Replacing minimum element in a numpy array subject to a condition

Question

I need to replace an element of a numpy array subject to the minimum of another numpy array verifying one condition. See the following minimal example:

arr = np.array([0, 1, 2, 3, 4])
label = np.array([0, 0, 1, 1, 2])
cond = (label == 1)
label[cond][np.argmin(arr[cond])] = 3

I would be expecting that label is now

 array([0,  0,  3,  1, 2])

instead I am getting

 array([0,  0,  1,  1, 2])

This is a consequence of the known fact that numpy arrays are not updated with double slicing.

Anyway, I can't figure out how to rewrite the above code in a simple way. Any hint?

Divakar · Accepted Answer · 2015-10-27 14:57:56Z

2

You are triggering NumPy's advanced indexing with that chaining of indexing, so the assigning doesn't go through. To solve this, one way would be to store the indices corresponding to the mask and then use indexing. Here's the implementation -

idx = np.where(cond)[0]
label[idx[arr[idx].argmin()]] = 3

Sample run -

In [51]: arr = np.array([5, 4, 5, 8, 9])
    ...: label = np.array([0, 0, 1, 1, 2])
    ...: cond = (label == 1)
    ...: 

In [52]: idx = np.where(cond)[0]
    ...: label[idx[arr[idx].argmin()]] = 3
    ...: 

In [53]: idx
Out[53]: array([2, 3])

In [54]: label
Out[54]: array([0, 0, 3, 1, 2])

edited Oct 27, 2015 at 14:57

answered Oct 27, 2015 at 14:50

Divakar

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