[Numpy-discussion] Use-case for np.choose
josef.pktd@gmai...
josef.pktd@gmai...
Sun Nov 8 22:19:43 CST 2009
On Sun, Nov 8, 2009 at 10:57 PM, David Goldsmith
<d.l.goldsmith@gmail.com> wrote:
> On Sun, Nov 8, 2009 at 7:23 PM, <josef.pktd@gmail.com> wrote:
>>
>> On Sun, Nov 8, 2009 at 10:03 PM, David Goldsmith
>> <d.l.goldsmith@gmail.com> wrote:
>> > OK, now I'm trying to wrap my brain around broadcasting in choose when
>> > both
>> > `a` *and* `choices` need to be (non-trivially) broadcast in order to
>> > arrive
>> > at a common shape, e.g.:
>> >
>> >>>> c=np.arange(4).reshape((2,1,2)) # shape is (2,1,2)
>> >>>> a=np.eye(2, dtype=int) # shape is (2,2)
>> >>>> np.choose(a,c)
>> > array([[2, 1],
>> > [0, 3]])
>> >
>> > (Unfortunately, the implementation is in C, so I can't easily insert
>> > print
>> > statements to see intermediate results.)
>> >
>> > First, let me confirm that the above is indeed an example of what I
>> > think it
>> > is, i.e., both `a` and `choices` are broadcast in order for this to
>> > work,
>> > correct? (And if incorrect, how is one broadcast to the shape of the
>> > other?) Second, both are broadcast to shape (2,2,2), correct?
>>
>> Not,
>>
>> You only have one choice array, so first it is interpreted as a
>> sequence with c[0], c[1], then c[0] and c[0] are broadcasted to each
>> other (1,2), then they are broadcasted to a (2,2), then a picks from
>>
>> 0,0
>> 1,1
>>
>> or
>> 2,2
>> 3,3
>>
>> which is
>> 2,0
>> 1,3
>
> Except that wasn't the result; is yours a typo?
No typo, my mistake, I mixed up rows and columns.
I thought c[0] and c[1] are column vectors when I retyped the answer
from the interpreter.
Sorry,
Josef
>
>> qed
>>
>> Josef
>>
>> >>> np.choose(a,[c[0],c[1]])
>> array([[2, 1],
>> [0, 3]])
>> >>> c[0]
>> array([[0, 1]])
>> >>> c[1]
>> array([[2, 3]])
>> >>> c[1].shape
>> (1, 2)
>> >>> c[0].shape
>> (1, 2)
>> >>> a
>> array([[1, 0],
>> [0, 1]])
>>
>>
>> But how,
>> > precisely, i.e., does c become
>> >
>> > [[[0, 1], [2, 3]], [[[0, 1], [0, 1]],
>> > [[0, 1], [2, 3]]] or [[2, 3], [2, 3]]]
>> >
>> > and same question for a? Then, once a is broadcast to a (2,2,2) shape,
>> > precisely how does it "pick and choose" from c to create a (2,2) result?
>> > For example, suppose a is broadcast to:
>> >
>> > [[[1, 0], [0, 1]],
>> > [[1, 0], [0, 1]]]
>> >
>> > (as indicated above, I'm uncertain at this point if this is indeed what
>> > a is
>> > broadcast to); how does this create the (2,2) result obtained above?
>> > (Obviously this depends in part on precisely how c is broadcast, I do
>> > recognize that much.)
>> >
>> > Finally, a seemingly relevant comment in the C source is:
>> >
>> > /* Broadcast all arrays to each other, index array at the end.*/
>> >
>> > This would appear to confirm that "co-broadcasting" is performed if
>> > necessary, but what does the "index array at the end" phrase mean?
>> >
>> > Thanks for your continued patience and tutelage.
>> >
>> > DG
>> >
>> > On Sun, Nov 8, 2009 at 5:36 AM, <josef.pktd@gmail.com> wrote:
>> >>
>> >> On Sun, Nov 8, 2009 at 5:00 AM, David Goldsmith
>> >> <d.l.goldsmith@gmail.com>
>> >> wrote:
>> >> > On Sun, Nov 8, 2009 at 12:57 AM, Anne Archibald
>> >> > <peridot.faceted@gmail.com>
>> >> > wrote:
>> >> >>
>> >> >> 2009/11/8 David Goldsmith <d.l.goldsmith@gmail.com>:
>> >> >> > On Sat, Nov 7, 2009 at 11:59 PM, Anne Archibald
>> >> >> > <peridot.faceted@gmail.com>
>> >> >> > wrote:
>> >> >> >>
>> >> >> >> 2009/11/7 David Goldsmith <d.l.goldsmith@gmail.com>:
>> >> >> >> > So in essence, at least as it presently functions, the shape of
>> >> >> >> > 'a'
>> >> >> >> > *defines* what the individual choices are within 'choices`, and
>> >> >> >> > if
>> >> >> >> > 'choices'
>> >> >> >> > can't be parsed into an integer number of such individual
>> >> >> >> > choices,
>> >> >> >> > that's
>> >> >> >> > when an exception is raised?
>> >> >> >>
>> >> >> >> Um, I don't think so.
>> >> >> >>
>> >> >> >> Think of it this way: you provide np.choose with a selector
>> >> >> >> array,
>> >> >> >> a,
>> >> >> >> and a list (not array!) [c0, c1, ..., cM] of choices. You
>> >> >> >> construct
>> >> >> >> an
>> >> >> >> output array, say r, the same shape as a (no matter how many
>> >> >> >> dimensions it has).
>> >> >> >
>> >> >> > Except that I haven't yet seen a working example with 'a' greater
>> >> >> > than
>> >> >> > 1-D,
>> >> >> > Josef's last two examples notwithstanding; or is that what you're
>> >> >> > saying
>> >> >> > is
>> >> >> > the bug.
>> >> >>
>> >> >> There's nothing magic about A being one-dimensional.
>> >> >>
>> >> >> C = np.random.randn(2,3,5)
>> >> >> A = (C>-1).astype(int) + (C>0).astype(int) + (C>1).astype(int)
>> >> >>
>> >> >> R = np.choose(A, (-1, -C, C, 1))
>> >> >
>> >> > OK, now I get it: np.choose(A[0,:,:], (-1,-C,C,-1)) and
>> >> > np.choose(A[0,:,0].reshape((3,1)), (-1,-C,C,1)), e.g., also work, but
>> >> > np.choose(A[0,:,0], (-1,-C,C,-1)) doesn't - what's necessary for
>> >> > choose's
>> >> > arguments is that both can be broadcast to a common shape (as you
>> >> > state
>> >> > below), but choose won't reshape the arguments for you to make this
>> >> > possible, you have to do so yourself first, if necessary. That does
>> >> > appear
>> >> > to be what's happening now; but do we want choose to be smarter than
>> >> > that
>> >> > (e.g., for np.choose(A[0,:,0], (-1,-C,C,-1)) to work, so that the
>> >> > user
>> >> > doesn't need to include the .reshape((3,1)))?
>> >>
>> >> No, I don't think we want to be that smart.
>> >>
>> >> If standard broadcasting rules apply, as I think they do, then I
>> >> wouldn't
>> >> want
>> >> any special newaxis or reshapes done automatically. It will be
>> >> confusing,
>> >> the function wouldn't know what to do if there are, e.g., as many rows
>> >> as
>> >> columns, and this looks like a big source of errors.
>> >> Standard broadcasting is pretty nice (once I got the hang of it), and
>> >> adding
>> >> a lot of np.newaxis (or some reshapes) to the code is only a small
>> >> price
>> >> to pay.
>> >>
>> >> Josef
>> >>
>> >>
>> >>
>> >> >
>> >> > DG
>> >> >
>> >> >>
>> >> >> Requv = np.minimum(np.abs(C),1)
>> >> >>
>> >> >> or:
>> >> >>
>> >> >> def wedge(*functions):
>> >> >> """Return a function whose value is the minimum of those of
>> >> >> functions"""
>> >> >> def wedgef(X):
>> >> >> fXs = [f(X) for f in functions]
>> >> >> A = np.argmin(fXs, axis=0)
>> >> >> return np.choose(A,fXs)
>> >> >> return wedgef
>> >> >>
>> >> >> so e.g. np.abs is -wedge(lambda X: X, lambda X: -X)
>> >> >>
>> >> >> This works no matter what shape of X the user supplies - so a wedged
>> >> >> function can be somewhat ufunclike - by making A the same shape.
>> >> >>
>> >> >> >> The (i0, i1, ..., iN) element of the output array
>> >> >> >> is obtained by looking at the (i0, i1, ..., iN) element of a,
>> >> >> >> which
>> >> >> >> should be an integer no larger than M; say j. Then r[i0, i1, ...,
>> >> >> >> iN]
>> >> >> >> = cj[i0, i1, ..., iN]. That is, each element of the selector
>> >> >> >> array
>> >> >> >> determines which of the choice arrays to pull the corresponding
>> >> >> >> element from.
>> >> >> >
>> >> >> > That's pretty clear (thanks for doing my work for me). ;-), Yet,
>> >> >> > see
>> >> >> > above.
>> >> >> >
>> >> >> >> For example, suppose that you are processing an array C, and have
>> >> >> >> constructed a selector array A the same shape as C in which a
>> >> >> >> value
>> >> >> >> is
>> >> >> >> 0, 1, or 2 depending on whether the C value is too small, okay,
>> >> >> >> or
>> >> >> >> too
>> >> >> >> big respectively. Then you might do something like:
>> >> >> >>
>> >> >> >> C = np.choose(A, [-inf, C, inf])
>> >> >> >>
>> >> >> >> This is something you might want to do no matter what shape A and
>> >> >> >> C
>> >> >> >> have. It's important not to require that the choices be an array
>> >> >> >> of
>> >> >> >> choices, because they often have quite different shapes (here,
>> >> >> >> two
>> >> >> >> are
>> >> >> >> scalars) and it would be wasteful to broadcast them up to the
>> >> >> >> same
>> >> >> >> shape as C, just to stack them.
>> >> >> >
>> >> >> > OK, that's a pretty generic use-case, thanks; let me see if I
>> >> >> > understand
>> >> >> > it
>> >> >> > correctly: A is some how created independently with a 0 everywhere
>> >> >> > C
>> >> >> > is
>> >> >> > too
>> >> >> > small, a 1 everywhere C is OK, and a 2 everywhere C is too big;
>> >> >> > then
>> >> >> > np.choose(A, [-inf, C, inf]) creates an array that is -inf
>> >> >> > everywhere
>> >> >> > C
>> >> >> > is
>> >> >> > too small, inf everywhere C is too large, and C otherwise (and
>> >> >> > since
>> >> >> > -inf
>> >> >> > and inf are scalars, this implies broadcasting of these is taking
>> >> >> > place).
>> >> >> > This is what you're asserting *should* be the behavior. So,
>> >> >> > unless
>> >> >> > there is
>> >> >> > disagreement about this (you yourself said the opposite viewpoint
>> >> >> > might
>> >> >> > rationally be held) np.choose definitely presently has a bug,
>> >> >> > namely,
>> >> >> > the
>> >> >> > index array can't be of arbitrary shape.
>> >> >>
>> >> >> There seems to be some disagreement between versions, but both Josef
>> >> >> and I find that the index array *can* be arbitrary shape. In numpy
>> >> >> 1.2.1 I find that all the choose items must be the same shape as it,
>> >> >> which I think is a bug.
>> >> >>
>> >> >> What I suggested might be okay was if the index array was not
>> >> >> broadcasted, so that the outputs always had exactly the same shape
>> >> >> as
>> >> >> the index array. But upon reflection it's useful to be able to use a
>> >> >> 1-d array to select rows from a set of matrices, so I now think that
>> >> >> all of A and the elements of choose should be broadcast to the same
>> >> >> shape. This seems to be what Josef observes in his version of numpy,
>> >> >> so maybe there's nothing to do.
>> >> >>
>> >> >> Anne
>> >> >>
>> >> >> > DG
>> >> >> >
>> >> >> >>
>> >> >> >> Anne
>> >> >> >> _______________________________________________
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>> >> >> >
>> >> >> >
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