Discussion:
[Numpy-discussion] Why do mgrid and meshgrid not return broadcast arrays?
Juan Nunez-Iglesias
2017-03-09 02:48:45 UTC
Permalink
I was a bit surprised to discover that both meshgrid nor mgrid return fully instantiated arrays, when simple broadcasting (ie with stride=0 for other axes) is functionally identical and happens much, much faster.

I wrote my own function to do this:


def broadcast_mgrid(arrays):
    shape = tuple(map(len, arrays))
    ndim = len(shape)
    result = []
    for i, arr in enumerate(arrays, start=1):
        reshaped = np.broadcast_to(arr[[...] + [np.newaxis] * (ndim - i)],
                                   shape)
        result.append(reshaped)
    return result


For even a modest-sized 512 x 512 grid, this version is close to 100x faster:


In [154]: %timeit th.broadcast_mgrid((np.arange(512), np.arange(512)))
10000 loops, best of 3: 25.9 µs per loop

In [156]: %timeit np.meshgrid(np.arange(512), np.arange(512))
100 loops, best of 3: 2.02 ms per loop

In [157]: %timeit np.mgrid[:512, :512]
100 loops, best of 3: 4.84 ms per loop


Is there a conscious design decision as to why this isn’t what meshgrid/mgrid do already? Or would a PR be welcome to do this?

Thanks,

Juan.
Warren Weckesser
2017-03-09 03:05:21 UTC
Permalink
Post by Juan Nunez-Iglesias
I was a bit surprised to discover that both meshgrid nor mgrid return
fully instantiated arrays, when simple broadcasting (ie with stride=0 for
other axes) is functionally identical and happens much, much faster.
Take a look at ogrid:
https://docs.scipy.org/doc/numpy/reference/generated/numpy.ogrid.html

Warren
Post by Juan Nunez-Iglesias
*def broadcast_mgrid(arrays):*
* shape = tuple(map(len, arrays))*
* ndim = len(shape)*
* result = []*
* for i, arr in enumerate(arrays, start=1):*
* reshaped = np.broadcast_to(arr[[...] + [np.newaxis] * (ndim -
i)],*
* shape)*
* result.append(reshaped)*
* return result*
*In [154]: %timeit th.broadcast_mgrid((np.arange(512), np.arange(512)))*
*10000 loops, best of 3: 25.9 µs per loop*
*In [156]: %timeit np.meshgrid(np.arange(512), np.arange(512))*
*100 loops, best of 3: 2.02 ms per loop*
*In [157]: %timeit np.mgrid[:512, :512]*
*100 loops, best of 3: 4.84 ms per loop*
Is there a conscious design decision as to why this isn’t what
meshgrid/mgrid do already? Or would a PR be welcome to do this?
Thanks,
Juan.
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Juan Nunez-Iglesias
2017-03-09 03:19:37 UTC
Permalink
Hi Warren,

ogrid doesn’t solve my problem. Note that my code returns arrays that would evaluate as equal to the mgrid output. It’s just that they are copied in mgrid into a giant array, instead of broadcast:


In [176]: a0, b0 = np.mgrid[:5, :5]

In [177]: a1, b1 = th.broadcast_mgrid((np.arange(5), np.arange(5)))

In [178]: a0
Out[178]:
array([[0, 0, 0, 0, 0],
       [1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2],
       [3, 3, 3, 3, 3],
       [4, 4, 4, 4, 4]])

In [179]: a1
Out[179]:
array([[0, 0, 0, 0, 0],
       [1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2],
       [3, 3, 3, 3, 3],
       [4, 4, 4, 4, 4]])

In [180]: a0.strides
Out[180]: (40, 8)

In [181]: a1.strides
Out[181]: (8, 0)
Post by Juan Nunez-Iglesias
Post by Juan Nunez-Iglesias
I was a bit surprised to discover that both meshgrid nor mgrid return fully instantiated arrays, when simple broadcasting (ie with stride=0 for other axes) is functionally identical and happens much, much faster.
Take a look at ogrid: https://docs.scipy.org/doc/numpy/reference/generated/numpy.ogrid.html
Warren
Post by Juan Nunez-Iglesias
    shape = tuple(map(len, arrays))
    ndim = len(shape)
    result = []
        reshaped = np.broadcast_to(arr[[...] + [np.newaxis] * (ndim - i)],
                                   shape)
        result.append(reshaped)
    return result
In [154]: %timeit th.broadcast_mgrid((np.arange(512), np.arange(512)))
10000 loops, best of 3: 25.9 µs per loop
In [156]: %timeit np.meshgrid(np.arange(512), np.arange(512))
100 loops, best of 3: 2.02 ms per loop
In [157]: %timeit np.mgrid[:512, :512]
100 loops, best of 3: 4.84 ms per loop
Is there a conscious design decision as to why this isn’t what meshgrid/mgrid do already? Or would a PR be welcome to do this?
Thanks,
Juan.
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P***@ffi.no
2017-03-09 07:28:56 UTC
Permalink
Hi, Juan.

Meshgrid can actually give what you want, but you must use the options: copy=False and indexing=’ij’.

In [7]: %timeit np.meshgrid(np.arange(512), np.arange(512))
1000 loops, best of 3: 1.24 ms per loop

In [8]: %timeit np.meshgrid(np.arange(512), np.arange(512), copy=False)
10000 loops, best of 3: 27 µs per loop

In [9]: %timeit np.meshgrid(np.arange(512), np.arange(512), copy=False, indexing='ij')
10000 loops, best of 3: 23 µs per loop

Best regards
Per A. Brodtkorb

From: NumPy-Discussion [mailto:numpy-discussion-***@scipy.org] On Behalf Of Juan Nunez-Iglesias
Sent: 9. mars 2017 04:20
To: Discussion of Numerical Python
Subject: Re: [Numpy-discussion] Why do mgrid and meshgrid not return broadcast arrays?

Hi Warren,

ogrid doesn’t solve my problem. Note that my code returns arrays that would evaluate as equal to the mgrid output. It’s just that they are copied in mgrid into a giant array, instead of broadcast:


In [176]: a0, b0 = np.mgrid[:5, :5]

In [177]: a1, b1 = th.broadcast_mgrid((np.arange(5), np.arange(5)))

In [178]: a0
Out[178]:
array([[0, 0, 0, 0, 0],
[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4]])

In [179]: a1
Out[179]:
array([[0, 0, 0, 0, 0],
[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4]])

In [180]: a0.strides
Out[180]: (40, 8)

In [181]: a1.strides
Out[181]: (8, 0)



On 9 Mar 2017, 2:05 PM +1100, Warren Weckesser <***@gmail.com<mailto:***@gmail.com>>, wrote:



On Wed, Mar 8, 2017 at 9:48 PM, Juan Nunez-Iglesias <***@gmail.com<mailto:***@gmail.com>> wrote:
I was a bit surprised to discover that both meshgrid nor mgrid return fully instantiated arrays, when simple broadcasting (ie with stride=0 for other axes) is functionally identical and happens much, much faster.


Take a look at ogrid: https://docs.scipy.org/doc/numpy/reference/generated/numpy.ogrid.html
Warren

I wrote my own function to do this:


def broadcast_mgrid(arrays):
shape = tuple(map(len, arrays))
ndim = len(shape)
result = []
for i, arr in enumerate(arrays, start=1):
reshaped = np.broadcast_to(arr[[...] + [np.newaxis] * (ndim - i)],
shape)
result.append(reshaped)
return result


For even a modest-sized 512 x 512 grid, this version is close to 100x faster:


In [154]: %timeit th.broadcast_mgrid((np.arange(512), np.arange(512)))
10000 loops, best of 3: 25.9 µs per loop

In [156]: %timeit np.meshgrid(np.arange(512), np.arange(512))
100 loops, best of 3: 2.02 ms per loop

In [157]: %timeit np.mgrid[:512, :512]
100 loops, best of 3: 4.84 ms per loop


Is there a conscious design decision as to why this isn’t what meshgrid/mgrid do already? Or would a PR be welcome to do this?

Thanks,

Juan.

_______________________________________________
NumPy-Discussion mailing list
NumPy-***@scipy.org<mailto:NumPy-***@scipy.org>
https://mail.scipy.org/mailman/listinfo/numpy-discussion

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NumPy-***@scipy.org<mailto:NumPy-***@scipy.org>
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Juan Nunez-Iglesias
2017-03-09 07:34:04 UTC
Permalink
Ah, fantastic, thanks Per!

I'd still be interested to hear from the core devs as to why this isn't the default, both with meshgrid and mgrid...

Juan.
Post by P***@ffi.no
Hi, Juan.
Meshgrid can actually give what you want, but you must use the options: copy=False  and indexing=’ij’.
In [7]: %timeit np.meshgrid(np.arange(512), np.arange(512))
1000 loops, best of 3: 1.24 ms per loop
In [8]: %timeit np.meshgrid(np.arange(512), np.arange(512), copy=False)
10000 loops, best of 3: 27 µs per loop
In [9]: %timeit np.meshgrid(np.arange(512), np.arange(512), copy=False, indexing='ij')
10000 loops, best of 3: 23 µs per loop
Best regards
Per A. Brodtkorb
Sent: 9. mars 2017 04:20
To: Discussion of Numerical Python
Subject: Re: [Numpy-discussion] Why do mgrid and meshgrid not return broadcast arrays?
Hi Warren,
In [176]: a0, b0 = np.mgrid[:5, :5]
In [177]: a1, b1 = th.broadcast_mgrid((np.arange(5), np.arange(5)))
In [178]: a0
array([[0, 0, 0, 0, 0],
       [1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2],
       [3, 3, 3, 3, 3],
       [4, 4, 4, 4, 4]])
In [179]: a1
array([[0, 0, 0, 0, 0],
       [1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2],
       [3, 3, 3, 3, 3],
       [4, 4, 4, 4, 4]])
In [180]: a0.strides
Out[180]: (40, 8)
In [181]: a1.strides
Out[181]: (8, 0)
I was a bit surprised to discover that both meshgrid nor mgrid return fully instantiated arrays, when simple broadcasting (ie with stride=0 for other axes) is functionally identical and happens much, much faster.
Take a look at ogrid: https://docs.scipy.org/doc/numpy/reference/generated/numpy.ogrid.html
Warren
Post by Juan Nunez-Iglesias
    shape = tuple(map(len, arrays))
    ndim = len(shape)
    result = []
        reshaped = np.broadcast_to(arr[[...] + [np.newaxis] * (ndim - i)],
                                   shape)
        result.append(reshaped)
    return result
In [154]: %timeit th.broadcast_mgrid((np.arange(512), np.arange(512)))
10000 loops, best of 3: 25.9 µs per loop
In [156]: %timeit np.meshgrid(np.arange(512), np.arange(512))
100 loops, best of 3: 2.02 ms per loop
In [157]: %timeit np.mgrid[:512, :512]
100 loops, best of 3: 4.84 ms per loop
Is there a conscious design decision as to why this isn’t what meshgrid/mgrid do already? Or would a PR be welcome to do this?
Thanks,
Juan.
_______________________________________________
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https://mail.scipy.org/mailman/listinfo/numpy-discussion
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P***@ffi.no
2017-03-09 08:01:06 UTC
Permalink
The reason for returning copies from meshgrid as default instead of views into to input arrays, was to not break backwards compatibility.
The old meshgrid returned copied arrays, which is safe if you need to write to those arrays.
If you use copy=False, a view into the original arrays are returned in order to
conserve memory, but will likely return non-contiguous arrays. Furthermore, more than one element of a broadcast array
may refer to a single memory location.

Per A

From: NumPy-Discussion [mailto:numpy-discussion-***@scipy.org] On Behalf Of Juan Nunez-Iglesias
Sent: 9. mars 2017 08:34
To: Discussion of Numerical Python
Subject: Re: [Numpy-discussion] Why do mgrid and meshgrid not return broadcast arrays?

Ah, fantastic, thanks Per!

I'd still be interested to hear from the core devs as to why this isn't the default, both with meshgrid and mgrid...

Juan.

On 9 Mar 2017, 6:29 PM +1100, ***@ffi.no<mailto:***@ffi.no>, wrote:

Hi, Juan.

Meshgrid can actually give what you want, but you must use the options: copy=False and indexing=’ij’.

In [7]: %timeit np.meshgrid(np.arange(512), np.arange(512))
1000 loops, best of 3: 1.24 ms per loop

In [8]: %timeit np.meshgrid(np.arange(512), np.arange(512), copy=False)
10000 loops, best of 3: 27 µs per loop

In [9]: %timeit np.meshgrid(np.arange(512), np.arange(512), copy=False, indexing='ij')
10000 loops, best of 3: 23 µs per loop

Best regards
Per A. Brodtkorb

From: NumPy-Discussion [mailto:numpy-discussion-***@scipy.org] On Behalf Of Juan Nunez-Iglesias
Sent: 9. mars 2017 04:20
To: Discussion of Numerical Python
Subject: Re: [Numpy-discussion] Why do mgrid and meshgrid not return broadcast arrays?

Hi Warren,

ogrid doesn’t solve my problem. Note that my code returns arrays that would evaluate as equal to the mgrid output. It’s just that they are copied in mgrid into a giant array, instead of broadcast:


In [176]: a0, b0 = np.mgrid[:5, :5]

In [177]: a1, b1 = th.broadcast_mgrid((np.arange(5), np.arange(5)))

In [178]: a0
Out[178]:
array([[0, 0, 0, 0, 0],
[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4]])

In [179]: a1
Out[179]:
array([[0, 0, 0, 0, 0],
[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4]])

In [180]: a0.strides
Out[180]: (40, 8)

In [181]: a1.strides
Out[181]: (8, 0)



On 9 Mar 2017, 2:05 PM +1100, Warren Weckesser <***@gmail.com<mailto:***@gmail.com>>, wrote:


On Wed, Mar 8, 2017 at 9:48 PM, Juan Nunez-Iglesias <***@gmail.com<mailto:***@gmail.com>> wrote:
I was a bit surprised to discover that both meshgrid nor mgrid return fully instantiated arrays, when simple broadcasting (ie with stride=0 for other axes) is functionally identical and happens much, much faster.


Take a look at ogrid: https://docs.scipy.org/doc/numpy/reference/generated/numpy.ogrid.html
Warren
I wrote my own function to do this:


def broadcast_mgrid(arrays):
shape = tuple(map(len, arrays))
ndim = len(shape)
result = []
for i, arr in enumerate(arrays, start=1):
reshaped = np.broadcast_to(arr[[...] + [np.newaxis] * (ndim - i)],
shape)
result.append(reshaped)
return result


For even a modest-sized 512 x 512 grid, this version is close to 100x faster:


In [154]: %timeit th.broadcast_mgrid((np.arange(512), np.arange(512)))
10000 loops, best of 3: 25.9 µs per loop

In [156]: %timeit np.meshgrid(np.arange(512), np.arange(512))
100 loops, best of 3: 2.02 ms per loop

In [157]: %timeit np.mgrid[:512, :512]
100 loops, best of 3: 4.84 ms per loop


Is there a conscious design decision as to why this isn’t what meshgrid/mgrid do already? Or would a PR be welcome to do this?

Thanks,

Juan.

_______________________________________________
NumPy-Discussion mailing list
NumPy-***@scipy.org<mailto:NumPy-***@scipy.org>
https://mail.scipy.org/mailman/listinfo/numpy-discussion

_______________________________________________
NumPy-Discussion mailing list
NumPy-***@scipy.org<mailto:NumPy-***@scipy.org>
https://mail.scipy.org/mailman/listinfo/numpy-discussion
_______________________________________________
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NumPy-***@scipy.org<mailto:NumPy-***@scipy.org>
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