Oh thanks, I would never have imagined a one-line solution...
Here are the benchmarks:
In [2]: %timeit stefan(S)
10 loops, best of 3: 10.8 µs per loop
In [3]: %timeit gregor(S)
1 loops, best of 3: 48.1 µs per loop
In [4]: %timeit gregor2(S)
10 loops, best of 3: 3.23 µs per loop
Am 07.08.2014 um 13:59 schrieb Gregor Thalhammer :
>
> Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier :
>
>>
>> Hi,
>>
>> I've a small problem for which I cannot find a solution and I'm quite sure
>> there is an obvious one:
>>
>> I've an array Z (any dtype) with some data.
>> I've a (so
Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier :
>
> Hi,
>
> I've a small problem for which I cannot find a solution and I'm quite sure
> there is an obvious one:
>
> I've an array Z (any dtype) with some data.
> I've a (sorted) array I (of integer, same size as Z) that tells me the index
Nice ! Thanks Stéfan.
I will add it to the numpy 100 problems.
Nicolas
On 07 Aug 2014, at 13:31, Stéfan van der Walt wrote:
> Hi Nicolas
>
> On Thu, Aug 7, 2014 at 1:16 PM, Nicolas P. Rougier
> wrote:
>> Here is a small example:
>>
>> Z = [(0,0), (1,1), (2,2), (3,3), (4,4))
>> I = [0,
Hi Nicolas
On Thu, Aug 7, 2014 at 1:16 PM, Nicolas P. Rougier
wrote:
> Here is a small example:
>
> Z = [(0,0), (1,1), (2,2), (3,3), (4,4))
> I = [0, 20, 23, 24, 37]
>
> S = [ 20,20,0,24]
> -> Result should be [(1,1), (1,1), (0,0),(3,3)]
>
> S = [15,15]
> -> Wrong (15 not in I) but ideally, I wo
Hi,
I've a small problem for which I cannot find a solution and I'm quite sure
there is an obvious one:
I've an array Z (any dtype) with some data.
I've a (sorted) array I (of integer, same size as Z) that tells me the index
of Z[i] (if necessary, the index can be stored in Z).
Now, I have
