Delete zeros from a sparse array [duplicate]
$begingroup$
This question already has an answer here:
How to prevent a SparseArray entry from being specified when it coincides with the default value?
1 answer
Why doesn't SparseArray automatically remove zero entries? The code
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0]
r //ArrayRules
returns
{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}
instead of
{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.
Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.
Of course I could use
A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]
but it seems clumsy to use this often.
Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?
matrix associations sparse-arrays rule
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
asked Dec 30 '18 at 2:44
LeonLeon
410212
$endgroup$
marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prevent a SparseArray entry from being specified when it coincides with the default value?
1 answer
Why doesn't SparseArray automatically remove zero entries? The code
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0]
r //ArrayRules
returns
{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}
instead of
{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.
Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.
Of course I could use
A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]
but it seems clumsy to use this often.
Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?
matrix associations sparse-arrays rule
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
asked Dec 30 '18 at 2:44
LeonLeon
410212
$endgroup$
marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just applySparseArrayagain.
$endgroup$
– Szabolcs
Dec 30 '18 at 9:11
add a comment |
$begingroup$
This question already has an answer here:
How to prevent a SparseArray entry from being specified when it coincides with the default value?
1 answer
Why doesn't SparseArray automatically remove zero entries? The code
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0]
r //ArrayRules
returns
{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}
instead of
{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.
Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.
Of course I could use
A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]
but it seems clumsy to use this often.
Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?
matrix associations sparse-arrays rule
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
asked Dec 30 '18 at 2:44
LeonLeon
410212
$endgroup$
This question already has an answer here:
How to prevent a SparseArray entry from being specified when it coincides with the default value?
1 answer
Why doesn't SparseArray automatically remove zero entries? The code
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0]
r //ArrayRules
returns
{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}
instead of
{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.
Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.
Of course I could use
A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]
but it seems clumsy to use this often.
Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?
This question already has an answer here:
How to prevent a SparseArray entry from being specified when it coincides with the default value?
1 answer
matrix associations sparse-arrays rule
matrix associations sparse-arrays rule
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
asked Dec 30 '18 at 2:44
LeonLeon
410212
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
asked Dec 30 '18 at 2:44
LeonLeon
410212
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
edited Dec 30 '18 at 16:51
m_goldberg
89k873200
89k873200
asked Dec 30 '18 at 2:44
LeonLeon
410212
asked Dec 30 '18 at 2:44
LeonLeon
410212
asked Dec 30 '18 at 2:44
LeonLeon
410212
410212
marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just applySparseArrayagain.
$endgroup$
– Szabolcs
Dec 30 '18 at 9:11
add a comment |
3
$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just applySparseArrayagain.
$endgroup$
– Szabolcs
Dec 30 '18 at 9:11
3
3
$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply
SparseArray again.$endgroup$
– Szabolcs
Dec 30 '18 at 9:11
$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply
SparseArray again.$endgroup$
– Szabolcs
Dec 30 '18 at 9:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can just apply SparseArray to a SparseArray object to normalize it. For your example:
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0];
SparseArray[r] //ArrayRules
{{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can just apply SparseArray to a SparseArray object to normalize it. For your example:
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0];
SparseArray[r] //ArrayRules
{{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
$endgroup$
add a comment |
$begingroup$
You can just apply SparseArray to a SparseArray object to normalize it. For your example:
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0];
SparseArray[r] //ArrayRules
{{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
$endgroup$
add a comment |
$begingroup$
You can just apply SparseArray to a SparseArray object to normalize it. For your example:
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0];
SparseArray[r] //ArrayRules
{{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
$endgroup$
You can just apply SparseArray to a SparseArray object to normalize it. For your example:
a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
r = DeleteCases[a-b,0];
SparseArray[r] //ArrayRules
{{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
answered Dec 30 '18 at 3:06
Carl WollCarl Woll
75.2k3100197
75.2k3100197
add a comment |
add a comment |
3
$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply
SparseArrayagain.$endgroup$
– Szabolcs
Dec 30 '18 at 9:11