Matlab to Python : returning an array instead of integer

Question

I have a statement in my Matlab code:

a = find(abs(ASE_lamda-YDFA_lam_s)<1e-15);

After to execution I get the output as:

octave:50> whos a
Variables in the current scope:

   Attr Name        Size                     Bytes  Class
   ==== ====        ====                     =====  ===== 
        a           1x1                          8  double

Total is 1 element using 8 bytes

octave:51> a
a =  33

I have migrated the code to Python using NumPy package in below method:

a = np.nonzero(np.abs(ASE_lamda-YDFA_lam_s)<1e-15)

But the type of variable a is tuple

(array([32]),)

What is the correct way to migrate the above code?

Hannes Ovrén · Accepted Answer · 2015-01-08 15:44:32Z

2

They are equivalent. As you can see your Matlab/Octave code returned the 1x1 dimensional matrix a = 33 while NumPy gave you the 1-dimensional vector a = [32].

The reason it is a tuple is because nonzero gives you a tuple of indices for each dimension. If you are dealing with 1-dimensional data (which is probably the case here) you could look into using numpy.flatnonzero instead.

I think the confusion is that Octave (and probably Matlab as well) treats the 1x1 matrix as a scalar, while NumPy does not.

So, to get the single entry you just take the first (and only) element by normal indexing:

a = a[0]

Something that will probably bite you when moving between Matlab and NumPy is that the former is a lot more lenient on indexing and matrix shapes. In NumPy there is a difference between an array of shape (N,) and (N, 1) which you will probably run into when e.g. multiplying matrices.

edited Jan 8, 2015 at 15:44

answered Jan 8, 2015 at 15:39

Hannes Ovrén

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3 Comments

hpaulj Over a year ago

It's not so much that MATLAB is more lenient, but that a 2d matrix is its base type. At one time (v 3.0) that's all there was.

Hannes Ovrén Over a year ago

@hpaulj, Yeah, sure. I was mostly aiming at the fact that NumPy sometimes gives you weird results if you are not careful. There is a difference between multiplying a (2,) array and a (2,) array and a (2,1) and (1,2) array. The former gives the expected (2,) array as result, while the latter gives you a (2,2) array, which is probably not what you wanted most times.

hpaulj Over a year ago

Though I've often done something like x=[1,2]; y=x.'*x. And now that Octave does numpy-like broadcasting x.' .* x produces the same thing.

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