using DeleteCases to delete vectors with particular index as 0
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 6 hours ago
cleanplaycleanplay
30918
add a comment |
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 6 hours ago
cleanplaycleanplay
30918
add a comment |
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 6 hours ago
cleanplaycleanplay
30918
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 6 hours ago
cleanplaycleanplay
30918
asked 6 hours ago
cleanplaycleanplay
30918
asked 6 hours ago
cleanplaycleanplay
30918
asked 6 hours ago
cleanplaycleanplay
30918
asked 6 hours ago
cleanplaycleanplay
30918
30918
add a comment |
add a comment |
2 Answers
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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 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
1
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
add a comment |
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
add a comment |
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2 Answers
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2 Answers
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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 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
1
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
add a comment |
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 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
1
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
add a comment |
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 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
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 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
edited 5 hours ago
answered 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
answered 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
answered 6 hours ago
Mr.Wizard♦Mr.Wizard
230k294741038
230k294741038
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
1
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
add a comment |
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
1
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
Thanks. Can you modify this to any index instead of just the last one?
– cleanplay
5 hours ago
1
1
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
@cleanplay Please see the addendum to my answer.
– Mr.Wizard♦
5 hours ago
add a comment |
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
add a comment |
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
add a comment |
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
Or...
Select[set, Last[#] != 0 &]
or...
Select[set, #[[-1]] != 0 &]
or...
DeleteCases[set, x_ /; Last[x] == 0]
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
edited 5 hours ago
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
answered 6 hours ago
David G. StorkDavid G. Stork
23.5k22051
23.5k22051
add a comment |
add a comment |
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