There are several commands that provide high-level information about the nonzero elements of a sparse matrix:
nnz
returns
the number of nonzero elements in a sparse matrix.
nonzeros
returns
a column vector containing all the nonzero elements of a sparse matrix.
nzmax
returns the amount of storage space
allocated for the nonzero entries of a sparse matrix.
To try some of these, load the supplied sparse matrix west0479
,
one of the Harwell-Boeing collection.
load west0479
whos
Name Size Bytes Class Attributes west0479 479x479 34032 double sparse
This matrix models an eight-stage chemical distillation column.
Try these commands.
nnz(west0479)
ans = 1887
format short e west0479
west0479 = (25,1) 1.0000e+00 (31,1) -3.7648e-02 (87,1) -3.4424e-01 (26,2) 1.0000e+00 (31,2) -2.4523e-02 (88,2) -3.7371e-01 (27,3) 1.0000e+00 (31,3) -3.6613e-02 (89,3) -8.3694e-01 (28,4) 1.3000e+02 . . .
nonzeros(west0479)
ans = 1.0000e+00 -3.7648e-02 -3.4424e-01 1.0000e+00 -2.4523e-02 -3.7371e-01 1.0000e+00 -3.6613e-02 -8.3694e-01 1.3000e+02 . . .
Note:
Use Ctrl+C to stop the |
Note that initially nnz
has the same value
as nzmax
by default. That is, the number of nonzero
elements is equivalent to the number of storage locations allocated
for nonzeros. However, MATLAB^{®} does not dynamically release memory
if you zero out additional array elements. Changing the value of some
matrix elements to zero changes the value of nnz
,
but not that of nzmax
.
However, you can add as many nonzero elements to the matrix
as desired. You are not constrained by the original value of nzmax
.
For any matrix, full or sparse, the find
function
returns the indices and values of nonzero elements. Its syntax is
[i,j,s] = find(S)
find
returns the row indices of nonzero values
in vector i
, the column indices in vector j
,
and the nonzero values themselves in the vector s
.
The example below uses find
to locate the indices
and values of the nonzeros in a sparse matrix. The sparse
function
uses the find
output, together with the size of
the matrix, to recreate the matrix.
[i,j,s] = find(S); [m,n] = size(S); S = sparse(i,j,s,m,n)
Because sparse matrices are stored in compressed sparse column format, there are different costs associated with indexing into a sparse matrix than there are with indexing into a full matrix. Such costs are negligible when you need to change only a few elements in a sparse matrix, so in those cases it's normal to use regular array indexing to reassign values:
B = speye(4); [i,j,s] = find(B); [i,j,s]
ans = 1 1 1 2 2 1 3 3 1 4 4 1
B(3,1) = 42; [i,j,s] = find(B); [i,j,s]
ans = 1 1 1 3 1 42 2 2 1 3 3 1 4 4 1
42
at (3,1)
, MATLAB inserts
an additional row into the nonzero values vector and subscript vectors,
then shifts all matrix values after (3,1)
.Using linear indexing to access or assign an element in a large
sparse matrix will fail if the linear index exceeds 2^48-1
,
which is the current upper bound for the number of elements allowed
in a matrix.
S = spalloc(2^30,2^30,2); S(end) = 1
Maximum variable size allowed by the program is exceeded.
To access an element whose linear index is greater than intmax
,
use array indexing:
S(2^30,2^30) = 1
S = (1073741824,1073741824) 1
While the cost of indexing into a sparse matrix to change a single element is negligible, it is compounded in the context of a loop and can become quite slow for large matrices. For that reason, in cases where many sparse matrix elements need to be changed, it is best to vectorize the operation instead of using a loop. For example, consider a sparse identity matrix:
n = 10000; A = 4*speye(n);
A
within
a loop takes is slower than a similar vectorized operation:tic; A(1:n-1,n) = -1; A(n,1:n-1) = -1; toc
Elapsed time is 0.003344 seconds.
tic; for k = 1:n-1, C(k,n) = -1; C(n,k) = -1; end, toc
Elapsed time is 0.448069 seconds.
A
during each pass through
the loop.Preallocating the memory for a sparse matrix and then filling it in an element-wise manner similarly causes a significant amount of overhead in indexing into the sparse array:
S1 = spalloc(1000,1000,100000); tic; for n = 1:100000 i = ceil(1000*rand(1,1)); j = ceil(1000*rand(1,1)); S1(i,j) = rand(1,1); end toc
Elapsed time is 2.577527 seconds.
Constructing the vectors of indices and values eliminates the need to index into the sparse array, and thus is significantly faster:
i = ceil(1000*rand(100000,1)); j = ceil(1000*rand(100000,1)); v = zeros(size(i)); for n = 1:100000 v(n) = rand(1,1); end tic; S2 = sparse(i,j,v,1000,1000); toc
Elapsed time is 0.017676 seconds.
For that reason, it's best to construct sparse matrices
all at once using a construction function, like the sparse
or spdiags
functions.
For example, suppose you wanted the sparse form of the coordinate
matrix C
:
$$\text{C}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\begin{array}{cccc}4& 0& 0& \begin{array}{cc}0& -1\end{array}\\ 0& 4& 0& \begin{array}{cc}0& -1\end{array}\\ 0& 0& 4& \begin{array}{cc}0& -1\end{array}\\ \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}0\\ 1\end{array}& \begin{array}{cc}\begin{array}{c}4\\ 1\end{array}& \begin{array}{c}-1\\ 4\end{array}\end{array}\end{array}\right)$$
Construct the five-column matrix directly with the sparse
function using the triplet pairs
for the row subscripts, column subscripts, and values:
i = [1 5 2 5 3 5 4 5 1 2 3 4 5]'; j = [1 1 2 2 3 3 4 4 5 5 5 5 5]'; s = [4 1 4 1 4 1 4 1 -1 -1 -1 -1 4]'; C = sparse(i,j,s)
C = (1,1) 4 (5,1) 1 (2,2) 4 (5,2) 1 (3,3) 4 (5,3) 1 (4,4) 4 (5,4) 1 (1,5) -1 (2,5) -1 (3,5) -1 (4,5) -1 (5,5) 4
It is often useful to use a graphical format to view the distribution of the nonzero elements within a sparse matrix. The MATLAB spy
function produces a template view of the sparsity structure, where each point on the graph represents the location of a nonzero array element.
For example:
Load the supplied sparse matrix west0479
, one of the Harwell-Boeing collection.
load west0479
View the sparsity structure.
spy(west0479)