Reputation: 1782
I have a 2D numpy array that represents a monochrome image from a CCD that has been binned 3x3 (that is, each value in the array represents 9 pixels (3x3) on the physical CCD).
I want to rescale it to match the original CCD layout (so I can easily overlay it with a non-binned image from the same CCD).
I saw Resampling a numpy array representing an image, but that doesn't seem to do what I want.
Suppose I have an array g:
import numpy as np
import scipy.ndimage
g = np.array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
When I try to scale it by a factor of 2:
o = scipy.ndimage.zoom(g, 2, order=0)
I get exactly what I expect - each value is now 2x2 identical values:
array([[0, 0, 1, 1, 2, 2],
[0, 0, 1, 1, 2, 2],
[3, 3, 4, 4, 5, 5],
[3, 3, 4, 4, 5, 5],
[6, 6, 7, 7, 8, 8],
[6, 6, 7, 7, 8, 8]])
But when I try to scale by a factor of 3, I get this:
o = scipy.ndimage.zoom(g, 3, order=0)
Gives me:
array([[0, 0, 1, 1, 1, 1, 2, 2, 2],
[0, 0, 1, 1, 1, 1, 2, 2, 2],
[3, 3, 4, 4, 4, 4, 5, 5, 5],
[3, 3, 4, 4, 4, 4, 5, 5, 5],
[3, 3, 4, 4, 4, 4, 5, 5, 5],
[3, 3, 4, 4, 4, 4, 5, 5, 5],
[6, 6, 7, 7, 7, 7, 8, 8, 8],
[6, 6, 7, 7, 7, 7, 8, 8, 8],
[6, 6, 7, 7, 7, 7, 8, 8, 8]])
I wanted each value in the original array to become a set of 3x3 values...that's not what I get.
How can I do it? (And why do I get this unintuitive result?)
Upvotes: 9
Views: 12259
Reputation: 114976
You can use np.kron
:
In [16]: g
Out[16]:
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
In [17]: np.kron(g, np.ones((3,3), dtype=int))
Out[17]:
array([[0, 0, 0, 1, 1, 1, 2, 2, 2],
[0, 0, 0, 1, 1, 1, 2, 2, 2],
[0, 0, 0, 1, 1, 1, 2, 2, 2],
[3, 3, 3, 4, 4, 4, 5, 5, 5],
[3, 3, 3, 4, 4, 4, 5, 5, 5],
[3, 3, 3, 4, 4, 4, 5, 5, 5],
[6, 6, 6, 7, 7, 7, 8, 8, 8],
[6, 6, 6, 7, 7, 7, 8, 8, 8],
[6, 6, 6, 7, 7, 7, 8, 8, 8]])
The output of zoom(g, 3, order=0)
is a bit surprising. Consider the first row: [0, 0, 1, 1, 1, 1, 2, 2, 2]
. Why are there four 1
s?
When order=0
zoom (in effect) computes np.linspace(0, 2, 9)
, which looks like
In [80]: np.linspace(0, 2, 9)
Out[80]: array([ 0. , 0.25, 0.5 , 0.75, 1. , 1.25, 1.5 , 1.75, 2. ])
and then rounds the values. If you use np.round()
, you get:
In [71]: np.round(np.linspace(0, 2, 9)).astype(int)
Out[71]: array([0, 0, 0, 1, 1, 1, 2, 2, 2])
Note that np.round(0.5)
gives 0
, but np.round(1.5)
gives 2
. np.round()
uses the "round half to even" tie-breaking rule. Apparently the rounding done in the zoom
code uses the "round half down" rule: it rounds 0.5
to 0
and 1.5
to 1
, as in the following
In [81]: [int(round(x)) for x in np.linspace(0, 2, 9)]
Out[81]: [0, 0, 1, 1, 1, 1, 2, 2, 2]
and that's why there are four 1
s in there.
Upvotes: 11
Reputation: 366133
And why do I get this unintuitive result?
Because zoom
is a spline interpolation function. In other words, it draws a cubic spline from the midpoint of that 1 to the midpoint of that 0, and the values in between get the values of the spline at the appropriate location.
If you want nearest, linear or quadratic interpolation instead of cubic, you can use the order=0
or order=1
or order=2
argument. But if you don't want interpolation at all—which you don't—don't use an interpolation function. This is like asking why using [int(i*2.3) for i in range(10)]
to get even numbers from 0 to 20 gives you some odd numbers. It's not a function to get even numbers from 0 to 20, so it doesn't do that, but it does exactly what you asked it to.
How can I do it?
Again, if you want non-interpolated scaling, don't use an interpolation function. The simplest way is probably to use np.kron
, to Kroenecker-multiply your array with np.ones((scale, scale))
.
Upvotes: 2