CODEWITHSUNDEEP
pythonarraysnumpy
theastronomist
theastronomist

Reputation: 1056

Setting Certain Values in Numpy Array to Zero

I am trying to set certain values of an array to 0. I have a larger array with many different values but want to set a random square/rectangle subset of values to zero.

a = np.array([[[1,1,1],[2,2,2],[3,3,3]],[[4,4,4],[5,5,5],[6,6,6]],[[7,7,7],[8,8,8],[9,9,9]]])
b = np.zeros((2,2,3)) 
#combination function

Expected Result:

combination = 
array([[[1, 1, 1],
        [2, 2, 2],
        [3, 3, 3]],

       [[4, 4, 4],
        [0, 0, 0],
        [0, 0, 0]],

       [[7, 7, 7],
        [0, 0, 0],
        [0, 0, 0]]])

I know this is wrong but, I tried just multiplying them like this but got an error:

masked = a*b
ValueError: operands could not be broadcast together with shapes (3,3,3) (2,2,3)

Upvotes: 0

Views: 1739

Answers (2)

theastronomist
theastronomist

Reputation: 1056

Thanks to @Naman and @bcr for hints. Slicing is the best method

>>> a = np.array([[[1,1,1],[2,2,2],[3,3,3]],[[4,4,4],[5,5,5],[6,6,6]],[[7,7,7],[8,8,8],[9,9,9]]])
>>> a[1:3,1:3,]=0 
>>> a
array([[[1, 1, 1],
        [2, 2, 2],
        [3, 3, 3]],

       [[4, 4, 4],
        [0, 0, 0],
        [0, 0, 0]],

       [[7, 7, 7],
        [0, 0, 0],
        [0, 0, 0]]])

Upvotes: 1

q9i
q9i

Reputation: 205

Numpy arrays use dimensions so two arrays can only be operated on if they have the same dimensions. You can find the dimension of an ndarray by using:

array.shape()

The dimension of an array is essentially its structure, for example:

>>> array = np.zeros((3,2))
>>> print(array)
>>> [[0, 0],
     [0, 0],
     [0, 0]]

As you can see in the previous example, the shape of the ndarray corresponds to its dimension--in this case it contains 3 subarrays with 2 elements each. Back to your question, in order for you to run an operation on 2 ndarrays, they need to have the same dimension, which in your example they do not. Try setting b to the same size as a with:

b = np.zeros((3, 3, 3))

Then your code should work as expected:

>>> a = np.array([[[1,1,1],[2,2,2],[3,3,3]],[[4,4,4],[5,5,5],[6,6,6]],[[7,7,7],[8,8,8],[9,9,9]]])
>>> b = np.zeros((3,3,3)) 
>>> print(a * b)
>>> [[[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]]]

Upvotes: 0

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