never allocate an output of numpy.ufunc












2















This question has info on using an input as an output to compute something in place with a numpy.ufunc:




  • Numpy passing input array as `out` argument to ufunc


Is it possible to avoid allocating space for an unwanted output of a numpy.ufunc? For example, say I only want one of the two outputs from modf. Can I ensure that the other, unwanted array is never allocated at all?



I thought passing _ to out might do it, but it throws an error:



import numpy as np

ar = np.arange(6)/3

np.modf(ar, out=(ar, _))    



TypeError: return arrays must be of ArrayType


As it says in the docs, passing None means that the output array is allocated in the function and returned. I can ignore the returned values, but it still has to be allocated and populated inside the function.






python numpy







share|improve this question


















  • 1





    Why worry about this? There aren't many ufunc that return more than one array. How about using np.remainder(arr, 1) instead?

    – hpaulj
    Nov 15 '18 at 23:21













  • This was more of a question about the limits of ufunc than a particular use case. My additional reading is making me pretty sure it cannot be done.

    – Paul T.
    Nov 16 '18 at 0:29


















2















This question has info on using an input as an output to compute something in place with a numpy.ufunc:




  • Numpy passing input array as `out` argument to ufunc


Is it possible to avoid allocating space for an unwanted output of a numpy.ufunc? For example, say I only want one of the two outputs from modf. Can I ensure that the other, unwanted array is never allocated at all?



I thought passing _ to out might do it, but it throws an error:



import numpy as np

ar = np.arange(6)/3

np.modf(ar, out=(ar, _))    



TypeError: return arrays must be of ArrayType


As it says in the docs, passing None means that the output array is allocated in the function and returned. I can ignore the returned values, but it still has to be allocated and populated inside the function.






python numpy







share|improve this question


















  • 1





    Why worry about this? There aren't many ufunc that return more than one array. How about using np.remainder(arr, 1) instead?

    – hpaulj
    Nov 15 '18 at 23:21













  • This was more of a question about the limits of ufunc than a particular use case. My additional reading is making me pretty sure it cannot be done.

    – Paul T.
    Nov 16 '18 at 0:29
















2












2








2


1






This question has info on using an input as an output to compute something in place with a numpy.ufunc:




  • Numpy passing input array as `out` argument to ufunc


Is it possible to avoid allocating space for an unwanted output of a numpy.ufunc? For example, say I only want one of the two outputs from modf. Can I ensure that the other, unwanted array is never allocated at all?



I thought passing _ to out might do it, but it throws an error:



import numpy as np

ar = np.arange(6)/3

np.modf(ar, out=(ar, _))    



TypeError: return arrays must be of ArrayType


As it says in the docs, passing None means that the output array is allocated in the function and returned. I can ignore the returned values, but it still has to be allocated and populated inside the function.






python numpy







share|improve this question














This question has info on using an input as an output to compute something in place with a numpy.ufunc:




  • Numpy passing input array as `out` argument to ufunc


Is it possible to avoid allocating space for an unwanted output of a numpy.ufunc? For example, say I only want one of the two outputs from modf. Can I ensure that the other, unwanted array is never allocated at all?



I thought passing _ to out might do it, but it throws an error:



import numpy as np

ar = np.arange(6)/3

np.modf(ar, out=(ar, _))    



TypeError: return arrays must be of ArrayType


As it says in the docs, passing None means that the output array is allocated in the function and returned. I can ignore the returned values, but it still has to be allocated and populated inside the function.






python numpy




python numpy






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 15 '18 at 23:12









Paul T.Paul T.

204210




204210








  • 1





    Why worry about this? There aren't many ufunc that return more than one array. How about using np.remainder(arr, 1) instead?

    – hpaulj
    Nov 15 '18 at 23:21













  • This was more of a question about the limits of ufunc than a particular use case. My additional reading is making me pretty sure it cannot be done.

    – Paul T.
    Nov 16 '18 at 0:29
















  • 1





    Why worry about this? There aren't many ufunc that return more than one array. How about using np.remainder(arr, 1) instead?

    – hpaulj
    Nov 15 '18 at 23:21













  • This was more of a question about the limits of ufunc than a particular use case. My additional reading is making me pretty sure it cannot be done.

    – Paul T.
    Nov 16 '18 at 0:29










1




1





Why worry about this? There aren't many ufunc that return more than one array. How about using np.remainder(arr, 1) instead?

– hpaulj
Nov 15 '18 at 23:21







Why worry about this? There aren't many ufunc that return more than one array. How about using np.remainder(arr, 1) instead?

– hpaulj
Nov 15 '18 at 23:21















This was more of a question about the limits of ufunc than a particular use case. My additional reading is making me pretty sure it cannot be done.

– Paul T.
Nov 16 '18 at 0:29







This was more of a question about the limits of ufunc than a particular use case. My additional reading is making me pretty sure it cannot be done.

– Paul T.
Nov 16 '18 at 0:29














1 Answer
1






active

oldest

votes


















4














You can minimize allocation by passing a "fake" array:



ar = np.arange(6) / 3

np.modf(ar, ar, np.broadcast_arrays(ar.dtype.type(0), ar)[0])


This dummy array is as big as a single double, and modf will not do allocation internally.



EDIT According to suggestions from @Eric and @hpaulj, a more general and long-term solution would be



np.lib.stride_tricks._broadcast_to(np.empty(1, ar.dtype), ar.shape, False, False)





share|improve this answer





















  • 1





    np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

    – Eric
    Nov 16 '18 at 6:56













  • np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

    – Paul T.
    Nov 16 '18 at 15:48











  • This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

    – Paul T.
    Nov 16 '18 at 15:57






  • 2





    Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

    – hpaulj
    Nov 17 '18 at 5:35













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4














You can minimize allocation by passing a "fake" array:



ar = np.arange(6) / 3

np.modf(ar, ar, np.broadcast_arrays(ar.dtype.type(0), ar)[0])


This dummy array is as big as a single double, and modf will not do allocation internally.



EDIT According to suggestions from @Eric and @hpaulj, a more general and long-term solution would be



np.lib.stride_tricks._broadcast_to(np.empty(1, ar.dtype), ar.shape, False, False)





share|improve this answer





















  • 1





    np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

    – Eric
    Nov 16 '18 at 6:56













  • np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

    – Paul T.
    Nov 16 '18 at 15:48











  • This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

    – Paul T.
    Nov 16 '18 at 15:57






  • 2





    Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

    – hpaulj
    Nov 17 '18 at 5:35


















4














You can minimize allocation by passing a "fake" array:



ar = np.arange(6) / 3

np.modf(ar, ar, np.broadcast_arrays(ar.dtype.type(0), ar)[0])


This dummy array is as big as a single double, and modf will not do allocation internally.



EDIT According to suggestions from @Eric and @hpaulj, a more general and long-term solution would be



np.lib.stride_tricks._broadcast_to(np.empty(1, ar.dtype), ar.shape, False, False)





share|improve this answer





















  • 1





    np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

    – Eric
    Nov 16 '18 at 6:56













  • np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

    – Paul T.
    Nov 16 '18 at 15:48











  • This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

    – Paul T.
    Nov 16 '18 at 15:57






  • 2





    Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

    – hpaulj
    Nov 17 '18 at 5:35
















4












4








4







You can minimize allocation by passing a "fake" array:



ar = np.arange(6) / 3

np.modf(ar, ar, np.broadcast_arrays(ar.dtype.type(0), ar)[0])


This dummy array is as big as a single double, and modf will not do allocation internally.



EDIT According to suggestions from @Eric and @hpaulj, a more general and long-term solution would be



np.lib.stride_tricks._broadcast_to(np.empty(1, ar.dtype), ar.shape, False, False)





share|improve this answer















You can minimize allocation by passing a "fake" array:



ar = np.arange(6) / 3

np.modf(ar, ar, np.broadcast_arrays(ar.dtype.type(0), ar)[0])


This dummy array is as big as a single double, and modf will not do allocation internally.



EDIT According to suggestions from @Eric and @hpaulj, a more general and long-term solution would be



np.lib.stride_tricks._broadcast_to(np.empty(1, ar.dtype), ar.shape, False, False)






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 19 '18 at 2:32

























answered Nov 16 '18 at 3:28









ZisIsNotZisZisIsNotZis

745620




745620








  • 1





    np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

    – Eric
    Nov 16 '18 at 6:56













  • np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

    – Paul T.
    Nov 16 '18 at 15:48











  • This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

    – Paul T.
    Nov 16 '18 at 15:57






  • 2





    Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

    – hpaulj
    Nov 17 '18 at 5:35
















  • 1





    np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

    – Eric
    Nov 16 '18 at 6:56













  • np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

    – Paul T.
    Nov 16 '18 at 15:48











  • This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

    – Paul T.
    Nov 16 '18 at 15:57






  • 2





    Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

    – hpaulj
    Nov 17 '18 at 5:35










1




1





np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

– Eric
Nov 16 '18 at 6:56







np.broadcast_to(np.empty((), dtype), shape) is a more general pattern, which works on non-numeric arrays too.

– Eric
Nov 16 '18 at 6:56















np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

– Paul T.
Nov 16 '18 at 15:48





np.modf(ar, out=(ar, np.broadcast_to(np.empty((), ar.dtype), ar.shape))) does not work because the second output is read only

– Paul T.
Nov 16 '18 at 15:48













This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

– Paul T.
Nov 16 '18 at 15:57





This is neat! The second output is still an array of the right length, but all elements point to the same location in memory.

– Paul T.
Nov 16 '18 at 15:57




2




2





Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

– hpaulj
Nov 17 '18 at 5:35







Both broadcast_to and broadcast_arrays use np.lib.stride_tricks._broadcast_to. One uses a readonly=True parameter, the other sets it to False. But there's a TODO comment in the code about changing it to readonly, so in the long term this answer may need modification. It's interesting that the underlying tool in all of this is nditer.

– hpaulj
Nov 17 '18 at 5:35






















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