numpy array where vectorized version [duplicate]

Question

This question already has an answer here:

Apply function to masked numpy array (1 answer)

Closed 8 years ago.

I am reading about vectorized expressions in cheat sheet. It is mentioned as below

Vectorized expressions

np.where(cond, x, y) is a vectorized version of the expression ‘x if condition else y’

example:

np.where([True, False], [1, 2], [2, 3]) => ndarray (1, 3)

I am not able understand above example. My understanding is that we should have expression but here we have list of [True, False].

Request to explain in break up and how we got output of ndarray(1,3)

Thanks

• Divakar: Did you read the official docs : docs.scipy.org/doc/numpy/reference/generated/numpy.where.html? Which specific part are you stuck with on its explanation?
• venkysmarty: my question is how do we go output ndarray(1,3) in above example
• Divakar: Possible duplicate of Apply function to masked numpy array. One more : stackoverflow.com/questions/35188508

Answer 1

I normally use np.where to convert a boolean array to an index array. Consider this example:

In [12]: a = np.random.rand((10))
In [13]: a
Out[13]: 
array([ 0.80785098, 0.49922039, 0.02018482, 0.69514821, 0.87127179,
 0.23235574, 0.73199572, 0.79935066, 0.46667908, 0.11330817])
In [14]: bool_array = a > 0.5
In [15]: bool_array
Out[15]: array([ True, False, False, True, True, False, True, True, False, False], dtype=bool)
In [16]: np.where(bool_array)
Out[16]: (array([0, 3, 4, 6, 7]),)

---

Explanation of your example. For every value in cond: if True, pick value from x, otherwise pick value from y.

cond: [True, False]
x : [1, 2]
y : [2, 3]
Result:
cond[0] == True -> out[0] == x[0]
cond[1] == False -> out[1] == y[1]
out == [1, 3]

Answer 2

The array [True, False] is what would be produced by an expression like x<y where x=np.array([1,1]) and y=np.array([2,0]). So cond is a boolean array that is often the result of an expression like the previous one. An example of a more real-world-ish usage would therfore be :

np.where(x<y, [1, 2], [2, 3])