>
>
> Furthermore, I find an interesting phenomenon for NumPy:
> %timeit x**2
> %timeit x**4
> %timeit (x**2)**2
>
> 100000 loops, best of 3: 4.54 µs per loop
> 1000 loops, best of 3: 774 µs per loop
> 100000 loops, best of 3: 9.85 µs per loop
>
> That suggests NumPy only optimized **2, not ** operator itself !
>
>
I don't think x**4 == (x**2)**2 is necessarily true for floating point 
numbers which means that you can in general not do that optimization.
 

> On Monday, March 16, 2015 at 3:31:41 PM UTC+1, Sisyphuss wrote:
>>
>> That's interesting!
>>
>> On Sunday, March 15, 2015 at 4:18:34 PM UTC+1, Dallas Morisette wrote:
>>>
>>> Thanks everyone for the suggestions! Here is my updated test:
>>>
>>> using TimeIt
>>> function vec!(x,y)
>>>     y = x.*x
>>> end
>>>
>>> function comp!(x,y)
>>>     y = [xi*xi for xi in x]
>>> end
>>>
>>> function forloop!(x,y,n)
>>>     for i = 1:n 
>>>         y[i] = x[i]*x[i]
>>>     end
>>> end
>>>
>>> function forloop2!(x,y,n)
>>>     @simd for i = 1:n 
>>>         @inbounds y[i] = x[i]*x[i] 
>>>     end
>>> end
>>>     
>>> function test()
>>>     n = 10000
>>>     x = linspace(0.0,1.0,n)
>>>     y = zeros(x)
>>>     @timeit vec!(x,y)
>>>     @timeit comp!(x,y)
>>>     @timeit forloop!(x,y,n)
>>>     @timeit forloop2!(x,y,n)
>>> end
>>> test();
>>>
>>> 10000 loops, best of 3: 87.82 µs per loop
>>> 1000 loops, best of 3: 62.73 µs per loop
>>> 10000 loops, best of 3: 12.66 µs per loop
>>> 100000 loops, best of 3: 3.54 µs per loop
>>>
>>>
>>> So the SIMD macros combined with a literal for loop give performance 
>>> essentially equivalent to a call to numpy. I switched to @time so I could 
>>> see the allocations:
>>>
>>> elapsed time: 2.467e-5 seconds (80512 bytes allocated)
>>> elapsed time: 2.1358e-5 seconds (80048 bytes allocated)
>>> elapsed time: 1.5124e-5 seconds (0 bytes allocated)
>>> elapsed time: 6.108e-6 seconds (0 bytes allocated)
>>>
>>>
>>> Looks like one temporary array has to be allocated in both vectorized 
>>> and comprehension forms, which reduced the performance by about 5-7X. I 
>>> suppose this would depend on the exact calculation being done and the size 
>>> of the arrays involved and would have to be tested on a case-by-case basis. 
>>>
>>> Thanks for the help - I'm sure I'll be back with more questions!
>>>
>>> Dallas
>>>
>>

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