KK Sasa wrote:
> Hi there,
>
> The list comprehension is results = [d2(t[k]) for k in xrange(1000)],
> where d2 is a function returning a list, say [x1,x2,x3,x4] for one
> example. So "results" is a list consisting of 1000 lists, each of length
> four. Here, what I want to get is the sum of 1000 lists, and then the
> result is a list of length four. Is there any efficient way to do this?
> Because I found it is slow in my case. I tried sum(d2(t[k]) for k in
> xrange(1000)), but it returned error: TypeError: unsupported operand
> type(s) for +: 'int' and 'list'. Thanks.
That's because sum() defaults to adding with a default value of 0:
py> sum([[1, 2], [3, 4]], 0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'int' and 'list'
py> sum([[1, 2], [3, 4]], [])
[1, 2, 3, 4]
But don't do that! It will be slow.
I don't completely understand your requirements. You should show some
typical data, and the expected result. The business with the xrange and
t[k] and even d2() is probably irrelevant. The important part is summing
the lists. That could mean either of these two things:
results = [ [1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]]
sum(results)
--> gives [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
The way to do this efficiently is:
results = []
for sublist in [d2(t[k]) for k in xrange(1000)]:
results.extend(sublist)
Or perhaps you mean this:
results = [ [1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]]
sum(results)
--> gives [15, 18, 21, 24]
I expect that the fastest, most efficient way to do this will be with the
third-party library numpy. But in pure Python, you can do this:
def add(alist, blist):
if len(blist) != len(alist): raise ValueError
for i, x in blist:
alist[i] += x
results = [0]*4 # Like [0,0,0,0]
for sublist in [d2(t[k]) for k in xrange(1000)]:
add(results, sublist)
--
Steven
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