to read it all once then search for
the various regex I need to match and replace?

I am not sure that regex helps, as that would match and replace every
line which had a "!". Perhaps if i explain more thoroughly?

So the input file looks something like this...

9*0.0                        !  canopy(1:ntiles)
12.100                       !  cs
0.0                          !  gs
9*50.0                       !  rgrain(1:ntiles)
 0.749, 0.743, 0.754, 0.759  !  stheta(1:sm_levels)(top to bottom)
9*0.46                       !  snow_tile(1:ntiles)
0.46                         !  snow_grnd
276.78,277.46,278.99,282.48  !  t_soil(1:sm_levels)(top to bottom)
9*276.78                     !  tstar_tile(1:ntiles)
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)

So for each of the strings following the "!" I may potentially want to
match them and replace some of the numbers. That is I might search for
the expression snow_grnd with the intention of substituting 0.46 for
another number. What i came up with was a way to match all the numbers
and pass the replacement string.