a = np.ones(30)
idx = np.array([2, 3, 2])
a += 2 * np.bincount(idx, minlength=len(a))
>>> a
array([ 1., 1., 5., 3., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
1., 1., 1., 1.])
As for speed:
def loop(a, idx):
for i in idx:
a[i] += 2
def count(a, idx):
a += 2 * np.bincount(idx, minlength=len(a))
%timeit loop(np.ones(30), np.array([2, 3, 2]))
10000 loops, best of 3: 19.9 us per loop
%timeit count(np.ones(30), np.array(2, 3, 2]))
100000 loops, best of 3: 19.2 us per loop
So no big difference here. But go to larger systems and you'll see a huge
difference:
%timeit loop(np.ones(10000), np.random.randint(10000, size=100000))
1 loops, best of 3: 260 ms per loop
%timeit count(np.ones(10000), np.random.randint(10000, size=100000))
100 loops, best of 3: 3.03 ms per loop.
~Brett
On Fri, Feb 22, 2013 at 8:38 PM, santhu kumar <mesan...@gmail.com> wrote:
> Sorry typo :
>
> a = np.ones(30)
> idx = np.array([2,3,2]) # there is a duplicate index of 2
> a[idx] += 2
>
> On Fri, Feb 22, 2013 at 8:35 PM, santhu kumar <mesan...@gmail.com> wrote:
>
>> Hi all,
>>
>> I dont want to run a loop for this but it should be possible using numpy
>> "smart" ways.
>>
>> a = np.ones(30)
>> idx = np.array([2,3,2]) # there is a duplicate index of 2
>> a += 2
>>
>> >>>a
>> array([ 1., 1., 3., 3., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
>> 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
>> 1., 1., 1., 1.])
>>
>>
>> But if we do this :
>> for i in range(idx.shape[0]):
>> a[idx[i]] += 2
>>
>> >>> a
>> array([ 1., 1., 5., 3., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
>> 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
>> 1., 1., 1., 1.])
>>
>> How to achieve the second result without looping??
>> Thanks
>> Santhosh
>>
>
>
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>
>
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