Porting IDL code, lindgen function to Python

Question

Afternoon everyone. I'm currently porting over an IDL code to python and it's been plain sailing up until this point so far. I'm stuck on this section of IDL code:

nsteps = 266    
ind2 = ((lindgen(nsteps+1,nsteps+1)) mod (nsteps+1))
dk2 = (k2arr((ind2+1) < nsteps) - k2arr(ind2-1) > 0)) / 2.

My version of this includes a rewritten lindgen function as follows:

def pylindgen(shape):
    nelem = numpy.prod(numpy.array(shape))
    out = numpy.arange(nelem,dtype=int)
    return numpy.reshape(out,shape)

... and the ported code where k2arr is an array of shape (267,):

ind2 = pylindgen((nsteps+1,nsteps+1)) % (nsteps+1)
dk2 = (k2arr[ (ind2+1) < nsteps ] - k2arr[ (ind2-1) > 0. ]) / 2.

Now, the problem is that my code makes ind2 an array where, by looking at the IDL code and the errors thrown in the python script, I'm sure it's meant to be a scalar. Am I missing some feature of these IDL functions?

Any thoughts would be greatly appreciated. Cheers.

By the definition in pylindgen() ind2 is a numpy array, so I don't think I need to define it as one again. — Carl M, Jul 23 '13 at 18:38
Please post the Python errors you are seeing. In the IDL code, `ind2` is an array, where the mod operation has been applied element-wise. The Python code you give for performing the same action is correct for applying `mod` to a NumPy `ndarray`. The error must be coming from a different section of your code. — ely, Jul 23 '13 at 18:44

score 0 · Answer 1 · answered Jul 23 '13 at 19:10

0

The code for pylindgen works perfectly for me. Produces an array of (267,267), though. IF k2array is a (267,) array, you should be getting an error like:

ValueError: boolean index array should have 1 dimension

Is that your problem? Cheers

answered Jul 23 '13 at 19:10

Jblasco

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From this, it seems you are trying to index a one-dimensional object `k2arr` using a two-dimensional (or more) index block, `ind2`. You have to make sure your index set is of the appropriate shape as the array you want to index upon. – ely Jul 23 '13 at 19:47
I see. With my limited knowledge of IDL, it looks like the IDL code is doing this! Obviously, this cannot be true. I will take a look into this line of investigation. Thanks @EMS. Much appreciated. – Carl M Jul 23 '13 at 19:53

Jblasco · Accepted Answer · 2013-07-23T22:52:48.290

My knowledge of IDL is not what it used to be, I had to research a little. The operator ">" in IDL is not an equivalent of python (or other languages). It stablishes a maximum, anything above it will be set to that value. Same goes for "<", obviously, it sets a minimum.

dk2 = (k2arr((ind2+1) < nsteps) - k2arr(ind2-1) > 0)) where k2arr is 266 and ind2 is (266,266) is equivalent to saying:

 - (ind2+1 < nsteps) take ind2+1 and, in any place that ind2+1 
   is greater than nsteps, replace by nsteps. 
 - (ind2-1 > 0) take ind2-1 and, in any place that ind2-1 is 
   less than zero, put zero instead.

The tricky part is now. k2arr (266,) is evaluated for each of the rows of (ind2+1) and (ind2-1), meaning that if (ind2+1 < nsteps) = [1,2,3,...,nsteps-1, nsteps, nsteps] the k2arr will be evaluated for exactly that 266 times, one on top of the other, with the result being (266,266) array.

And NOW I remember why I stopped programming in IDL!

Oh hell. The operator was the only thing I didn't look up as it was so ingrained in me that it means LT/GT. I will implement this tomorrow and let you know the result. Thanks so much! — Carl M, Jul 23 '13 at 22:17

Porting IDL code, lindgen function to Python

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