"Jeff Epler" wrote:
> On Sun, Jun 12, 2005 at 04:55:38PM -0700, Xah Lee wrote:
> > if i have
> > mytext.replace(a,b)
> > how to find out many many occurances has been replaced?
>
> The count isn't returned by the replace method. You'll have to count
> and then replace.
>
> def count_replace(a, b, c):
> count = a.count(b)
> return count, s.replace(b, c)
>
> >>> count_replace("a car and a carriage", "car", "bat")
> (2, 'a bat and a batriage')
I thought naively that scanning a long string twice would be almost
twice as slow compared to when counting was done along with replacing.
Although it can done with a single scan, it is almost 9-10 times
slower, mainly because of the function call overhead; the code is also
longer:
import re
def count_replace_slow(aString, old, new):
count = [0]
def counter(match):
count[0] += 1
return new
replaced = re.sub(old,counter,aString)
return count[0], replaced
A good example of trying to be smart and failing :)
George
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