using DeleteCases to delete vectors with particular index as 0
$begingroup$
Assume I have a set of vectors
set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.
I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.
Can I use DeleteCases in a better way than following?
Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]],
{j, 1, Length[set]}];
set = DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];
matrix vector column deletecases
asked Jan 9 at 1:15
cleanplaycleanplay
31918
$endgroup$
add a comment |
$begingroup$
Assume I have a set of vectors
set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.
I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.
Can I use DeleteCases in a better way than following?
Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]],
{j, 1, Length[set]}];
set = DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];
matrix vector column deletecases
asked Jan 9 at 1:15
cleanplaycleanplay
31918
$endgroup$
add a comment |
$begingroup$
Assume I have a set of vectors
set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.
I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.
Can I use DeleteCases in a better way than following?
Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]],
{j, 1, Length[set]}];
set = DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];
matrix vector column deletecases
asked Jan 9 at 1:15
cleanplaycleanplay
31918
$endgroup$
Assume I have a set of vectors
set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.
I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.
Can I use DeleteCases in a better way than following?
Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]],
{j, 1, Length[set]}];
set = DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];
matrix vector column deletecases
matrix vector column deletecases
asked Jan 9 at 1:15
cleanplaycleanplay
31918
asked Jan 9 at 1:15
cleanplaycleanplay
31918
asked Jan 9 at 1:15
cleanplaycleanplay
31918
asked Jan 9 at 1:15
cleanplaycleanplay
31918
asked Jan 9 at 1:15
cleanplaycleanplay
31918
31918
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
DeleteCases[set, {___, 0}]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
In this case Pick also works:
Pick[set, Last /@ set, 1]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.
You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.
Picking according to the second column:
Pick[set, set[[All, 2]], 1]
{{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}
Suppose however that your keep elements are not all the same, like 1 in this example above:
set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
Pick[set2, Unitize @ set2[[All, 3]], 1]
{{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}
Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:
(* Not recommended where performance matters *)
Pick[set2, Positive @ set2[[All, 3]]]
Pick[set2, set2[[All, 3]], _?Positive]
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
$endgroup$
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
1
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
add a comment |
$begingroup$
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
add a comment |
$begingroup$
Another way is to use Delete in conjunction with Position:
Delete[set, Position[set[[All, -1]], 0]]
If I cared about speed for long lists, I'd probably use
Delete[set, Partition[Random`Private`PositionsOf[set[[All, -1]], 0], 1]]
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
$endgroup$
2
$begingroup$
In version 10.1 under Windows x64 yourPositionsOfcode is significantly slower thanPick, at least the way I tested it, e.g.set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?
$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
1
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me thatPickshould be faster. OP asked for usingDeleteCaseswhich sort of drove me to usingDelete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS,Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]]requires0.030seconds whilePick[set, set[[All, -1]], 1]needs only0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.
$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
DeleteCases[set, {___, 0}]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
In this case Pick also works:
Pick[set, Last /@ set, 1]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.
You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.
Picking according to the second column:
Pick[set, set[[All, 2]], 1]
{{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}
Suppose however that your keep elements are not all the same, like 1 in this example above:
set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
Pick[set2, Unitize @ set2[[All, 3]], 1]
{{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}
Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:
(* Not recommended where performance matters *)
Pick[set2, Positive @ set2[[All, 3]]]
Pick[set2, set2[[All, 3]], _?Positive]
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
$endgroup$
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
1
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
add a comment |
$begingroup$
DeleteCases[set, {___, 0}]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
In this case Pick also works:
Pick[set, Last /@ set, 1]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.
You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.
Picking according to the second column:
Pick[set, set[[All, 2]], 1]
{{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}
Suppose however that your keep elements are not all the same, like 1 in this example above:
set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
Pick[set2, Unitize @ set2[[All, 3]], 1]
{{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}
Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:
(* Not recommended where performance matters *)
Pick[set2, Positive @ set2[[All, 3]]]
Pick[set2, set2[[All, 3]], _?Positive]
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
$endgroup$
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
1
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
add a comment |
$begingroup$
DeleteCases[set, {___, 0}]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
In this case Pick also works:
Pick[set, Last /@ set, 1]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.
You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.
Picking according to the second column:
Pick[set, set[[All, 2]], 1]
{{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}
Suppose however that your keep elements are not all the same, like 1 in this example above:
set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
Pick[set2, Unitize @ set2[[All, 3]], 1]
{{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}
Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:
(* Not recommended where performance matters *)
Pick[set2, Positive @ set2[[All, 3]]]
Pick[set2, set2[[All, 3]], _?Positive]
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
$endgroup$
DeleteCases[set, {___, 0}]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
In this case Pick also works:
Pick[set, Last /@ set, 1]
{{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}
You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.
You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.
Picking according to the second column:
Pick[set, set[[All, 2]], 1]
{{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}
Suppose however that your keep elements are not all the same, like 1 in this example above:
set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
Pick[set2, Unitize @ set2[[All, 3]], 1]
{{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}
Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:
(* Not recommended where performance matters *)
Pick[set2, Positive @ set2[[All, 3]]]
Pick[set2, set2[[All, 3]], _?Positive]
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
edited Jan 9 at 1:44
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
answered Jan 9 at 1:20
Mr.Wizard♦Mr.Wizard
231k294741042
231k294741042
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
1
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
add a comment |
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
1
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
$begingroup$
Thanks. Can you modify this to any index instead of just the last one?
$endgroup$
– cleanplay
Jan 9 at 1:29
1
1
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
@cleanplay Please see the addendum to my answer.
$endgroup$
– Mr.Wizard♦
Jan 9 at 1:45
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
$begingroup$
Wizard, thanks again. Can I use Pick over multiple indices in the columns of set. For example, to find vectors with both 2nd and 3rd index as 1, I tried Pick[set, set[[All, 2 ;; 3]], {1, 1}] but that doesn't recognize the pattern in the last argument
$endgroup$
– cleanplay
Jan 10 at 19:26
add a comment |
$begingroup$
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
add a comment |
$begingroup$
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
add a comment |
$begingroup$
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
edited Jan 9 at 2:02
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
answered Jan 9 at 1:23
David G. StorkDavid G. Stork
23.9k22153
23.9k22153
add a comment |
add a comment |
$begingroup$
Another way is to use Delete in conjunction with Position:
Delete[set, Position[set[[All, -1]], 0]]
If I cared about speed for long lists, I'd probably use
Delete[set, Partition[Random`Private`PositionsOf[set[[All, -1]], 0], 1]]
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
$endgroup$
2
$begingroup$
In version 10.1 under Windows x64 yourPositionsOfcode is significantly slower thanPick, at least the way I tested it, e.g.set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?
$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
1
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me thatPickshould be faster. OP asked for usingDeleteCaseswhich sort of drove me to usingDelete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS,Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]]requires0.030seconds whilePick[set, set[[All, -1]], 1]needs only0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.
$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
add a comment |
$begingroup$
Another way is to use Delete in conjunction with Position:
Delete[set, Position[set[[All, -1]], 0]]
If I cared about speed for long lists, I'd probably use
Delete[set, Partition[Random`Private`PositionsOf[set[[All, -1]], 0], 1]]
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
$endgroup$
2
$begingroup$
In version 10.1 under Windows x64 yourPositionsOfcode is significantly slower thanPick, at least the way I tested it, e.g.set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?
$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
1
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me thatPickshould be faster. OP asked for usingDeleteCaseswhich sort of drove me to usingDelete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS,Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]]requires0.030seconds whilePick[set, set[[All, -1]], 1]needs only0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.
$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
add a comment |
$begingroup$
Another way is to use Delete in conjunction with Position:
Delete[set, Position[set[[All, -1]], 0]]
If I cared about speed for long lists, I'd probably use
Delete[set, Partition[Random`Private`PositionsOf[set[[All, -1]], 0], 1]]
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
$endgroup$
Another way is to use Delete in conjunction with Position:
Delete[set, Position[set[[All, -1]], 0]]
If I cared about speed for long lists, I'd probably use
Delete[set, Partition[Random`Private`PositionsOf[set[[All, -1]], 0], 1]]
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
answered Jan 9 at 8:19
Henrik SchumacherHenrik Schumacher
50.4k469144
50.4k469144
2
$begingroup$
In version 10.1 under Windows x64 yourPositionsOfcode is significantly slower thanPick, at least the way I tested it, e.g.set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?
$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
1
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me thatPickshould be faster. OP asked for usingDeleteCaseswhich sort of drove me to usingDelete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS,Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]]requires0.030seconds whilePick[set, set[[All, -1]], 1]needs only0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.
$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
add a comment |
2
$begingroup$
In version 10.1 under Windows x64 yourPositionsOfcode is significantly slower thanPick, at least the way I tested it, e.g.set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?
$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
1
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me thatPickshould be faster. OP asked for usingDeleteCaseswhich sort of drove me to usingDelete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS,Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]]requires0.030seconds whilePick[set, set[[All, -1]], 1]needs only0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.
$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
2
2
$begingroup$
In version 10.1 under Windows x64 your
PositionsOf code is significantly slower than Pick, at least the way I tested it, e.g. set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
$begingroup$
In version 10.1 under Windows x64 your
PositionsOf code is significantly slower than Pick, at least the way I tested it, e.g. set = RandomInteger[3, {1*^6, 5}]. Are you seeing different behavior?$endgroup$
– Mr.Wizard♦
Jan 9 at 9:11
1
1
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me that
Pick should be faster. OP asked for using DeleteCases which sort of drove me to using Delete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS, Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]] requires 0.030 seconds while Pick[set, set[[All, -1]], 1] needs only 0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
$begingroup$
@Mr.Wizard Now that you say it, it becomes clear to me that
Pick should be faster. OP asked for using DeleteCases which sort of drove me to using Delete. But significance lies somewhat in the eye of the beholder. On my Haswell QuadCore with Mathematica 11.3 under macOS, Delete[set, Partition[RandomPrivatePositionsOf[set[[All, -1]], 0], 1]] requires 0.030 seconds while Pick[set, set[[All, -1]], 1] needs only 0.022. This difference is really small if you compare the timings to the timings of pattern-matching based methods.$endgroup$
– Henrik Schumacher
Jan 9 at 20:42
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