To answer the part about the most efficient way to do that, In [1]: a = array([0,1,4,76,3,0,4,67,9,5,3,9,0,5,23,3,0,5,3,3,0,5,0]) In [8]: %timeit len(where(a!=0)[0]) 100000 loops, best of 3: 6.54 us per loop In [9]: %timeit (a!=0).sum() 100000 loops, best of 3: 9.81 us per loop Seems like the where option is faster. Now I create a large array In [13]: a = hstack([a,a,a,a,a,a,a,a,a,a,a,a]) In [14]: %timeit len(where(a!=0)[0]) 100000 loops, best of 3: 12.3 us per loop In [15]: %timeit (a!=0).sum() 100000 loops, best of 3: 11 us per loop Now the fastest way is using the sum. The where function is not vectorized because it doesn't know in advance the size of the final array. In the case of a big array, there will be a lot of copy in the memory, as it grows. And the difference increases fast... In [20]: a = hstack([a,a,a,a,a,a,a,a,a,a,a,a]) In [21]: %timeit len(where(a!=0)[0]) 10000 loops, best of 3: 79.1 us per loop In [22]: %timeit (a!=0).sum() 10000 loops, best of 3: 24.5 us per loop Regards, Jonathan On Wed, Dec 22, 2010 at 11:43 AM, Thomas K Gamble <tkg@lanl.gov> wrote:

On Wednesday, December 22, 2010 07:16:17 am Ian Stokes-Rees wrote:

...

What is the most efficient way to do the Matlab equivalent of nnz(M) (nnz = number-of-non-zeros function)?

I've tried Google, but no luck.

My assumption is that something like

a != 0

will be used, but I'm not sure then how to "count" the number of "True" entries.

TIA.

Ian

one possibility:

len(where(a != 0)[0])

-- Thomas K. Gamble Research Technologist, System/Network Administrator Chemical Diagnostics and Engineering (C-CDE) Los Alamos National Laboratory MS-E543,p:505-665-4323 f:505-665-4267

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