Fernando Perez wrote:

Hi all,

A couple of times I've been confused by numpy.where(), and I think part of it comes from the docstring. Searching my gmail archive seems to indicate I'm not the only one bitten by this.

Compare:

In [14]: pdoc numpy.where Class Docstring: where(condition, | x, y)

The result is shaped like condition and has elements of x and y where
condition is respectively true or false.  If x or y are not given,
then it is equivalent to condition.nonzero().

To group the indices by element, rather than dimension, use

    transpose(where(condition, | x, y))

instead. This always results in a 2d array, with a row of indices for
each element that satisfies the condition.

with (b is just any array):

In [17]: pdoc b.nonzero Class Docstring: a.nonzero() returns a tuple of arrays

Returns a tuple of arrays, one for each dimension of a,
containing the indices of the non-zero elements in that
dimension.  The corresponding non-zero values can be obtained
with
    a[a.nonzero()].

To group the indices by element, rather than dimension, use
    transpose(a.nonzero())
instead. The result of this is always a 2d array, with a row for
each non-zero element.;

The sentence "The result is shaped like condition" in the where() docstring is misleading, since the behavior is really that of nonzero(). Where() *always* returns a tuple, not an array shaped like condition. If this were more clearly explained, along with a simple example for the usual case that seems to trip everyone:

In [21]: a=arange(10)

In [22]: N.where(a>5) Out[22]: (array([6, 7, 8, 9]),)

In [23]: N.where(a>5)[0] Out[23]: array([6, 7, 8, 9])

I think we'd get a lot less confusion.

Or am I missing something, or just being dense (quite likely)?

That sentence applies to the 3-argument form, which has nothing to do with nonzero() and does not yield a tuple. But in general, yes, the docstring leaves much to be desired.