Sparsity of a sparse array without converting it to a regular one
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
asked Nov 27 '18 at 19:13
mavzolej
38019
add a comment |
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
asked Nov 27 '18 at 19:13
mavzolej
38019
add a comment |
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
asked Nov 27 '18 at 19:13
mavzolej
38019
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 '18 at 19:13
mavzolej
38019
asked Nov 27 '18 at 19:13
mavzolej
38019
asked Nov 27 '18 at 19:13
mavzolej
38019
asked Nov 27 '18 at 19:13
mavzolej
38019
asked Nov 27 '18 at 19:13
mavzolej
38019
38019
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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 '18 at 19:18
Coolwater
14.7k32553
add a comment |
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
answered Nov 27 '18 at 19:22
kglr
177k9198406
add a comment |
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"]]]]
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
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
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 '18 at 19:18
Coolwater
14.7k32553
add a comment |
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 '18 at 19:18
Coolwater
14.7k32553
add a comment |
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 '18 at 19:18
Coolwater
14.7k32553
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 '18 at 19:18
Coolwater
14.7k32553
answered Nov 27 '18 at 19:18
Coolwater
14.7k32553
answered Nov 27 '18 at 19:18
Coolwater
14.7k32553
answered Nov 27 '18 at 19:18
Coolwater
14.7k32553
14.7k32553
add a comment |
add a comment |
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
answered Nov 27 '18 at 19:22
kglr
177k9198406
add a comment |
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
answered Nov 27 '18 at 19:22
kglr
177k9198406
add a comment |
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
answered Nov 27 '18 at 19:22
kglr
177k9198406
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
answered Nov 27 '18 at 19:22
kglr
177k9198406
edited Nov 27 '18 at 19:35
answered Nov 27 '18 at 19:22
kglr
177k9198406
answered Nov 27 '18 at 19:22
kglr
177k9198406
answered Nov 27 '18 at 19:22
kglr
177k9198406
177k9198406
add a comment |
add a comment |
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"]]]]
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
add a comment |
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"]]]]
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
add a comment |
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"]]]]
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
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"]]]]
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
edited Nov 27 '18 at 20:13
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
answered Nov 27 '18 at 19:51
Henrik Schumacher
49.2k467139
49.2k467139
add a comment |
add a comment |
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