Replace diagonal elements in sparse matrix
You can multiply a binary matrix by your Sparse Array, and that matrix may itself be a SparseArray
.
spA = SparseArray[{{1, 1} -> 1, {2, 2} -> 2, {3, 3} -> 3, {1, 3} -> 4}];
zeros = 1 - IdentityMatrix[Dimensions @ spA, SparseArray];
spA*zeros // Grid
$\begin{array}{ccc} 0 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}$
See also Band
and DiagonalMatrix
for other tools to construct the zeros array.
Just subtract the diagonal matrix:
SparseArray[spA - DiagonalMatrix[Diagonal[spA]]]
As Mr. Wizard pointed out, if spA
is not a square matrix, one can use
k = 0;
SparseArray[spA - DiagonalMatrix[Diagonal[spA, k], k, Dimensions[spA]]]
Here, k
stands for the number of the diagonal you what to delete. k = 0
stands for the main diagonal, k = 1
for the first diagonal above the main diagonal, k = -1
for the first diagonal below the main diagonal etc.
removeDiagonal =
# SparseArray[SparseArray`SparseArrayRemoveDiagonal[#]["NonzeroPositions"] -> 1,
Dimensions[#]] &;
removeDiagonal[spA]
SparseArray[<4>,{3,3}]
removeDiagonal[spA] // MatrixForm // TeXForm
$\left( \begin{array}{ccc} 0 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$
removeDiagonal[spA]["NonzeroPositions"]
{{1,3}}
Some timings:
f1 = removeDiagonal;
f2 = SparseArray[# - DiagonalMatrix[Diagonal[#, 0], 0,
Dimensions[#]]] &; (* Henrik Shumacher *)
f3 = With[{zeros = 1 - IdentityMatrix[Dimensions@#, SparseArray]}, #*
zeros] &; (* Mr. Wizard *)
f4 = ReplacePart[#, {i_, i_} :> 0] &; (* tomd *)
m1 = SparseArray[Tuples[RandomSample[Range[101], 100], {2}] -> 1, {101, 101}];
m2 = SparseArray[Tuples[RandomSample[Range[100000], 100], {2}] -> 1, {100000, 100000}];
m = m1;
t11 = First[RepeatedTiming[r11 = f1 @ m ;]];
t21 = First[RepeatedTiming[r21 = f2 @ m;]];
t31 = First[RepeatedTiming[ r31 = f3 @ m;]];
t41 = First[RepeatedTiming[ r41 = f4 @ m;]];
r11 == r21 == r31 == r41
True
m = m2;
t12 = First[RepeatedTiming[r12 = f1 @ m ;]];
t22 = First[RepeatedTiming[r22 = f2 @ m;]];
t32 = First[RepeatedTiming[ r32 = f3 @ m;]];
t42 = "n/a"; (* b/c computation exceeded the limitation of free plan on Wolfram Cloud *)
r12 == r22 == r32
True
{{"dimensions", "non-zero elements", "f1", "f2", "f3", "f4"},
{{101, 101}, 100, t11, t21, t31, t41},
{{100000, 100000}, 100, t12, t22, t32, t42}} // Grid[#, Dividers -> All] & // TeXForm
$\small\begin{array}{|c|c|c|c|c|c|} \hline \text{dimensions} & \text{non-zero elements} & \text{f1} & \text{f2} & \text{f3} & \text{f4} \\ \hline \{101,101\} & 100 & 0.000322 & 0.000279 & 0.000131 & 0.00487 \\ \hline \{100000,100000\} & 100 & 0.0017 & 0.0053 & 0.004 & \text{n/a} \\ \hline \end{array}$