Sparsity of a sparse array without converting it to a regular one
up vote
4
down vote
favorite
My goal is to find such properties of a sparse matrix as the maximum/average number of non-zero elements per row.
The brute-force way of doing this is via converting the sparse array into a regular one:
MaxSpar[matr_] := Module[{curr, ms = 0},
Do[
curr = Length[Cases[matr[[k]], 0]];
If[curr > ms, ms = curr];
, {k, 1, Length[matr]}
];
Return[ms];
];
MaxSpar[Normal[SomeSparseMatrix]]
How can we do the same without using Normal
?
sparse-arrays
add a comment |
up vote
4
down vote
favorite
My goal is to find such properties of a sparse matrix as the maximum/average number of non-zero elements per row.
The brute-force way of doing this is via converting the sparse array into a regular one:
MaxSpar[matr_] := Module[{curr, ms = 0},
Do[
curr = Length[Cases[matr[[k]], 0]];
If[curr > ms, ms = curr];
, {k, 1, Length[matr]}
];
Return[ms];
];
MaxSpar[Normal[SomeSparseMatrix]]
How can we do the same without using Normal
?
sparse-arrays
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
My goal is to find such properties of a sparse matrix as the maximum/average number of non-zero elements per row.
The brute-force way of doing this is via converting the sparse array into a regular one:
MaxSpar[matr_] := Module[{curr, ms = 0},
Do[
curr = Length[Cases[matr[[k]], 0]];
If[curr > ms, ms = curr];
, {k, 1, Length[matr]}
];
Return[ms];
];
MaxSpar[Normal[SomeSparseMatrix]]
How can we do the same without using Normal
?
sparse-arrays
My goal is to find such properties of a sparse matrix as the maximum/average number of non-zero elements per row.
The brute-force way of doing this is via converting the sparse array into a regular one:
MaxSpar[matr_] := Module[{curr, ms = 0},
Do[
curr = Length[Cases[matr[[k]], 0]];
If[curr > ms, ms = curr];
, {k, 1, Length[matr]}
];
Return[ms];
];
MaxSpar[Normal[SomeSparseMatrix]]
How can we do the same without using Normal
?
sparse-arrays
sparse-arrays
asked Nov 27 at 19:13
mavzolej
38019
38019
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
To obtain the number of nonzero entry of the row with fewest zeros:
Max[Length /@ SomeSparseMatrix["AdjacencyLists"]]
There are other useful strings. "Methods"
shows which are availble:
SomeSparseMatrix["Methods"]
{"AdjacencyLists", "Background", "ColumnIndices", "Density",
"MatrixColumns", "MethodInformation", "Methods", "NonzeroPositions",
"NonzeroValues", "PatternArray", "PatternValues", "Properties",
"RowPointers"}
add a comment |
up vote
3
down vote
maxNonZero = Max[Length /@ #["MatrixColumns"]] &;
aveNonZero = Mean[Length /@ #["MatrixColumns"] ] &
SeedRandom[1]
sa = SparseArray[RandomInteger[3, {7, 10}]];
sa // MatrixForm // TeXForm
$left(
begin{array}{cccccccccc}
3 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 3 \
0 & 0 & 0 & 0 & 2 & 0 & 1 & 2 & 0 & 0 \
3 & 3 & 3 & 1 & 1 & 0 & 0 & 1 & 3 & 0 \
2 & 0 & 1 & 1 & 3 & 3 & 3 & 2 & 3 & 2 \
0 & 1 & 3 & 3 & 0 & 1 & 0 & 1 & 0 & 3 \
0 & 2 & 3 & 0 & 2 & 2 & 0 & 1 & 3 & 2 \
1 & 2 & 0 & 0 & 0 & 2 & 1 & 2 & 1 & 0 \
end{array}
right)$
maxNonZero[sa]
9
N @ aveNonZero[sa]
6.285714285714
add a comment |
up vote
1
down vote
m = 100000;
n = 2000000;
A = SparseArray[
RandomInteger[{1, m}, {n, 2}] -> RandomReal[{-1, 1}, n],
{m, m}, 0.
];
Maximum number of nonempty elements per row:
a = Max[Unitize[A].ConstantArray[1, Dimensions[A][[2]]]]; // RepeatedTiming // First
b = Max[Length /@ A["AdjacencyLists"]]; // RepeatedTiming // First
0.122
0.053
A faster way (that works only for rows) is
c = Max[Differences[A["RowPointers"]]]; // RepeatedTiming // First
a == b == c
0.000642
True
Analogously, the mean of the numbers of nonempty elements per row can be obtain as follows:
Mean[N[Differences[A["RowPointers"]]]]
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
To obtain the number of nonzero entry of the row with fewest zeros:
Max[Length /@ SomeSparseMatrix["AdjacencyLists"]]
There are other useful strings. "Methods"
shows which are availble:
SomeSparseMatrix["Methods"]
{"AdjacencyLists", "Background", "ColumnIndices", "Density",
"MatrixColumns", "MethodInformation", "Methods", "NonzeroPositions",
"NonzeroValues", "PatternArray", "PatternValues", "Properties",
"RowPointers"}
add a comment |
up vote
6
down vote
accepted
To obtain the number of nonzero entry of the row with fewest zeros:
Max[Length /@ SomeSparseMatrix["AdjacencyLists"]]
There are other useful strings. "Methods"
shows which are availble:
SomeSparseMatrix["Methods"]
{"AdjacencyLists", "Background", "ColumnIndices", "Density",
"MatrixColumns", "MethodInformation", "Methods", "NonzeroPositions",
"NonzeroValues", "PatternArray", "PatternValues", "Properties",
"RowPointers"}
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
To obtain the number of nonzero entry of the row with fewest zeros:
Max[Length /@ SomeSparseMatrix["AdjacencyLists"]]
There are other useful strings. "Methods"
shows which are availble:
SomeSparseMatrix["Methods"]
{"AdjacencyLists", "Background", "ColumnIndices", "Density",
"MatrixColumns", "MethodInformation", "Methods", "NonzeroPositions",
"NonzeroValues", "PatternArray", "PatternValues", "Properties",
"RowPointers"}
To obtain the number of nonzero entry of the row with fewest zeros:
Max[Length /@ SomeSparseMatrix["AdjacencyLists"]]
There are other useful strings. "Methods"
shows which are availble:
SomeSparseMatrix["Methods"]
{"AdjacencyLists", "Background", "ColumnIndices", "Density",
"MatrixColumns", "MethodInformation", "Methods", "NonzeroPositions",
"NonzeroValues", "PatternArray", "PatternValues", "Properties",
"RowPointers"}
answered Nov 27 at 19:18
Coolwater
14.4k32452
14.4k32452
add a comment |
add a comment |
up vote
3
down vote
maxNonZero = Max[Length /@ #["MatrixColumns"]] &;
aveNonZero = Mean[Length /@ #["MatrixColumns"] ] &
SeedRandom[1]
sa = SparseArray[RandomInteger[3, {7, 10}]];
sa // MatrixForm // TeXForm
$left(
begin{array}{cccccccccc}
3 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 3 \
0 & 0 & 0 & 0 & 2 & 0 & 1 & 2 & 0 & 0 \
3 & 3 & 3 & 1 & 1 & 0 & 0 & 1 & 3 & 0 \
2 & 0 & 1 & 1 & 3 & 3 & 3 & 2 & 3 & 2 \
0 & 1 & 3 & 3 & 0 & 1 & 0 & 1 & 0 & 3 \
0 & 2 & 3 & 0 & 2 & 2 & 0 & 1 & 3 & 2 \
1 & 2 & 0 & 0 & 0 & 2 & 1 & 2 & 1 & 0 \
end{array}
right)$
maxNonZero[sa]
9
N @ aveNonZero[sa]
6.285714285714
add a comment |
up vote
3
down vote
maxNonZero = Max[Length /@ #["MatrixColumns"]] &;
aveNonZero = Mean[Length /@ #["MatrixColumns"] ] &
SeedRandom[1]
sa = SparseArray[RandomInteger[3, {7, 10}]];
sa // MatrixForm // TeXForm
$left(
begin{array}{cccccccccc}
3 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 3 \
0 & 0 & 0 & 0 & 2 & 0 & 1 & 2 & 0 & 0 \
3 & 3 & 3 & 1 & 1 & 0 & 0 & 1 & 3 & 0 \
2 & 0 & 1 & 1 & 3 & 3 & 3 & 2 & 3 & 2 \
0 & 1 & 3 & 3 & 0 & 1 & 0 & 1 & 0 & 3 \
0 & 2 & 3 & 0 & 2 & 2 & 0 & 1 & 3 & 2 \
1 & 2 & 0 & 0 & 0 & 2 & 1 & 2 & 1 & 0 \
end{array}
right)$
maxNonZero[sa]
9
N @ aveNonZero[sa]
6.285714285714
add a comment |
up vote
3
down vote
up vote
3
down vote
maxNonZero = Max[Length /@ #["MatrixColumns"]] &;
aveNonZero = Mean[Length /@ #["MatrixColumns"] ] &
SeedRandom[1]
sa = SparseArray[RandomInteger[3, {7, 10}]];
sa // MatrixForm // TeXForm
$left(
begin{array}{cccccccccc}
3 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 3 \
0 & 0 & 0 & 0 & 2 & 0 & 1 & 2 & 0 & 0 \
3 & 3 & 3 & 1 & 1 & 0 & 0 & 1 & 3 & 0 \
2 & 0 & 1 & 1 & 3 & 3 & 3 & 2 & 3 & 2 \
0 & 1 & 3 & 3 & 0 & 1 & 0 & 1 & 0 & 3 \
0 & 2 & 3 & 0 & 2 & 2 & 0 & 1 & 3 & 2 \
1 & 2 & 0 & 0 & 0 & 2 & 1 & 2 & 1 & 0 \
end{array}
right)$
maxNonZero[sa]
9
N @ aveNonZero[sa]
6.285714285714
maxNonZero = Max[Length /@ #["MatrixColumns"]] &;
aveNonZero = Mean[Length /@ #["MatrixColumns"] ] &
SeedRandom[1]
sa = SparseArray[RandomInteger[3, {7, 10}]];
sa // MatrixForm // TeXForm
$left(
begin{array}{cccccccccc}
3 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 3 \
0 & 0 & 0 & 0 & 2 & 0 & 1 & 2 & 0 & 0 \
3 & 3 & 3 & 1 & 1 & 0 & 0 & 1 & 3 & 0 \
2 & 0 & 1 & 1 & 3 & 3 & 3 & 2 & 3 & 2 \
0 & 1 & 3 & 3 & 0 & 1 & 0 & 1 & 0 & 3 \
0 & 2 & 3 & 0 & 2 & 2 & 0 & 1 & 3 & 2 \
1 & 2 & 0 & 0 & 0 & 2 & 1 & 2 & 1 & 0 \
end{array}
right)$
maxNonZero[sa]
9
N @ aveNonZero[sa]
6.285714285714
edited Nov 27 at 19:35
answered Nov 27 at 19:22
kglr
174k9197402
174k9197402
add a comment |
add a comment |
up vote
1
down vote
m = 100000;
n = 2000000;
A = SparseArray[
RandomInteger[{1, m}, {n, 2}] -> RandomReal[{-1, 1}, n],
{m, m}, 0.
];
Maximum number of nonempty elements per row:
a = Max[Unitize[A].ConstantArray[1, Dimensions[A][[2]]]]; // RepeatedTiming // First
b = Max[Length /@ A["AdjacencyLists"]]; // RepeatedTiming // First
0.122
0.053
A faster way (that works only for rows) is
c = Max[Differences[A["RowPointers"]]]; // RepeatedTiming // First
a == b == c
0.000642
True
Analogously, the mean of the numbers of nonempty elements per row can be obtain as follows:
Mean[N[Differences[A["RowPointers"]]]]
add a comment |
up vote
1
down vote
m = 100000;
n = 2000000;
A = SparseArray[
RandomInteger[{1, m}, {n, 2}] -> RandomReal[{-1, 1}, n],
{m, m}, 0.
];
Maximum number of nonempty elements per row:
a = Max[Unitize[A].ConstantArray[1, Dimensions[A][[2]]]]; // RepeatedTiming // First
b = Max[Length /@ A["AdjacencyLists"]]; // RepeatedTiming // First
0.122
0.053
A faster way (that works only for rows) is
c = Max[Differences[A["RowPointers"]]]; // RepeatedTiming // First
a == b == c
0.000642
True
Analogously, the mean of the numbers of nonempty elements per row can be obtain as follows:
Mean[N[Differences[A["RowPointers"]]]]
add a comment |
up vote
1
down vote
up vote
1
down vote
m = 100000;
n = 2000000;
A = SparseArray[
RandomInteger[{1, m}, {n, 2}] -> RandomReal[{-1, 1}, n],
{m, m}, 0.
];
Maximum number of nonempty elements per row:
a = Max[Unitize[A].ConstantArray[1, Dimensions[A][[2]]]]; // RepeatedTiming // First
b = Max[Length /@ A["AdjacencyLists"]]; // RepeatedTiming // First
0.122
0.053
A faster way (that works only for rows) is
c = Max[Differences[A["RowPointers"]]]; // RepeatedTiming // First
a == b == c
0.000642
True
Analogously, the mean of the numbers of nonempty elements per row can be obtain as follows:
Mean[N[Differences[A["RowPointers"]]]]
m = 100000;
n = 2000000;
A = SparseArray[
RandomInteger[{1, m}, {n, 2}] -> RandomReal[{-1, 1}, n],
{m, m}, 0.
];
Maximum number of nonempty elements per row:
a = Max[Unitize[A].ConstantArray[1, Dimensions[A][[2]]]]; // RepeatedTiming // First
b = Max[Length /@ A["AdjacencyLists"]]; // RepeatedTiming // First
0.122
0.053
A faster way (that works only for rows) is
c = Max[Differences[A["RowPointers"]]]; // RepeatedTiming // First
a == b == c
0.000642
True
Analogously, the mean of the numbers of nonempty elements per row can be obtain as follows:
Mean[N[Differences[A["RowPointers"]]]]
edited Nov 27 at 20:13
answered Nov 27 at 19:51
Henrik Schumacher
46.5k466133
46.5k466133
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f186810%2fsparsity-of-a-sparse-array-without-converting-it-to-a-regular-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown