create an x-dim array using numpy.fromfunction from a function with more than x parameters

Question

I would like to create a 2D (x,y) array from a 5d function, say some kernel KMID:

import numpy as np
def KMID(x,y,mumid,delta,cmid):
    rsq=(x-float(len(x))/2.+0.5)**2+(y-float(len(y))/2.+0.5)**2
    return cmid*np.exp(-mumid*np.sqrt(rsq))/(rsq+delta**2)

by something like this:

shape=256,256
midscatterkernel=np.fromfunction(KMID(:,:,0.1,0.2,0.3),shape)

This gives:

SyntaxError: invalid syntax

i.e. I want to make a 2D array by iterating over just the first two indices. What is the most elegant way to do this?

unutbu · Accepted Answer · 2012-01-11 14:33:46Z

4

Don't use np.fromfunction for this, since KMID can accept numpy arrays as arguments:

import numpy as np

def KMID(x, y, mumid, delta, cmid):
    rsq = (x-len(x)/2.+0.5)**2+(y-len(y)/2.+0.5)**2
    return cmid*np.exp(-mumid*np.sqrt(rsq))/(rsq+delta**2)

lenx, leny = 256, 256
midscatterkernel = KMID(
    np.arange(lenx),
    np.arange(leny)[:, np.newaxis],
    0.1, 0.2, 0.3)

(np.fromfunction is syntactic sugar for a slow Python loop. If you can do the same thing with numpy array operations, use the numpy array operations. It will be faster.)

To answer you question, however, if you did need to use np.fromfunction, but wanted to supply some of the arguments as constants, then you could use functools.partial:

import functools
def KMID(x, y, mumid, delta, cmid):
    rsq = (x-len(x)/2.+0.5)**2+(y-len(y)/2.+0.5)**2
    return cmid*np.exp(-mumid*np.sqrt(rsq))/(rsq+delta**2)

shape = 256, 256
midscatterkernel = np.fromfunction(functools.partial(KMID,mumid=0.1,delta=0.2,cmid=0.3),shape)

edited Jan 11, 2012 at 14:33

answered Jan 11, 2012 at 14:18

unutbu

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1 Comment

Alex van Houten Over a year ago

Thanks for the tip on np.fromfunction and how to avoid using it, I didn't know it was slow.

Fred Foo · Accepted Answer · 2012-01-11 14:14:24Z

3

KMID is a function, not an array, so you can't index it with :. Do

midscatterkernel=np.fromfunction(lambda x, y: KMID(x,y,0.1,0.2,0.3), shape)

answered Jan 11, 2012 at 14:14

Fred Foo

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