Matt McCredie wrote:
> That certainly is fast, unfortunately it doesn't pass all of the tests.
> I came up with those tests so I don't know how important they are to the
> original poster. I modified it and came up with a generator and a
> non-generator version based (roughly) on your algorithm, that are almost
> as quick, and pass all of the tests. Some of the modifications were done
> just to make it quicker, so it would be fair when comparing against the
> other methods. I hard-coded the comparison instead of using a function
> and created a function that directly generates and returns a list
> instead of a generator. I would probably use the generator version in my
> code, but wrapping `list' around a generator adds about 4us (on my
> machine). Anyway, getgroups7 passes all of the tests I mentioned and it
> was timed at 10.37usec/pass. The down side: the code doesn't seem nearly
> as elegant.
>
> Matt
In most cases you wouldn't wrap the generator version in a list(), but use
it directly as a loop iterator.
A little renaming of variables helps it be a bit more elegant I think ...
def getgroups8(seq):
groups = []
iseq = iter(xrange(len(seq)))
for start in iseq:
if seq[start] & 0x80:
for stop in iseq:
if seq[stop] & 0x80:
groups.append(seq[start:stop])
start = stop
groups.append(seq[start:])
return groups
This passes all the tests and runs about the same speed.
Cheers,
Ron
> <code>
> def gengroups7(seq):
> iseq = iter(xrange(len(seq)))
> start = 0
> for i in iseq:
> if seq[i]&0x80:
> start = i
> break
> else:
> return
> for i in iseq:
> if seq[i]&0x80:
> yield seq[start:i]
> start = i
> yield seq[start:]
>
>
> def getgroups7(seq):
> groups = []
> iseq = iter(xrange(len(seq)))
> start = 0
> for i in iseq:
> if seq[i]&0x80:
> start = i
> break
> else:
> return groups
> for i in iseq:
> if seq[i]&0x80:
> groups.append(seq[start:i])
> start = i
> groups.append(seq[start:])
> return groups
>
> </code>
>
>
>
>
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