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dlib/examples/matrix_ex.cpp

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/*
This is an example illustrating the use of the matrix object
from the dlib C++ Library.
*/
#include <iostream>
#include "dlib/matrix.h"
using namespace dlib;
using namespace std;
// ----------------------------------------------------------------------------------------
int main()
{
// Lets begin this example by using the library to solve a simple
// linear system.
//
// We will find the value of x such that y = M*x where
//
// 3.5
// y = 1.2
// 7.8
//
// and M is
//
// 54.2 7.4 12.1
// M = 1 2 3
// 5.9 0.05 1
// First lets declare these 3 matrices.
// This declares a matrix that contains doubles and has 3 rows and 1 column.
matrix<double,3,1> y;
// Make a 3 by 3 matrix of doubles for the M matrix.
matrix<double,3,3> M;
// Make a matrix of doubles that has unknown dimensions (the dimensions are
// decided at runtime unlike the above two matrices which are bound at compile
// time). We could declare x the same way as y but I'm doing it differently
// for the purposes of illustration.
matrix<double> x;
// You may be wondering why someone would want to specify the size of a matrix
// at compile time when you don't have to. The reason is two fold. First,
// there is often a substantial performance improvement, especially for small
// matrices, because the compiler is able to perform loop unrolling if it knows
// the sizes of matrices. Second, the dlib::matrix object checks these compile
// time sizes to ensure that the matrices are being used correctly. For example,
// if you attempt to compile the expression y = M; or x = y*y; you will get
// a compiler error on those lines since those are not legal matrix operations.
// So if you know the size of a matrix at compile time then it is always a good
// idea to let the compiler know about it.
// now we need to initialize the y and M matrices and we can do so like this:
M = 54.2, 7.4, 12.1,
1, 2, 3,
5.9, 0.05, 1;
y = 3.5,
1.2,
7.8;
// the solution can be obtained now by multiplying the inverse of M with y
x = inv(M)*y;
cout << "x: \n" << x << endl;
// We can check that it really worked by plugging x back into the original equation
// and subtracting y to see if we get a column vector with values all very close
// to zero (Which is what happens. Also, the values may not be exactly zero because
// there may be some numerical error and round off).
cout << "M*x - y: \n" << M*x - y << endl;
// Also note that we can create run-time sized column or row vectors like so
matrix<double,0,1> runtime_sized_column_vector;
matrix<double,1,0> runtime_sized_row_vector;
// and then they are sized by saying
runtime_sized_column_vector.set_size(3);
// Similarly, the x matrix can be resized by calling set_size(num rows, num columns). For example
x.set_size(3,4); // x now has 3 rows and 4 columns.
// The elements of a matrix are accessed using the () operator like so
cout << M(0,1) << endl;
// The above expression prints out the value 7.4. That is, the value of
// the element at row 0 and column 1.
// Let's compute the sum of elements in the M matrix.
double M_sum = 0;
// loop over all the rows
for (long r = 0; r < M.nr(); ++r)
{
// loop over all the columns
for (long c = 0; c < M.nc(); ++c)
{
M_sum += M(r,c);
}
}
cout << "sum of all elements in M is " << M_sum << endl;
// The above code is just to show you how to loop over the elements of a matrix. An
// easier way to find this sum is to do the following:
cout << "sum of all elements in M is " << sum(M) << endl;
// If we have a matrix that is a row or column vector. That is, it contains either
// a single row or a single column then we know that any access is always either
// to row 0 or column 0 so we can omit that 0 and use the following syntax.
cout << y(1) << endl;
// The above expression prints out the value 1.2
// --------------------------------- Comparison with MATLAB ---------------------------------
// Here I list a set of Matlab commands and their equivalent expressions using the dlib matrix.
matrix<double> A, B, C, D, E;
matrix<int> Aint;
matrix<long> Blong;
// MATLAB: A = eye(3)
A = identity_matrix<double>(3);
// MATLAB: B = ones(3,4)
B = uniform_matrix<double>(3,4, 1);
// MATLAB: C = 1.4*A
C = 1.4*A;
// MATLAB: D = A.*C
D = pointwise_multiply(A,C);
// MATLAB: E = A * B
E = A*B;
// MATLAB: E = A + B
E = A + C;
// MATLAB: E = E'
E = trans(E); // Note that if you want a conjugate transpose then you need to say conj(trans(E))
// MATLAB: E = B' * B
E = trans(B)*B;
double var;
// MATLAB: var = A(1,2)
var = A(0,1); // dlib::matrix is 0 indexed rather than starting at 1 like Matlab.
// MATLAB: C = round(C)
C = round(C);
// MATLAB: C = floor(C)
C = floor(C);
// MATLAB: C = ceil(C)
C = ceil(C);
// MATLAB: C = diag(B)
C = diag(B);
// MATLAB: B = cast(A, "int32")
Aint = matrix_cast<int>(A);
// MATLAB: A = B(1,:)
A = rowm(B,0);
// MATLAB: A = B(:,1)
A = colm(B,0);
// MATLAB: A = [1:5]'
Blong = range(1,5);
// MATLAB: A = [1:5]
Blong = trans(range(1,5));
// MATLAB: A = [1:2:5]
Blong = trans(range(1,2,5));
// MATLAB: A = B([1:3], [1:2])
A = subm(B, range(0,2), range(0,1));
// or equivalently
A = subm(B, rectangle(0,0,1,2));
// MATLAB: A = B([1:3], [1:2:4])
A = subm(B, range(0,2), range(0,2,3));
// MATLAB: B(:,:) = 5
B = 5;
// or equivalently
set_all_elements(B,5);
// or equivalently
set_subm(B,get_rect(B)) = 5;
// MATLAB: B([1:2],[1,2]) = 7
set_subm(B,range(0,1), range(0,1)) = 7;
// MATLAB: B([1:3],[2:3]) = A
set_subm(B,range(0,2), range(1,2)) = A;
// MATLAB: B(:,1) = 4
set_colm(B,0) = 4;
// MATLAB: B(:,1) = B(:,2)
set_colm(B,0) = colm(B,1);
// MATLAB: B(1,:) = 4
set_rowm(B,0) = 4;
// MATLAB: B(1,:) = B(2,:)
set_rowm(B,0) = rowm(B,1);
// MATLAB: var = det(E' * E)
var = det(trans(E)*E);
// MATLAB: C = pinv(E)
C = pinv(E);
// MATLAB: C = inv(E)
C = inv(E);
// MATLAB: [A,B,C] = svd(E)
svd(E,A,B,C);
// MATLAB: A = chol(E,'lower')
A = cholesky_decomposition(E);
// MATLAB: var = min(min(A))
var = min(A);
// ------------------------- Template Expressions -----------------------------
// Now I will discuss the "template expressions" technique and how it is
// used in the matrix object. First consider the following expression:
x = y + y;
/*
Normally this expression results in machine code that looks, at a high
level, like the following:
temp = y + y;
x = temp
Temp is a temporary matrix returned by the overloaded + operator.
temp then contains the result of adding y to itself. The assignment
operator copies the value of temp into x and temp is then destroyed while
the blissful C++ user never sees any of this.
This is, however, totally inefficient. In the process described above
you have to pay for the cost of constructing a temporary matrix object
and allocating its memory. Then you pay the additional cost of copying
it over to x. It also gets worse when you have more complex expressions
such as x = round(y + y + y + M*y) which would involve the creation and copying
of 5 temporary matrices.
All these inefficiencies are removed by using the template expressions
technique. The exact details of how the technique is performed are well
outside the scope of this example but the basic idea is as follows. Instead
of having operators and functions return temporary matrix objects you
return a special object that represents the expression you wish to perform.
So consider the expression x = y + y again. With dlib::matrix what happens
is the expression y+y returns a matrix_exp object instead of a temporary matrix.
The construction of a matrix_exp does not allocate any memory or perform any
computations. The matrix_exp however has an interface that looks just like a
dlib::matrix object and when you ask it for the value of one of its elements
it computes that value on the spot. Only in the assignment operator does
someone ask the matrix_exp for these values so this avoids the use of any
temporary matrices. Thus the statement x = y + y is equivalent to the following
code:
// loop over all elements in y matrix
for (long r = 0; r < y.nr(); ++r)
for (long c = 0; c < y.nc(); ++c)
x(r,c) = y(r,c) + y(r,c);
This technique works for expressions of arbitrary complexity. So if you
typed x = round(y + y + y + M*y) it would involve no temporary matrices being
created at all. Each operator takes and returns only matrix_exp objects.
Thus, no computations are performed until the assignment operator requests
the values from the matrix_exp it receives as input.
There is only one caveat in all of this. It is for statements that involve
the multiplication of a complex matrix_exp such as the following:
*/
x = M*(M+M+M+M+M+M+M);
/*
This statement computes the value of M*(M+M+M+M+M+M+M) totally without
any temporary matrix objects. This sounds good but we should take
a closer look. Consider that the + operator is invoked 6 times. This
means we have something like this:
x = M * (matrix_exp representing M+M+M+M+M+M+M);
M is being multiplied with a quite complex matrix_exp. Now recall that when
you ask a matrix_exp what the value of any of its elements are it computes
their values *right then*.
If you think on what is involved in performing a matrix multiply you will
realize that each element of a matrix is accessed M.nr() times. In the
case of our above expression the cost of accessing an element of the
matrix_exp on the right hand side is the cost of doing 6 addition operations.
Thus, it would be faster to assign M+M+M+M+M+M+M to a real matrix and then
multiply that by M.
So do something like this:
*/
matrix<double,3,3> Mtemp;
Mtemp = M+M+M+M+M+M+M;
x = M*Mtemp;
// Or alternatively you can use the tmp() function like so.
x = M*tmp(M+M+M+M+M+M+M);
/*
tmp() just evaluates a matrix_exp and returns a real matrix object. So it
does the same thing as the above code that uses Mtemp.
Another example of this would be chains of matrix multiplies. For example:
*/
x = M*M*M*M;
// A much faster version of this expression would be
x = tmp(M*M)*tmp(M*M);
/*
Anyway, the point of the above discussion is that you shouldn't multiply
complex matrix expressions. You should instead assign the expression to
a matrix object and then use that object in the multiply. This will ensure
that your multiplies are always fast.
Note however, that the following two expressions are not afflicted with the
above problem:
*/
double value1 = trans(y)*M*y;
double value2 = trans(y)*M*M*y;
/*
These expressions can be evaluated without using temporaries or
needlessly recalculating things as in the case of the above
examples.
!*/
}
// ----------------------------------------------------------------------------------------