On Sun, 16 Sep 2007 09:56:04 +0000, cesco wrote:
> Hi,
>
> I have the following list:
> l = [1, 2, 3, 4]
> and I'd like to obtain a list like the following:
> l_partial_sum = [1, 3, 6, 10]
> (that is [1, 1+2, 1+2+3, 1+2+3+4])
>
> Is there a simple way to accomplish this?
Yes, use a loop. Put it into a function:
def partial_sums(L):
if not L:
return []
ps = [None]*len(L)
ps[0] = L[0]
for i in xrange(1, len(L)):
ps[i] = ps[i-1] + L[i]
return ps
>>> l = [1, 2, 3, 4]
>>> partial_sums(l)
[1, 3, 6, 10]
Here's a one-liner that give the same result:
[sum(l[0:i]) for i in xrange(len(l)+1)][1:]
Which is better, the one-liner using a super-fast built-in function
written in C, or the multi-line function using just slow Python code?
(Hint: I wouldn't be asking the question if the "obvious" answer was
right.)
Measure, don't guess!
>>> import timeit
>>>
>>> timeit.Timer("partial_sums(l)",
... "from __main__ import partial_sums; l = range(1000)").timeit(1000)
1.3528358936309814
>>> timeit.Timer("[sum(l[0:i]) for i in xrange(len(l)+1)][1:]",
... "l = range(1000)").timeit(1000)
41.276183843612671
The one-liner is thirty times slower than the function. Can you explain
why?
(Hint: how many times does it add numbers? How many times does it copy
parts of the list?)
--
Steven.
--
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