Save time processing netcdf

[vorticitywolfe]

Hello,

I am working with a netcdf file and trying to remove a long string
from it e.g.
"KAST BLAH BLAH BLAH BLAH DATA BLAH DATA BLAH BLAH BLAH BLAH BLAH DATA
BLAH DATA BLAH"

Right now what I have works, but takes a long time to run... I think
this could be cut down to a quick 2 second run, but now takes around 5
minutes primarily because of the problem described below...

Why can't I use something like line 9 to get the entire string (which
returns [0], not
["KAST BLAH BLAH BLAH BLAH DATA BLAH DATA BLAH BLAH BLAH BLAH BLAH
DATA BLAH DATA BLAH"])? Rather I have to loop through it grabbing one
character at a time which drastically increases the processing time
and is very cumbersome.

Any one have any ideas or suggestions?


0 #OPEN NetCDF FILE TO READ
1 allDimNames=file.dimensions.keys()
2 max_string_length=file.dimensions['maxstringLen']
3 variableNames=file.variables.keys()
4 globalAttList=dir(file)
5 globalAttValue=getattr(file,'filePeriod')
6
7#GRABS DATA FROM STRING (2 DIMENSIONAL ARRAY) WITH THE CHARACTERS AS
INDIVIDUAL 8ELEMENTS
9 #YOU CAN'T JUST SPECIFY DATA VALUES DATA[0][0:max_string_length]
10 data=file.variables['rawstring'].getValue()
11 num_stations=data.shape
12 station_id=[str(mz)]
13
14 for m in station_id:
15 station_id=m
16 prec=[]
17 temps=[]
18
19 #Cycles through all of the stations in the file (LARGE ARRAY!!)
20 for k in range(num_stations[0]):
21 #Finds a certain station
22 if data[k][6]=='K' and data[k][7]==station_id[0:1] and data[k]
[8]==station_id[1:2] and 23data[k][9]==station_id[2:3] and data[k]
[15]=='5' or data[k][6]=='K' and data[k][7]==station_id[0:1] and
24data[k][8]==station_id[1:2] and data[k][9]==station_id[2:3] and
data[k][15]=='4':
25
26 #SEPARATES STRING CHARACTER BY CHARACTER ONLY WAY I'VE 27BEEN
ABLE TO FIGURE OUT HOW TO READ ENTIRE STRING INTO A STRING
28 for j in range(max_string_length):
29 prec.append(str(data[k][j]))
30 S= ''.join(prec)
31 # # # ##THEN RIP OFF THE WHITESPACE AT THE RIGHT OF THE
STRING
32 code=S.rstrip("\0")
33

Thanks for any of your help!