How to assign values to a numpy array as a function of index?

Question

I can only find results that use tuples of coordinates to assign certain values, like this one.

I want to assign values to a 2-dimensional array as a function of their coordinates. The simplest case would be to set the value of each element to their second index (e.g. x-coordinate). Such that,

x[0][0] = 0
x[1][0] = 0
...
x[2][0] = 0
x[0][1] = 1
x[0][2] = 2
...

More complex case would be to set these values to the Euclidean distance to a certain point (x, y).

My current solution is to use a for-loop, which is definitely not efficient. A vectorized implementation would be nice.

My current implementation:

x_mask = np.zeros((256, 256))
for i in range(256):
    for j in range(256):
        x_mask[i][j] = j

Your simple example is trivial. For more complex cases, you should try a vectorised solution yourself and tell us where you got stuck. There's no "generalized way" to vectorise an operation. If there was, we wouldn't need programmers :). — jpp
– jpp, Commented Jul 27, 2018 at 9:34
@jpp was expecting something like a lambda expression for each position. I guess it works equally well once I have x/y coordinates as helper matrix to compute a lot of other stuff. — hkchengrex
– hkchengrex, Commented Jul 27, 2018 at 9:37

jpp · Accepted Answer · 2018-07-27 09:27:59Z

4

You can assign a range and take advantage of NumPy broadcasting:

A = np.zeros((256, 256))
A[:] = range(A.shape[1])
# or A[:] = np.arange(A.shape[1])

The method you choose will be dependent on the function you wish to apply.

answered Jul 27, 2018 at 9:27

jpp

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hkchengrex Over a year ago

Works perfectly. Getting Euclidean/Manhattan distance would be easy once I have the x/y coordinates too, thank you!

xdze2 · Accepted Answer · 2018-07-27 13:06:58Z

2

numpy.mgrid seems to be the right tool for this (especially if you want both x and y).

Example copied from the documentation:

>>> np.mgrid[0:5,0:5]
array([[[0, 0, 0, 0, 0],
        [1, 1, 1, 1, 1],
        [2, 2, 2, 2, 2],
        [3, 3, 3, 3, 3],
        [4, 4, 4, 4, 4]],
       [[0, 1, 2, 3, 4],
        [0, 1, 2, 3, 4],
        [0, 1, 2, 3, 4],
        [0, 1, 2, 3, 4],
        [0, 1, 2, 3, 4]]])

answered Jul 27, 2018 at 13:06

xdze2

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Comments

Nivedh · Accepted Answer · 2018-07-27 09:27:46Z

-1

Try this :

a=[]
for i in range(4):
    b=[]
    for j in range(4):
        b.append(j)
    a.append(b)

answered Jul 27, 2018 at 9:27

Nivedh

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1 Comment

hkchengrex Over a year ago

..I don't want to use for-loop as stated.

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