output of numpy.where(condition) is not an array, but a tuple of arrays: why?

Question

I am experimenting with the numpy.where(condition[, x, y]) function.
From the numpy documentation, I learn that if you give just one array as input, it shou

Answer 1

In Python (1) means just 1. () can be freely added to group numbers and expressions for human readability (e.g. (1+3)*3 v (1+3,)*3). Thus to denote a 1 element tuple it uses (1,) (and requires you to use it as well).

Thus

(array([4, 5, 6, 7, 8]),)

is a one element tuple, that element being an array.

If you applied where to a 2d array, the result would be a 2 element tuple.

The result of where is such that it can be plugged directly into an indexing slot, e.g.

a[where(a>0)]
a[a>0]

should return the same things

as would

I,J = where(a>0)   # a is 2d
a[I,J]
a[(I,J)]

Or with your example:

In [278]: a=np.array([1,2,3,4,5,6,7,8,9])
In [279]: np.where(a>4)
Out[279]: (array([4, 5, 6, 7, 8], dtype=int32),)  # tuple

In [280]: a[np.where(a>4)]
Out[280]: array([5, 6, 7, 8, 9])

In [281]: I=np.where(a>4)
In [282]: I
Out[282]: (array([4, 5, 6, 7, 8], dtype=int32),)
In [283]: a[I]
Out[283]: array([5, 6, 7, 8, 9])

In [286]: i, = np.where(a>4)   # note the , on LHS
In [287]: i
Out[287]: array([4, 5, 6, 7, 8], dtype=int32)  # not tuple
In [288]: a[i]
Out[288]: array([5, 6, 7, 8, 9])
In [289]: a[(i,)]
Out[289]: array([5, 6, 7, 8, 9])

======================

np.flatnonzero shows the correct way of returning just one array, regardless of the dimensions of the input array.

In [299]: np.flatnonzero(a>4)
Out[299]: array([4, 5, 6, 7, 8], dtype=int32)
In [300]: np.flatnonzero(a>4)+10
Out[300]: array([14, 15, 16, 17, 18], dtype=int32)

It's doc says:

This is equivalent to a.ravel().nonzero()[0]

In fact that is literally what the function does.

By flattening a removes the question of what to do with multiple dimensions. And then it takes the response out of the tuple, giving you a plain array. With flattening it doesn't have make a special case for 1d arrays.

===========================

@Divakar suggests np.argwhere:

In [303]: np.argwhere(a>4)
Out[303]: 
array([[4],
       [5],
       [6],
       [7],
       [8]], dtype=int32)

which does np.transpose(np.where(a>4))

Or if you don't like the column vector, you could transpose it again

In [307]: np.argwhere(a>4).T
Out[307]: array([[4, 5, 6, 7, 8]], dtype=int32)

except now it is a 1xn array.

We could just as well have wrapped where in array:

In [311]: np.array(np.where(a>4))
Out[311]: array([[4, 5, 6, 7, 8]], dtype=int32)

Lots of ways of taking an array out the where tuple ([0], i,=, transpose, array, etc).

Answer 2

Just use np.asarray function. In your case:

>>> import numpy as np
>>> array = np.array([1,2,3,4,5,6,7,8,9])
>>> pippo = np.asarray(np.where(array>4))
>>> pippo + 1
array([[5, 6, 7, 8, 9]])

Answer 3

Short answer: np.where is designed to have consistent output regardless of the dimension of the array.

A two-dimensional array has two indices, so the result of np.where is a length-2 tuple containing the relevant indices. This generalizes to a length-3 tuple for 3-dimensions, a length-4 tuple for 4 dimensions, or a length-N tuple for N dimensions. By this rule, it is clear that in 1 dimension, the result should be a length-1 tuple.