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 <mesanthu@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 <mesanthu@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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