Maintain 2D structure after logical indexing a numpy array

Question

I am tracking a dynamically changing mask that rolls using some input shift. This mask stores values that determine where I can trust values in another array of the same shape. An example of how the mask changes over each iteration is below. I have a large stack of logical checks that determine how to set the rolled parts of the mask to zero based on whether the x and y values of the shift are equal to 0 or are positive or negative. Here I just hardcoded it all for clarity.

import numpy as np
mask = np.full((8,8), 10)

#Iteration 1
mask = np.roll(mask, (0, 1), axis = (0,1))
mask[:, :1] = 0
#logical indexing happens here
mask += 1
print (mask)

#Iteration 2
mask = np.roll(mask, (1, 0), axis = (0,1))
mask[:1, :] = 0
#logical indexing happens here
mask +=1
print (mask)

#Iteration 3
mask = np.roll(mask, (2, -1), axis = (0,1))
mask[:, -1:] = 0
mask[:2, :] = 0
#logical indexing happens here
mask +=1
print (mask)

After each iteration and before the mask is increased by one, I need to index into and pull the values of a second array where the mask is above some threshold (10 in this case). Since I am rolling and setting values, I always know that the part of the mask that fulfills this condition can be broadcast into a 2d array. A simplified example of what I am doing now is below where arr2 is a flattened array.

import numpy as np

arr1 = np.arange(0, 64, 1).reshape((8,8))
mask = np.full((8,8), 10)
mask[:, 0] = 0

arr2 = arr1[mask >= 10]

How can I keep arr2 as a 2d array where the mask is above the set threshold?

I do not know a priori what the shift will be that is applied to the mask so I have to rely on the values in the mask to determine the shape of the resulting array. My arrays are much larger than this example and the shifts are between -5 and 5 so I know I won't get close to setting the entire array below the threshold. The idea is that after ~10 iterations, some parts of the array become trustworthy again and can be useful information after the logical index.

applying a 2D mask on a 2D array will always broadcast to a flat array, unless you manually set it not to. This is (among other things) because the number of Trues (hence the number of resulting elements) in rows or column may be different. From your data, it would seem you can apply a 1D mask instead (if the mask is the same row-wise or column-wise). — Tarifazo
– Tarifazo, Commented Apr 15, 2019 at 16:49
Can you elaborate on how to "manually set it not to"? If you mean reshape the array after the fact, I can't a prioiri know what shape the part of the mask that is above the threshold besides to say it conforms to being a 2d array. — Will.Evo
– Will.Evo, Commented Apr 16, 2019 at 20:52
reshape is generally the simplest way. You can also use a np.ma.MaskedArray. However, at least in the code you posted, it doesn't make sense to use a 2D mask. — Tarifazo
– Tarifazo, Commented Apr 17, 2019 at 12:23

Will.Evo · Accepted Answer · 2019-06-04 19:29:45Z

1

The answer here is a work around and was obvious now that it has simmered in my mind for a while. Basically, since I know that the resulting area will be square, I can just count across a row and column where each index meets my condition. So continuing my example from before I just add in a couple lines to determine how many values in a row and column meet my condition.

import numpy as np
#Initializing array
arr1 = np.arange(0, 64, 1).reshape((8,8))
#mask array
mask = np.full((8,8), 10)

#Setting some rows and cols to zero to simulate my roll functionality
mask[:, 0] = 0
mask[:2, :] = 0

#Summing across a row and col where condition is met
sizex = np.sum(mask[4, :] >= 10)
sizey = np.sum(mask[:, 4] >= 10)

#Using the mask to index into the original array and reshaping
arr2 = arr1[mask >= 10].reshape((sizey, sizex))

answered Jun 4, 2019 at 19:29

Will.Evo

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