python - Why the performance difference between numpy.zeros and numpy.zeros_like?

Question

I finally found a performance bottleneck in my code but am confused as to what the reason is. To solve it I changed all my calls of numpy.zeros_like to instead use numpy.zeros. But why is zeros_like sooooo much slower?

For example (note e-05 on the zeros call):

>>> timeit.timeit('np.zeros((12488, 7588, 3), np.uint8)', 'import numpy as np', number = 10)
5.2928924560546875e-05
>>> timeit.timeit('np.zeros_like(x)', 'import numpy as np; x = np.zeros((12488, 7588, 3), np.uint8)', number = 10)
1.4402990341186523

But then strangely writing to an array created with zeros is noticeably slower than an array created with zeros_like:

>>> timeit.timeit('x[100:-100, 100:-100] = 1', 'import numpy as np; x = np.zeros((12488, 7588, 3), np.uint8)', number = 10)
0.4310588836669922
>>> timeit.timeit('x[100:-100, 100:-100] = 1', 'import numpy as np; x = np.zeros_like(np.zeros((12488, 7588, 3), np.uint8))', number = 10)
0.33325695991516113

My guess is zeros is using some CPU trick and not actually writing to the memory to allocate it. This is done on the fly when it's written to. But that still doesn't explain the massive discrepancy in array creation times.

I'm running Mac OS X Yosemite with the current numpy version:

>>> numpy.__version__
'1.9.1'

Answer 1

Modern OS allocate memory virtually, ie., memory is given to a process only when it is first used. zeros obtains memory from the operating system so that the OS zeroes it when it is first used. zeros_like on the other hand fills the alloced memory with zeros by itself. Both ways require about same amount of work --- it's just that with zeros_like the zeroing is done upfront, whereas zeros ends up doing it on the fly.

Technically, in C the difference is calling calloc vs. malloc+memset.