Numpy element-wise in operation

Question

Suppose I have a column vector y with length n, and I have a matrix X of size n*m. I want to check for each element i in y, whether the element is in the corresponding row in X. What is the most efficient way of doing this?

For example:

y = [1,2,3,4].T and

X =[[1, 2, 3],[3, 4, 5],[4, 3, 2],[2, 2, 2]]

Then the output should be

[1, 0, 1, 0] or [True, False, True, False]

which ever is easier.

Of course we can use a for loop to iterate through both y and X, but is there any more efficient way of doing this?

Are you sure you don't want the output to be np.array([True, False, True, False])? — Eric
– Eric, Commented Oct 21, 2016 at 11:30

Divakar · Accepted Answer · 2016-10-21 11:21:17Z

4

Vectorized approach using broadcasting -

((X == y[:,None]).any(1)).astype(int)

Sample run -

In [41]: X        # Input 1
Out[41]: 
array([[1, 2, 3],
       [3, 4, 5],
       [4, 3, 2],
       [2, 2, 2]])

In [42]: y        # Input 2
Out[42]: array([1, 2, 3, 4])

In [43]: X == y[:,None] # Broadcasted  comparison
Out[43]: 
array([[ True, False, False],
       [False, False, False],
       [False,  True, False],
       [False, False, False]], dtype=bool)

In [44]: (X == y[:,None]).any(1) # Check for any match along each row
Out[44]: array([ True, False,  True, False], dtype=bool)

In [45]: ((X == y[:,None]).any(1)).astype(int) # Convert to 1s and 0s
Out[45]: array([1, 0, 1, 0])

answered Oct 21, 2016 at 11:21

Divakar

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1 Comment

Eric Over a year ago

I'd argue that any(1) should be spelt .any(axis=1) for clarity

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