numpy: concise conditional syntax for array division

Question

Assume a simple 1-dimensional numpy array:

>>> x = np.array([1,3,5,0,3,2])

Now assume I want to perform the operation 1.0/x. I can do this with numpy:

>>> 1.0/x
array([ 1.        ,  0.33333333,  0.2       ,         inf,  0.33333333,
    0.5       ])

The problem here is the infinity (inf) result for the original element value 0, because 1.0/0 seems to return infinity in place of undefined behaviour.

Instead of infinity, I would like to provide my own custom value where these divide by 0 scenarios arise. While I know this can be accomplished using a loop, I would like to know whether there is any kind of idiomatic syntax for this kind of operation.

There's a related question here, but it only deals with the if something: (do this) else: (do nothing) scenario whereas my question is a if something: (do this) else: (do that) scenario.

tacaswell · Accepted Answer · 2013-01-24 03:40:56Z

4

You can always patch it up later:

a = 1.0/x
inf_ind = np.isinf(a)
a[inf_ind] = your_value

or

a[inf_ind] = f(x[inf_ind])

Which has the advantage of not getting in the way of the nice optimized numpy methods.

answered Jan 24, 2013 at 3:40

tacaswell

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gbronner · Accepted Answer · 2013-01-24 03:57:57Z

3

Building on the previous answer, you can also set floating point callback modes to detect when you need to apply the inf transform.

I can't find a callback that gets called on each floating point error, however.

See also: http://docs.scipy.org/doc/numpy/reference/generated/numpy.seterr.html

import numpy
class IsDivideError(object):
    def __init__(self):
        self.hasDivideError=False

    def callback(self, x,y):
        self.hasDivideError=True

ide=IsDivideError()
numpy.seterr(divide='call')
numpy.seterrcall(lambda x,y: ide.callback(x,y) )
x = numpy.array([1,3,5,0,3,2])
val=1.0/x
if(ide.hasDivideError):
    val[numpy.isinf(val)]=5

answered Jan 24, 2013 at 3:57

gbronner

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