[Numpy-discussion] Smart way to do this?

Brett Olsen brett.olsen@gmail....
Sat Feb 23 00:45:55 CST 2013


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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