Incrementing with advanced indexing: why don't repeated indexes repeatedly increment?
Hello,
I've noticed that If you try to increment elements of an array with advanced indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only increment by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python loop through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd like to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look for the code responsible for this behavior?
Hi,
I get across the numpy.put[1] function. I'm not sure, but maybe it do what you want. My memory are fuzy about this and they don't tell about this in the doc of this function.
Fred
[1] http://docs.scipy.org/doc/numpy/reference/generated/numpy.put.html
On Wed, Jun 6, 2012 at 4:48 AM, John Salvatier jsalvati@u.washington.edu wrote:
Hello,
I've noticed that If you try to increment elements of an array with advanced indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only increment by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python loop through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd like to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look for the code responsible for this behavior?
NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Thank you for the suggestion, but it looks like that has the same behavior too:
In [43]: x = zeros(5)
In [44]: idx = array([1,1,1,3,4])
In [45]: put(x,idx, [2,4,8,10,30])
In [46]: x Out[46]: array([ 0., 8., 0., 10., 30.])
On Wed, Jun 6, 2012 at 6:07 AM, Frédéric Bastien nouiz@nouiz.org wrote:
Hi,
I get across the numpy.put[1] function. I'm not sure, but maybe it do what you want. My memory are fuzy about this and they don't tell about this in the doc of this function.
Fred
[1] http://docs.scipy.org/doc/numpy/reference/generated/numpy.put.html
On Wed, Jun 6, 2012 at 4:48 AM, John Salvatier jsalvati@u.washington.edu wrote:
Hello,
I've noticed that If you try to increment elements of an array with
advanced
indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only
increment
by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python
loop
through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd
like
to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look
for
the code responsible for this behavior?
NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
On Wed, Jun 6, 2012 at 9:48 AM, John Salvatier jsalvati@u.washington.edu wrote:
Hello,
I've noticed that If you try to increment elements of an array with advanced indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only increment by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python loop through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd like to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look for the code responsible for this behavior?
Strictly speaking, it isn't actually in the numpy codebase at all -- what's happening is that the Python interpreter sees this code:
x[idx] += vals
and then it translates it into this code before running it:
tmp = x.__getitem__(idx) tmp = tmp.__iadd__(vals) x.__setitem__(idx, tmp)
So you can find the implementations of the ndarray methods __getitem__, __iadd__, __setitem__ (they're called array_subscript_nice, array_inplace_add, and array_ass_sub in the C code), but there's no way to fix them so that this works the way you want it to, because there's no way for __iadd__ to know that the temporary values that it's working with are really duplicate copies of "the same" value in the original array.
It would be nice if numpy had some sort of standard API for doing what you want, but not sure what a good API would look like, and someone would have to implement it.
-n
On 06/06/2012 05:06 PM, Nathaniel Smith wrote:
On Wed, Jun 6, 2012 at 9:48 AM, John Salvatier jsalvati@u.washington.edu wrote:
Hello,
I've noticed that If you try to increment elements of an array with advanced indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only increment by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python loop through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd like to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look for the code responsible for this behavior?
Strictly speaking, it isn't actually in the numpy codebase at all -- what's happening is that the Python interpreter sees this code:
x[idx] += vals
and then it translates it into this code before running it:
tmp = x.__getitem__(idx) tmp = tmp.__iadd__(vals) x.__setitem__(idx, tmp)
So you can find the implementations of the ndarray methods __getitem__, __iadd__, __setitem__ (they're called array_subscript_nice, array_inplace_add, and array_ass_sub in the C code), but there's no way to fix them so that this works the way you want it to, because there's no way for __iadd__ to know that the temporary values that it's working with are really duplicate copies of "the same" value in the original array.
It would be nice if numpy had some sort of standard API for doing what you want, but not sure what a good API would look like, and someone would have to implement it.
This operation is also heavily used for the finite element assembling, and a similar question has been raised already several times (e.g. http://old.nabble.com/How-to-assemble-large-sparse-matrices-effectively-td33...). So why not adding a function np.assemble()?
r.
On Wed, Jun 6, 2012 at 4:30 PM, Robert Cimrman cimrman3@ntc.zcu.cz wrote:
On 06/06/2012 05:06 PM, Nathaniel Smith wrote:
On Wed, Jun 6, 2012 at 9:48 AM, John Salvatier jsalvati@u.washington.edu wrote:
Hello,
I've noticed that If you try to increment elements of an array with advanced indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only increment by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python loop through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd like to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look for the code responsible for this behavior?
Strictly speaking, it isn't actually in the numpy codebase at all -- what's happening is that the Python interpreter sees this code:
x[idx] += vals
and then it translates it into this code before running it:
tmp = x.__getitem__(idx) tmp = tmp.__iadd__(vals) x.__setitem__(idx, tmp)
So you can find the implementations of the ndarray methods __getitem__, __iadd__, __setitem__ (they're called array_subscript_nice, array_inplace_add, and array_ass_sub in the C code), but there's no way to fix them so that this works the way you want it to, because there's no way for __iadd__ to know that the temporary values that it's working with are really duplicate copies of "the same" value in the original array.
It would be nice if numpy had some sort of standard API for doing what you want, but not sure what a good API would look like, and someone would have to implement it.
This operation is also heavily used for the finite element assembling, and a similar question has been raised already several times (e.g. http://old.nabble.com/How-to-assemble-large-sparse-matrices-effectively-td33...). So why not adding a function np.assemble()?
I read that message, but I don't see what it has to do with this discussion? It seemed to be about fast ways to assign dense matrices into sparse matrices, not fast ways of applying in-place arithmetic to specific spots in a dense matrix.
-n
On 06/06/2012 05:34 PM, Nathaniel Smith wrote:
On Wed, Jun 6, 2012 at 4:30 PM, Robert Cimrmancimrman3@ntc.zcu.cz wrote:
On 06/06/2012 05:06 PM, Nathaniel Smith wrote:
On Wed, Jun 6, 2012 at 9:48 AM, John Salvatier jsalvati@u.washington.edu wrote:
Hello,
I've noticed that If you try to increment elements of an array with advanced indexing, repeated indexes don't get repeatedly incremented. For example:
In [30]: x = zeros(5)
In [31]: idx = array([1,1,1,3,4])
In [32]: x[idx] += [2,4,8,10,30]
In [33]: x Out[33]: array([ 0., 8., 0., 10., 30.])
I would intuitively expect the output to be array([0,14, 0,10,30]) since index 1 is incremented by 2+4+8=14, but instead it seems to only increment by 8. What is numpy actually doing here?
The authors of Theano noticed this behavior a while ago so they python loop through the values in idx (this kind of calculation is necessary for calculating gradients), but this is a bit slow for my purposes, so I'd like to figure out how to get the behavior I expected, but faster.
I'm also not sure how to navigate the numpy codebase, where would I look for the code responsible for this behavior?
Strictly speaking, it isn't actually in the numpy codebase at all -- what's happening is that the Python interpreter sees this code:
x[idx] += valsand then it translates it into this code before running it:
tmp = x.__getitem__(idx) tmp = tmp.__iadd__(vals) x.__setitem__(idx, tmp)So you can find the implementations of the ndarray methods __getitem__, __iadd__, __setitem__ (they're called array_subscript_nice, array_inplace_add, and array_ass_sub in the C code), but there's no way to fix them so that this works the way you want it to, because there's no way for __iadd__ to know that the temporary values that it's working with are really duplicate copies of "the same" value in the original array.
It would be nice if numpy had some sort of standard API for doing what you want, but not sure what a good API would look like, and someone would have to implement it.
This operation is also heavily used for the finite element assembling, and a similar question has been raised already several times (e.g. http://old.nabble.com/How-to-assemble-large-sparse-matrices-effectively-td33...). So why not adding a function np.assemble()?
I read that message, but I don't see what it has to do with this discussion? It seemed to be about fast ways to assign dense matrices into sparse matrices, not fast ways of applying in-place arithmetic to specific spots in a dense matrix.
Yes (in that thread), but it applies also adding/assembling vectors into a global vector - this is just x[idx] += vals. I linked that discussion as that was recent enough for me to recall it, but there were other.
Anyway, my point was that a having a function with the "adding" semantics in NumPy would be handy.
r.
On Wed, Jun 6, 2012 at 4:52 PM, Robert Cimrman cimrman3@ntc.zcu.cz wrote:
Yes (in that thread), but it applies also adding/assembling vectors into a global vector - this is just x[idx] += vals. I linked that discussion as that was recent enough for me to recall it, but there were other.
Anyway, my point was that a having a function with the "adding" semantics in NumPy would be handy.
x += numpy.bincount(idx, vals, minlength=len(x))
That does seem like it should work well if len(unique(idx)) is close to len(x). Thanks!
On Wed, Jun 6, 2012 at 9:35 AM, Robert Kern robert.kern@gmail.com wrote:
On Wed, Jun 6, 2012 at 4:52 PM, Robert Cimrman cimrman3@ntc.zcu.cz wrote:
Yes (in that thread), but it applies also adding/assembling vectors into
a
global vector - this is just x[idx] += vals. I linked that discussion as
that
was recent enough for me to recall it, but there were other.
Anyway, my point was that a having a function with the "adding"
semantics in
NumPy would be handy.
x += numpy.bincount(idx, vals, minlength=len(x))
-- Robert Kern _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
On 06/06/2012 06:35 PM, Robert Kern wrote:
On Wed, Jun 6, 2012 at 4:52 PM, Robert Cimrmancimrman3@ntc.zcu.cz wrote:
Yes (in that thread), but it applies also adding/assembling vectors into a global vector - this is just x[idx] += vals. I linked that discussion as that was recent enough for me to recall it, but there were other.
Anyway, my point was that a having a function with the "adding" semantics in NumPy would be handy.
x += numpy.bincount(idx, vals, minlength=len(x))
Nice! Looking at the C source, it seems it should be pretty efficient for this task.
r.
participants (5)
-
Frédéric Bastien -
John Salvatier -
Nathaniel Smith -
Robert Cimrman -
Robert Kern