Delete with multiple indices is extremely slow--workaround?
up vote
13
down vote
favorite
Delete
is unbelievably slow when deleting multiple elements from a non-packed array.
Is there a robust workaround that will work on any non-packed array?
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
(* {17.8957, Null} *)
On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Range[100000], inds]; // AbsoluteTiming
(* {0.005767, Null} *)
I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete
will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.
list-manipulation performance-tuning
add a comment |
up vote
13
down vote
favorite
Delete
is unbelievably slow when deleting multiple elements from a non-packed array.
Is there a robust workaround that will work on any non-packed array?
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
(* {17.8957, Null} *)
On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Range[100000], inds]; // AbsoluteTiming
(* {0.005767, Null} *)
I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete
will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.
list-manipulation performance-tuning
add a comment |
up vote
13
down vote
favorite
up vote
13
down vote
favorite
Delete
is unbelievably slow when deleting multiple elements from a non-packed array.
Is there a robust workaround that will work on any non-packed array?
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
(* {17.8957, Null} *)
On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Range[100000], inds]; // AbsoluteTiming
(* {0.005767, Null} *)
I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete
will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.
list-manipulation performance-tuning
Delete
is unbelievably slow when deleting multiple elements from a non-packed array.
Is there a robust workaround that will work on any non-packed array?
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Developer`FromPackedArray@Range[100000], inds]; // AbsoluteTiming
(* {17.8957, Null} *)
On packed arrays it performs as expected, but my array cannot be packed. It does not necessarily contain numbers.
inds = List /@ RandomSample[Range[100000], 50000];
Delete[Range[100000], inds]; // AbsoluteTiming
(* {0.005767, Null} *)
I did not try to test for this, but one possible explanation is that even when given multiple indices, Delete
will delete elements one-by-one, re-allocating the array after each step. If someone feels like testing it, you can try to see if the timing is quadratic in the number of elements deleted.
list-manipulation performance-tuning
list-manipulation performance-tuning
edited Dec 4 at 6:56
xzczd
25.7k469245
25.7k469245
asked Dec 3 at 9:40
Szabolcs
158k13432926
158k13432926
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
14
down vote
Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000], 50000];
Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
(* {0.006371, Null} *)
1
Embarrassing forDelete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
– b3m2a1
Dec 3 at 9:52
3
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad thatDelete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
– Henrik Schumacher
Dec 3 at 10:00
add a comment |
up vote
7
down vote
Why not use Part
assignment (to Sequence
) instead?
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000],50000];
r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
(r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming
r1 === r2
{0.0059, Null}
{0.0019, Null}
True
Nothing
is slightly faster thanSequence
on my laptop.
– Sjoerd C. de Vries
Dec 3 at 20:11
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed byr2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this:r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000], 50000];
Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
(* {0.006371, Null} *)
1
Embarrassing forDelete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
– b3m2a1
Dec 3 at 9:52
3
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad thatDelete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
– Henrik Schumacher
Dec 3 at 10:00
add a comment |
up vote
14
down vote
Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000], 50000];
Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
(* {0.006371, Null} *)
1
Embarrassing forDelete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
– b3m2a1
Dec 3 at 9:52
3
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad thatDelete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
– Henrik Schumacher
Dec 3 at 10:00
add a comment |
up vote
14
down vote
up vote
14
down vote
Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000], 50000];
Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
(* {0.006371, Null} *)
Exploiting the fact that Delete works fine on packed arrays, we can first construct an index vector, delete the unneeded indices, then finally use the remaining ones to index into the main array.
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000], 50000];
Part[arr, Delete[Range@Length[arr], inds]]; // AbsoluteTiming
(* {0.006371, Null} *)
answered Dec 3 at 9:42
Szabolcs
158k13432926
158k13432926
1
Embarrassing forDelete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
– b3m2a1
Dec 3 at 9:52
3
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad thatDelete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
– Henrik Schumacher
Dec 3 at 10:00
add a comment |
1
Embarrassing forDelete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?
– b3m2a1
Dec 3 at 9:52
3
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad thatDelete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.
– Henrik Schumacher
Dec 3 at 10:00
1
1
Embarrassing for
Delete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?– b3m2a1
Dec 3 at 9:52
Embarrassing for
Delete
that this method which ought to have terrible time complexity is so much faster... Have you tested some of the other things like delete on unpacked arrays?– b3m2a1
Dec 3 at 9:52
3
3
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that
Delete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.– Henrik Schumacher
Dec 3 at 10:00
@b3m2a1 I don't think that this has larger complexity... Still it is pretty bad that
Delete
is not clever enough to do that automatically. I'd suggest to inform Wolfram Support.– Henrik Schumacher
Dec 3 at 10:00
add a comment |
up vote
7
down vote
Why not use Part
assignment (to Sequence
) instead?
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000],50000];
r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
(r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming
r1 === r2
{0.0059, Null}
{0.0019, Null}
True
Nothing
is slightly faster thanSequence
on my laptop.
– Sjoerd C. de Vries
Dec 3 at 20:11
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed byr2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this:r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
add a comment |
up vote
7
down vote
Why not use Part
assignment (to Sequence
) instead?
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000],50000];
r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
(r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming
r1 === r2
{0.0059, Null}
{0.0019, Null}
True
Nothing
is slightly faster thanSequence
on my laptop.
– Sjoerd C. de Vries
Dec 3 at 20:11
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed byr2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this:r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
add a comment |
up vote
7
down vote
up vote
7
down vote
Why not use Part
assignment (to Sequence
) instead?
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000],50000];
r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
(r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming
r1 === r2
{0.0059, Null}
{0.0019, Null}
True
Why not use Part
assignment (to Sequence
) instead?
arr = Developer`FromPackedArray@Range[100000];
inds = List /@ RandomSample[Range[100000],50000];
r1 = Part[arr, Delete[Range@Length[arr], inds]]; //RepeatedTiming
(r2 = arr; r2[[Flatten @ inds]] = Sequence;) //RepeatedTiming
r1 === r2
{0.0059, Null}
{0.0019, Null}
True
answered Dec 3 at 19:23
Carl Woll
66.7k385174
66.7k385174
Nothing
is slightly faster thanSequence
on my laptop.
– Sjoerd C. de Vries
Dec 3 at 20:11
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed byr2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this:r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
add a comment |
Nothing
is slightly faster thanSequence
on my laptop.
– Sjoerd C. de Vries
Dec 3 at 20:11
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed byr2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this:r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
Nothing
is slightly faster than Sequence
on my laptop.– Sjoerd C. de Vries
Dec 3 at 20:11
Nothing
is slightly faster than Sequence
on my laptop.– Sjoerd C. de Vries
Dec 3 at 20:11
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by
r2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
@SjoerdC.deVries and Carl, to fully reduce the array the procedure needs to be followed by
r2;
. Both Sequence and Nothing will now produce same timings as Part+Delete. p.s. to see what I mean try this: r = {1, 2}; r[[1]] = Nothing; Information@r
– Kuba♦
Dec 4 at 7:28
add a comment |
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