How to efficiently set values of numpy array using indices and boolean indexing?

Question

What is the most efficient way to use a mask to select elements of a multidimensional numpy array, when the mask is to be applied with an offset? For example:

import numpy as np

# in real application, following line would read an image
figure = np.random.uniform(size=(4, 4))  # used as a mask
canvas = np.zeros((10, 10))

# The following doesn't do anything, because a copy is modified
canvas[np.ix_(np.arange(4) + 3, range(4))][figure > 0.5] = 1.0

print np.mean(figure > 0.5)  # should be ~ 0.5
print canvas.max()  # prints 0.0

A similar question is posted here: Setting values of Numpy array when indexing an indexed array but I'm using a mask and I'm not asking why it doesn't work.

Would you always have the open mesh arrays as arrays with sequential numbers? — Divakar
– Divakar, Commented Jan 17, 2017 at 17:02

javidcf · Accepted Answer · 2017-01-17 16:36:25Z

2

The problem, it seems, is that using the arrays returned by np.ix_ as index means you are doing advanced indexing, and, as the documentation of NumPy states:

Advanced indexing always returns a copy of the data (contrast with basic slicing that returns a view).

But in this case, if the real application is similar to the code you have posted (that is, if you really just need an offset), you can get away with basic slicing:

import numpy as np

figure = np.random.uniform(size=(4, 4))
canvas = np.zeros((10, 10))

# Either of the following works fine
canvas[3:(3 + 4), :4][figure > 0.5] = 1.0
canvas[slice(3, 3 + 4), slice(4)][figure > 0.5] = 1.0

print np.mean(figure > 0.5)  # ~ 0.5
print canvas.max()  # Prints 1.0 now

answered Jan 17, 2017 at 16:36

javidcf

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Divakar · Accepted Answer · 2017-01-17 16:57:11Z

1

One approach would be to work with the linear indices. So, we would get the row and column indices from np.ix_, get the linear index equivalents from those. Then, use the mask to select the valid ones and finally assign new values to the data array with the valid linear indices.

Thus, the implementation would be -

# Get the open mesh arrays from np.ix_ corresponding to row, col indices
row, col = np.ix_(np.arange(4) + 3, range(4))

# Get the linear indices from those row and column index arrays 
linear_index = (row*canvas.shape[1] + col)[figure>0.5]

# Finally, assign values
np.put(canvas, linear_index, 1.0) # Or canvas.ravel()[linear_index] = 1.0

edited Jan 17, 2017 at 16:57

answered Jan 17, 2017 at 16:51

Divakar

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Comments

MSeifert · Accepted Answer · 2017-01-17 17:02:07Z

1

I generally use a helper function that creates an appropriatly shaped part (view) of the array:

arr = np.ones((10, 10)) * 10
mask = np.random.uniform(size=(4, 4))

def get_part(arr, shape, offs_x, offs_y):
    # This example just does 2D but can easily be expanded for ND-arrays
    return arr[offs_x : (offs_x + shape[0]), 
               offs_y : (offs_y + shape[1])]

get_part(arr, mask.shape, offs_x=3, offs_y=4)[mask > 0.5] = 1.0

A ND implementation would look like this:

def get_part(arr, shape, offsets):
    slices = tuple(slice(offs, offs+length) for offs, length in zip(offsets, shape))
    return arr[slices]

get_part(arr, mask.shape, (3, 4))

edited Jan 17, 2017 at 17:02

answered Jan 17, 2017 at 16:56

MSeifert

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Comments

hpaulj · Accepted Answer · 2017-01-17 17:35:19Z

mask=figure>0.5

If ix indices are really ranges, they can be replaced with slices as @jdehesa shows:

canvas[3:3+4,:4][mask]=1

If the use of arange is just an example convenience, we could use a two stage assignment

In [277]: idx=np.ix_(np.arange(4) + 3, range(4))
In [278]: canvas = np.zeros((10, 10))
In [279]: subcanvas=np.zeros_like(figure)
In [280]: subcanvas[mask] = 1
In [281]: subcanvas
Out[281]: 
array([[ 0.,  1.,  1.,  1.],
       [ 0.,  0.,  0.,  1.],
       [ 1.,  0.,  0.,  1.],
       [ 1.,  0.,  0.,  1.]])
In [282]: canvas[idx]=subcanvas
In [283]: canvas
Out[283]: 
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

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