Put zeros in the given index values of a 2D array in Python

Question

I have arrays

A = np.array([
  [1,6,5],
  [2,3,4],
  [8,3,0]
])

B = np.array([
  [3,2,1],
  [9,2,8],
  [2,1,8]
])

Doing a = np.argwhere(A > 4) gives me an array of position/indexes of values in array A that are greater than 4 i.e. [[0 1], [0 2], [2 0]].

I need to use these indexes/position from a = np.argwhere(A > 4) to replace the values in array B to zero at these specific position i.e. array B should now be

B = np.array([
  [3,0,0],
  [9,2,8],
  [0,1,8]
])

I am big time stuck any help with this will be really appreciated.

Thank You :)

I don't want to threshold B, I want to threshold A and positions/indexes that are has value less than the threshold has to be made zero in array B. — Laveena
– Laveena, Commented Jan 14, 2018 at 2:35
That's why you take the mask for A and index it with B. What's the confusion? It gives you what you want. — cs95
– cs95, Commented Jan 14, 2018 at 2:36
Apologies for the simple question .. I am pretty new to python. Thank you.. — Laveena
– Laveena, Commented Jan 14, 2018 at 2:40

John Zwinck · Accepted Answer · 2018-01-14 02:24:50Z

2

It should be as simple as:

B[A > 4] = 0

answered Jan 14, 2018 at 2:24

John Zwinck

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juanpa.arrivillaga · Accepted Answer · 2018-01-14 02:50:04Z

2

In general, though, note that the indices returned by np.where are meant to be applied to numpy.ndarray objects, so you could have done:

B[np.where(A > 4)] = 0

Generally I don't use np.where with a condition like this, I just use the boolean mask directly, as in John Zwinck's answer. But it is probably important to understand that you could

>>> B[np.where(A > 4)] = 0
>>> B
array([[3, 0, 0],
       [9, 2, 8],
       [0, 1, 8]])

answered Jan 14, 2018 at 2:50

juanpa.arrivillaga

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