On Sat, Apr 27, 2002 at 12:24:46AM -0400, Tim Peters wrote:
> [holger krekel]
> > measured with your driver script the following code
> > is about 10-20% faster than 'g2' and it gets close to
> >
> > > > Challenge 3: do it faster and with less code.
> >
> > def g3(filename, lineno):
> > if lineno>0:
> > f = file(filename)
> > while lineno>0:
> > read = f.read(1024)
> > count = read.count('\n')
> > lineno-=count or lineno
> > f.close()
> > if lineno<count:
> > return read.split('\n')[lineno+count-1]
> > return ''
> >
> > vertically it's one line less but it stretches
> > a bit horizontally. But it is more portable :-)
>
> What if (the simplest example of what can go wrong) the first line is more
> than 1024 characters long, and the caller asks for lineno 1?
right. i noticed after sending it that you also need to
change
lineno-= count or lineno
into
lineno-= count or read and lineno
and the problem of lines longer>1024 and around the 1024 boundary remains.
Solving this probably blows the code by several lines.
But anyway, isn't reading and counting still a faster technique than
using readlines() esp. for larger files?
holger
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