Bernard wrote:
On 12 août, 12:43, Martin <mdeka...@gmail.com> wrote:
On Aug 12, 1:42 pm, Martin <mdeka...@gmail.com> wrote:
On Aug 12, 1:23 pm, Steven D'Aprano <st...@remove-this-
cybersource.com.au> wrote:
On Wed, 12 Aug 2009 05:12:22 -0700, Martin wrote:
I tried
re.findall((\w+COORDINATE).*\s+VALUE\s+=\s([\d\.\w-]+),s)
You need to put quotes around strings.
In this case, because you're using regular expressions, you should use a
raw string:
re.findall(r"(\w+COORDINATE).*\s+VALUE\s+=\s([\d\.\w-]+)",s)
will probably work.
--
Steven
Thanks I see.
so I tried it and if I use it as it is, it matches the first instance:
I
n [594]: re.findall(r"(\w+COORDINATE).*\s+VALUE\s+=\s([\d\.\w-]+)",s)
Out[594]: [('NORTHBOUNDINGCOORDINATE', '1')]
So I adjusted the first part of the regex, on the basis I could sub
NORTH for SOUTH etc.
In [595]: re.findall(r"(NORTHBOUNDINGCOORDINATE).*\s+VALUE\s+=\s([\d\.
\w-]+)",s)
Out[595]: [('NORTHBOUNDINGCOORDINATE', '1')]
But in both cases it doesn't return the decimal value rather the value
that comes after NUM_VAL = , rather than VALUE = ?
I think I kind of got that to work...but I am clearly not quite
understanding how it works as I tried to use it again to match
something else.
In this case I want to print the values 0.000000 and 2223901.039333
from a string like this...
YDim=1200\n\t\tUpperLeftPointMtrs=(0.000000,2223901.039333)\n\t\t
I tried which I though was matching the statement and printing the
decimal number after the equals sign??
re.findall(r"(\w+UpperLeftPointMtrs)*=\s([\d\.\w-]+)", s)
where s is the string
Many thanks for the help
You have to do it with 2 matches in the same regex:
regex = r"UpperLeftPointMtrs=\(([\d\.]+),([\d\.]+)"
The first match is before the , and the second one is after the , :)
You should probably learn how to play with regexes.
I personnaly use a visual tool called RX Toolkit[1] that comes with
Komodo IDE.
[1] http://docs.activestate.com/komodo/4.4/regex.html
Haven't tried it myself but how about this?
http://re-try.appspot.com/
--
Kindest regards.
Mark Lawrence.
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