im working on octave not sure if it would run in matlab, im trying to do a simple lagrange polynomial that plot the points, the function and the lagrange aproximation, so i did the following,
clear all
clc
function [y] = lagrange(x, x0, y0)
n = size(x0,1);
y = 0;
for i=1:n
p = 1;
for j=1:n
if j == i % avoiding division by 0
continue;
endif;
p *= (x-x0(j)) / (x0(i)-x0(j));
endfor;
y += y0(i) * p;
endfor;
endfunction;
x=[0:0.1:5];
x0=[2;2.75;4];
y0=[1/2;1/2.75;1/4];
y=lagrange(x,x0,y0);
I'm having the following problem, "operator *: nonconformant arguments (op1 is 1x41, op2 is 1x41)" which only appears when using a vector in x, if i try and evaluate for example lagrange(3,x0,y0) in a specific point, the function works correctly and there is no problem with it, the problem is when you use a larger vector for x,
So my question is, is there a way i can modify the function so it will work with a larger vector x, or there is a way i can plot the function without using the larger vector directly in the function?
The line
p *= (x-x0(j)) / (x0(i)-x0(j));
means
p = p * (x-x0(j)) / (x0(i)-x0(j));
This * means matrix multiplication, and one can't multiply two matrices (1, 41) and (1, 41): inner dimensions do not match.
What you need there is elementwise multiplication,
p = p .* (x-x0(j)) / (x0(i)-x0(j));
(See docs for an explanation of the difference).
It could be written as
p .*= (x-x0(j)) / (x0(i)-x0(j));
But I suggest avoiding such contraction; in Matlab/Octave it's not used as much as in some other languages, possibly because placing two operands side by side makes the errors in binary operation choice more apparent.
Related
Andrew Ng's course in Coursera, which Stanford's Machine Learning course, features programming assignments that deal with implementing the algorithms taught in class. The goal of this assignment is to implement linear regression through gradient descent with an input set of X, y, theta, alpha (learning rate), and number of iterations.
I implemented this solution in Octave, the prescribed language in the course.
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
numJ = size(theta, 1);
for iter = 1:num_iters
for i = 1:m
for j = 1:numJ
temp = theta(j) - alpha /m * X(i, j) * (((X * theta)(i, 1)) - y(i, 1));
theta(j) = temp
end
prediction = X * theta;
J_history(iter, 1) = computeCost(X,y,theta)
end
end
On the other hand, here is the cost function:
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
prediction = X * theta;
error = (prediction - y).^2;
J = 1/(2 * m) .* sum(error);
end
This does not pass the submit() function. The submit() function simply validates the data through passing an unknown test case.
I have checked other questions on StackOverflow but I really don't get it. :)
Thank you very much!
Your gradient seems to be correct and as already pointed out in the answer given by #Kasinath P, it is likely that the problem is that the code is too slow. You just need to vectorize it. In Matlab/Octave, you usually need to avoid for loops (note that although you have parfor in Matlab, it is not yet available in octave). So it is always better, performance-wise, to write something like A*x instead of iterating over each row of A with a for loop. You can read about vectorization here.
If I understand correctly, X is a matrix of size m*numJ where m is the number of examples, and numJ is the number of features (or the dimension of the space where each point lies. In that case, you can rewrite your cost function as
(1/(2*m)) * (X*theta-y)'*(X*theta-y);%since ||v||_2^2=v'*v for any vector v in Euclidean space
Now, we know from basic matrix calculus that for any two vectors s and v that are functions from R^{num_J} to R^m, the Jacobian of s^{t}v is given by
s^{t}Jacobian(v)+v^{t}*Jacobian(s) %this Jacobian will have size 1*num_J.
Applying that to your cost function, we obtain
jacobian=(1/m)*(theta'*X'-y')*X;
So if you just replace
for i = 1:m
for j = 1:numJ
%%% theta(j) updates
end
end
with
%note that the gradient is the transpose of the Jacobian we've computed
theta-=alpha*(1/m)*X'*(X*theta-y)
you should see a great increase in performance.
your computecost code is correct
and Better follow the vectorized implementation of Gradient Descent.
You are just iterating and it is slow and may have error.
That course aims you to do vectorized implementation as it is simple and handy at the same time.
I knew this because I did that after sweating a lot.
Vectorization is good:)
Question
Trying to follow Gradients, Gradient Plots and Tangent Planes.
The gradient vectors of (X^2 + Y^2) do not show up correctly in Octave 4.2.0 on Windows. With the code, expected the gradients of a circle diverge from center outwards. However the actual is diagonal.
Please assist to understand what is wrong.
syms x y
f1 = x^2 + y^2;
gradf1 = jacobian(f1,[x,y]);
f1fun = function_handle(f1);
f1xfun = function_handle(gradf1(1));
f1yfun = function_handle(gradf1(2));
[xx, yy] = meshgrid(-1:.1:1,-1:.1:1);
hold on
contour(xx, yy, f1fun(xx, yy), 10)
quiver(xx, yy, f1xfun(xx, yy), f1yfun(xx, yy), 0.5)
axis equal tight
hold off
Expected
Actual
When you perform:
f1xfun = function_handle(gradf1(1));
f1yfun = function_handle(gradf1(2));
The output is:
f1xfun =
#(x) 2 * x % note: single-argument function
f1yfun =
#(y) 2 * y % note: single-argument function
that is AS OPPOSED TO
f1xfun =
#(x,y) 2 * x % two-argument function
f1yfun =
#(x,y) 2 * y % two-argument function
which is what you seem to think was happening. (i.e. the resulting functions actually only take a single input, not both x and y).
Therefore later on when you call f1yfun with two inputs, the second input (i.e. y) is simply silently discarded, and you are essentially calculating 2*x in both axes, hence the diagonal arrows.
tl;dr Your call to quiver should be:
quiver(xx, yy, f1xfun(xx), f1yfun(yy), 0.5);
I think you have a bug in your code and the call to quiver should be
quiver(xx, yy, f1xfun(xx), f1yfun(yy), 0.5)
which then gives (with colormap("jet"))
I'm implementing the system in this paper and I've come a little unstuck correctly implementing the radial tensor field.
All tensors in this system are of the form given on page 3, section 4
R [ cos(2t), sin(2t); sin(2t), -cos(2t) ]
The radial tensor field is defined as:
R [ yy - xx, -2xy; -2xy, -(yy-xx) ]
In my system I'm only storing R and Theta, since I can calculate the tensor based off just that information. This means I need to calculate R and Theta for the radial tensor. Unfortunately, my attempts at this have failed. Although it looks correct, my solution fails in the top left and bottom right quadrants.
Addendum: Following on from discussion in the comments about the image of the system not working, I'll put some hard numbers here too.
The entire tensor field is 800x480, the center point is at { 400, 240 }, and we're using the standard graphics coordinate system with a negative y axis (ie. origin in the top left).
At { 400, 240 }, the tensor is R = 0, T = 0
At { 200, 120 }, the tensor is R = 2.95936E+9, T = 2.111216
At { 600, 120 }, the tensor is R = 2.95936E+9, T = 1.03037679
I can easily sample any more points which you think may help.
The code I'm using to calculate values is:
float x = i - center.X;
float xSqr = x * x;
float y = j - center.Y;
float ySqr = y * y;
float r = (float)Math.Pow(xSqr + ySqr, 2);
float theta = (float)Math.Atan2((-2 * x * y), (ySqr - xSqr)) / 2;
if (theta < 0)
theta += MathHelper.Pi;
Evidently you are comparing formulas (1) and (2) of the paper. Note the scalar multiple l = || (u_x,u_y) || in formula (1), and identify that with R early in the section. This factor is implicit in formula (2), so to make them match we have to factor R out.
Formula (2) works with an offset from the "center" (x0,y0) of the radial map:
x = xp - x0
y = yp - y0
to form the given 2x2 matrix:
y^2 - x^2 -2xy
-2xy -(y^2 - x^2)
We need to factor out a scalar R from this matrix to get a traceless orthogonal 2x2 matrix as in formula (1):
cos(2t) sin(2t)
sin(2t) -cos(2t)
Since cos^2(2t) + sin^2(2t) = 1 the factor R can be identified as:
R = (y^2 - x^2)^2 + (-2xy)^2 = (x^2 + y^2)^2
leaving a traceless orthogonal 2x2 matrix:
C S
S -C
from which the angle 'tan(2t) = S/C` can be extracted by an inverse trig function.
Well, almost. As belisarius warns, we need to check that angle t is in the correct quadrant. The authors of the paper write at the beginning of Sec. 4 that their "t" (which refers to the tensor) depends on R >= 0 and theta (your t) lying in [0,2pi) according to the formula R [ cos(2t), sin(2t); sin(2t) -cos(2t) ].
Since sine and cosine have period 2pi, t (theta) is only uniquely determined up to an interval of length pi. I suspect the authors meant to write either that 2t lies in [0,2pi) or more simply that t lies in [0,pi). belisarius suggestion to use "the atan2 equivalent" will avoid any division by zero. We may (if the function returns a negative value) need to add pi so that t >= 0. This amounts to adding 2pi to 2t, so it doesn't affect the signs of the entries in the traceless orthogonal matrix (since 'R >= 0` the pattern of signs should agree in formulas (1) and (2) ).
Hi
I am trying to sum two function handles, but it doesn't work.
for example:
y1=#(x)(x*x);
y2=#(x)(x*x+3*x);
y3=y1+y2
The error I receive is "??? Undefined function or method 'plus' for input arguments of type 'function_handle'."
This is just a small example, in reality I actually need to iteratively sum about 500 functions that are dependent on each other.
EDIT
The solution by Clement J. indeed works but I couldn't manage to generalize this into a loop and ran into a problem. I have the function s=#(x,y,z)((1-exp(-x*y)-z)*exp(-x*y)); And I have a vector v that contains 536 data points and another vector w that also contains 536 data points. My goal is to sum up s(v(i),y,w(i)) for i=1...536 Thus getting one function in the variable y which is the sum of 536 functions. The syntax I tried in order to do this is:
sum=#(y)(s(v(1),y,z2(1)));
for i=2:536
sum=#(y)(sum+s(v(i),y,z2(i)))
end
The solution proposed by Fyodor Soikin works.
>> y3=#(x)(y1(x) + y2(x))
y3 =
#(x) (y1 (x) + y2 (x))
If you want to do it on multiple functions you can use intermediate variables :
>> f1 = y1;
>> f2 = y2;
>> y3=#(x)(f1(x) + f2(x))
EDIT after the comment:
I'm not sure to understand the problem. Can you define your vectors v and w like that outside the function :
v = [5 4]; % your 536 data
w = [4 5];
y = 8;
s=#(y)((1-exp(-v*y)-w).*exp(-v*y))
s_sum = sum(s(y))
Note the dot in the multiplication to do it element-wise.
I think the most succinct solution is given in the comment by Mikhail. I'll flesh it out in more detail...
First, you will want to modify your anonymous function s so that it can operate on vector inputs of the same size as well as scalar inputs (as suggested by Clement J.) by using element-wise arithmetic operators as follows:
s = #(x,y,z) (1-exp(-x.*y)-z).*exp(-x.*y); %# Note the periods
Then, assuming that you have vectors v and w defined in the given workspace, you can create a new function sy that, for a given scalar value of y, will sum across s evaluated at each set of values in v and w:
sy = #(y) sum(s(v,y,w));
If you want to evaluate this function using an array of values for y, you can add a call to the function ARRAYFUN like so:
sy = #(y) arrayfun(#(yi) sum(s(v,yi,w)),y);
Note that the values for v and w that will be used in the function sy will be fixed to what they were when the function was created. In other words, changing v and w in the workspace will not change the values used by sy. Note also that I didn't name the new anonymous function sum, since there is already a built-in function with that name.
I have a question i know a line i just know its slope(m) and a point on it A(x,y) How can i calculate the points(actually two of them) on this line with a distance(d) from point A ???
I m asking this for finding intensity of pixels on a line that pass through A(x,y) with a distance .Distance in this case will be number of pixels.
I would suggest converting the line to a parametric format instead of point-slope. That is, a parametric function for the line returns points along that line for the value of some parameter t. You can represent the line as a reference point, and a vector representing the direction of the line going through that point. That way, you just travel d units forward and backward from point A to get your other points.
Since your line has slope m, its direction vector is <1, m>. Since it moves m pixels in y for every 1 pixel in x. You want to normalize that direction vector to be unit length so you divide by the magnitude of the vector.
magnitude = (1^2 + m^2)^(1/2)
N = <1, m> / magnitude = <1 / magnitude, m / magnitude>
The normalized direction vector is N. Now you are almost done. You just need to write the equation for your line in parameterized format:
f(t) = A + t*N
This uses vector math. Specifically, scalar vector multiplication (of your parameter t and the vector N) and vector addition (of A and t*N). The result of the function f is a point along the line. The 2 points you are looking for are f(d) and f(-d). Implement that in the language of your choosing.
The advantage to using this method, as opposed to all the other answers so far, is that you can easily extend this method to support a line with "infinite" slope. That is, a vertical line like x = 3. You don't really need the slope, all you need is the normalized direction vector. For a vertical line, it is <0, 1>. This is why graphics operations often use vector math, because the calculations are more straight-forward and less prone to singularities.
It may seem a little complicated at first, but once you get the hang of vector operations, a lot of computer graphics tasks get a lot easier.
Let me explain the answer in a simple way.
Start point - (x0, y0)
End point - (x1, y1)
We need to find a point (xt, yt) at a distance dt from start point towards end point.
The distance between Start and End point is given by d = sqrt((x1 - x0)^2 + (y1 - y0)^2)
Let the ratio of distances, t = dt / d
Then the point (xt, yt) = (((1 - t) * x0 + t * x1), ((1 - t) * y0 + t * y1))
When 0 < t < 1, the point is on the line.
When t < 0, the point is outside the line near to (x0, y0).
When t > 1, the point is outside the line near to (x1, y1).
Here's a Python implementation to find a point on a line segment at a given distance from the initial point:
import numpy as np
def get_point_on_vector(initial_pt, terminal_pt, distance):
v = np.array(initial_pt, dtype=float)
u = np.array(terminal_pt, dtype=float)
n = v - u
n /= np.linalg.norm(n, 2)
point = v - distance * n
return tuple(point)
Based on the excellent answer from #Theophile here on math stackexchange.
Let's call the point you are trying to find P, with coordinates px, py, and your starting point A's coordinates ax and ay. Slope m is just the ratio of the change in Y over the change in X, so if your point P is distance s from A, then its coordinates are px = ax + s, and py = ay + m * s. Now using Pythagoras, the distance d from A to P will be d = sqrt(s * s + (m * s) * (m * s)). To make P be a specific D units away from A, find s as s = D/sqrt(1 + m * m).
I thought this was an awesome and easy to understand solution:
http://www.physicsforums.com/showpost.php?s=f04d131386fbd83b7b5df27f8da84fa1&p=2822353&postcount=4