I have an equation that is used to make an isosurface which is then saved into a file and I need to keep track of which equations belong to which file. Therefore I want to label the files my Octave script produces with the equation that produced them without labeling them all by hand.
This is my code right now:
clf;
function [f, v] = doiso(dodraw)
m = 3;
dim = -m:0.1:m;
if (dodraw > 0)
dim = -m:0.6:m;
endif
[x,y,z] = meshgrid(dim, dim, dim);
func = cos(x) .* sin(y) + cos(y) .* sin(z) + cos(z) .* sin(x);
if (dodraw > 0)
isosurface(func, 0);
else
[f, v] = isosurface(func, 0);
endif
endfunction
#draw
doiso(1);
axis equal;
title("isosurface() of the function");
#saveq
[f, v] = doiso(0);
vertface2obj(v, f, strcat("objs/", int2str(time * 1000), "out.obj"));
The saved file should have names like cos(x) . sin(y) + cos(y) . sin(z) + cos(z) . sin(x) 1513441860368.obj where the long number is a timestamp and the expression containing sin and cos is the equation that produced the file (same as in the code). Invalid chars will have to be removed or replaced in the file name string.
No online resource seems to mention printing an equation; only printing numbers or solving equations.
One way you could do this is using func2str():
func2str (fcn_handle)
Return a string containing the name of the function referenced by the function handle fcn_handle.
You will have to create an anonymous function for your equation. For example,
> f = #(x,y,z) cos(x) .* sin(y) + cos(y) .* sin(z) + cos(z) .* sin(x);
> eqn = func2str(f);
> fprintf(stdout, '%s\n', eqn)
#(x, y, z) cos (x) .* sin (y) + cos (y) .* sin (z) + cos (z) .* sin (x)
As you can see, the above code creates the string eqn containing the expression of the function f.
You can then manipulate the string to get something more reasonable as a file name. Here's a simple example:
> fname = regexprep(strjoin(strsplit(eqn(11:end)), ''), '[().*+]', '_')
fname = cos_x___sin_y__cos_y___sin_z__cos_z___sin_x_
Here strjoin(strsplit(str), '') removes all whitespace from the string str. The function regexprep() uses regex substitutions to replace the "undesired" characters with an underscore.
You can of course have more elaborate manipulations, such as changing * to _TIMES_ or whatever you prefer.
More about manipulating strings here.
Related
I have a function:
int = input("Introduce the intensity: ")
wave = input("Introduce the wavelength: ");
e = input("Introduce e: ");
function [brighttemp] = brightness_temp_function(int, wave, e)
A = 1.19 .* (10 .^ 8);
B = 1.441 .* (10 .^ 4);
brighttemp = sprintf("%.2f",(B ./ (wave .* (log (1 .+ ((e .* A) ./ (int .* (power(wave,5)))))))) .- 273.15);
disp(brighttemp);
endfunction
[brighttemp] = brightness_temp_function(wave, e, int);`
When I enter a single value for each variable, it outputs a single answer for brighttemp. But when I enter in a vector for one of the variables, such as [8, 9; 7, 8.5] for the int variable and single values for the others, I get back an output like this for brighttemp: 20.727.812.924.3 instead of a vector similar in format to the vector inputted for the int variable, like this [20.7, 27.8; 12.9, 24.3] . What do I have to do to get an output like the latter vector?
I have a simple model where I want to minimize the RMSE between my dependent variable y and my model values. The model is: y = alpha + beta'*x.
For minimization, I am using Matlab's fmincon function and am struggling with multiplying my parameter p(2) by x.
MWE:
% data
y = [5.072, 7.1588, 7.263, 4.255, 6.282, 6.9118, 4.044, 7.2595, 6.898, 4.8744, 6.5179, 7.3434, 5.4316, 3.38, 5.464, 5.90, 6.80, 6.193, 6.070, 5.737]
x = [18.3447, 79.86538, 85.09788, 10.5211, 44.4556, 69.567, 8.960, 86.197, 66.857, 16.875, 52.2697, 93.971, 24.35, 5.118, 25.126, 34.037, 61.4445, 42.704, 39.531, 29.988]
% initial values
p_initial = [0, 0];
% function: SEE BELOW
objective = #(p) sqrt(mean((y - y_mod(p)).^2));
% optimization
[param_opt, fval] = fmincon(objective, p_initial)
If I specify my function as follows then it works.
y_mod = #(p) p(1) + p(2).*x
However, it does not work if I use the following code. How can I multiply p(2) with x? Where x is not optimized, because the values are given.
function f = y_mod(p)
f = p(1) + p(2).*x
end
Here is the output from a script that has the function declaration:
>> modelFitExample2a
RMS Error=0.374, intercept=4.208, slope=0.0388
And here is code for the above. It has many commented lines because it includes alternate ways to fit the data: an inline declaration of y_mod(), or a multi-line declaration of y_mod(), or no y_mod() at all. This version uses the multi-line declaration of y_mod().
%modelFitExample2a.m WCR 2021-01-19
%Reply to stack exchange question on parameter fitting
clear;
global x %need this if define y_mod() separately, and in that case y_mod() must declare x global
% data
y = [5.0720, 7.1588, 7.2630, 4.2550, 6.2820, 6.9118, 4.0440, 7.2595, 6.8980, 4.8744...
6.5179, 7.3434, 5.4316, 3.3800, 5.4640, 5.9000, 6.8000, 6.1930, 6.0700, 5.7370];
x = [18.3447,79.8654,85.0979,10.5211,44.4556,69.5670, 8.9600,86.1970,66.8570,16.8750,...
52.2697,93.9710,24.3500, 5.1180,25.1260,34.0370,61.4445,42.7040,39.5310,29.9880];
% initial values
p_initial = [0, 0];
%predictive model with parameter p
%y_mod = #(p) p(1) + p(2)*x;
% objective function
%If you use y_mod(), then you must define it somewhere
objective = #(p) sqrt(mean((y - y_mod(p)).^2));
%objective = #(p) sqrt(mean((y-p(1)-p(2)*x).^2));
% optimization
options = optimset('Display','Notify');
[param_opt, fval] = fmincon(objective,p_initial,[],[],[],[],[],[],[],options);
% display results
fprintf('RMS Error=%.3f, intercept=%.3f, slope=%.4f\n',...
fval,param_opt(1),param_opt(2));
%function declaration: predictive model
%This is an alternative to the inline definition of y_mod() above.
function f = y_mod(p)
global x
f = p(1) + p(2)*x;
end
carl,
The second method, in which you declare y_mod() explicitly (at the end of your script, or in a separate file y_mod.m), does not work because y_mod() does not know what x is. Fix it by declaring x global in the main program at the top, and declare x global in y_mod().
%function declaration
function f = y_mod(p)
global x
f = p(1) + p(2)*x;
end
Of course you don't need y_mod() at all. The code also works if you use the following, and in this case, no global x is needed:
% objective function
objective = #(p) sqrt(mean((y-p(1)-p(2)*x).^2));
By the way, you don't need to multiply with .* in y_mod. You may use *, because you are multiplying a scalar by a vector.
Goal: Plot the graph using a non-linear function.
Function and graph
This is my first time working at Octave. To plot the graph, I need to calculate a function in the range Fx (0.1 ... 10).
I tried to implement this by looping the function through the for loop, writing the results to an array (x-axis - Fn, y-axis - function value), then loading the arrays into the plot() function.
Fn = 1
Ln = 5
Q = 0.5
function retval = test (Fn, Ln, Q)
# Fn squared (for common used)
Fn = Fn^2
# Node A + Node B
nodeA = Fn * (Ln - 1)
nodeB = (Ln * Fn - 1)^2 + Fn * (Fn - 1)^2 * (Ln - 1)^2 * Q^2
nodeB = sqrt(nodeB)
# Result
result = nodeA / nodeB
retval = result
return;
endfunction
frequencyArray = {}
gainArray = {}
fCount = 1
gCount = 1
for i = 0:0.5:5
# F
Fn = i
frequencyArray{fCount} = Fn
fCount = fCount + 1
# G
gainArray{gCount} = test(Fn, Ln, Q)
gCount = gCount + 1
end
plot(frequencyArray, gainArray);
As a result, I get an error about the format of the arrays.
>> plot(frequencyArray, gainArray);
error: invalid value for array property "xdata"
error: __go_line__: unable to create graphics handle
error: called from
__plt__>__plt2vv__ at line 495 column 10
__plt__>__plt2__ at line 242 column 14
__plt__ at line 107 column 18
plot at line 223 column 10
In addition to the error, I believe that these tasks are solved in more correct ways, but I did not quite understand what to look for.
Questions:
Did I choose the right way to solve the problem? Are there any more elegant ways?
How can I fix this error?
Thank you!
If I have correctly interpreted what you are trying to do, the following should work. Firstly, you need to use the term-by-term versions of all arithmetic operators that act on Fn. These are the same as the normal operators except preceded by a dot. Next, you need to put Fn equal to a vector containing the x-values of all the points you wish to plot and put Q equal to a vector containing the values of Q for which you want to draw curves. Use a for-loop to loop through the values of Q and plot a single curve in each iteration of the loop. You don't need a loop to plot each curve because Octave will apply your "test" function to the whole Fn vector and return the result as a vector of the same size. To plot the curves on a log axis, use the function "semilogx(x, y)" insetad of "plot(x, y)". To make the plots appear on the same figure, rather than separate ones put "hold on" before the loop and "hold off" afterwards. You used cell arrays instead of vectors in your for-loop, which the plotting functions don't accept. Also, you don't need an explicit return statement in an Octave function.
The following code produces a set of curves that look like the ones in the figure you pasted in your question:
Ln = 5
function result = test (Fn, Ln, Q)
# Fn squared (for common used)
Fn = Fn.^2;
# Node A + Node B
nodeA = Fn .* (Ln - 1);
nodeB = (Ln .* Fn .- 1).^2 + Fn .* (Fn .- 1).^2 .* (Ln - 1)^2 * Q^2;
nodeB = sqrt(nodeB);
# Result
result = nodeA ./ nodeB;
endfunction
Fn = linspace(0.1, 10, 500);
Q = [0.1 0.2 0.5 0.8 1 2 5 8 10];
hold on
for q = Q
K = test(Fn, Ln, q);
semilogx(Fn, K);
endfor
hold off
I am trying to make a function in Octave where you give octave a function f(x,y) as a string, a change in X, a change in Y, a starting point, and the size of a matrix, the function will create a matrix populated with the values of f(x,y) at each point in the matrix.
This is for an application that displays a 3d graph, using the matrix to map each value to a block
# funcStr: The function whose Z values are being calculated
# dx: the change in x that each block in the x direction represents
# dy: the change in y that each block in the y direction represents
# startPt: the point (in an array of x, y) that center block represents
# res: the side length (in blocks) of the plane
pkg load symbolic
syms x y
function[zValues] = calculateZValues(funcStr, dx, dy, startPt, res)
zValues = zeros(res);
eqn = #(x, y) inline(funcStr);
startX = startPt{1};
startY = startPt{2};
for yOffset = 1:res
for xOffset = 1:res
xCoord = startX + dx * xOffset;
yCoord = startY + dy * yOffset;
zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
endfor
endfor
endfunction
The error I am getting is:
>> calculateZValues("x*y", 1, 1, {0,0}, 10)
parse error near line 20 of file /home/rahul/Documents/3dGraph/graph/calculateZValues.m
anonymous function bodies must be single expressions
>>> zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
I have no idea what the issue is. I have replaced the #(x,y) part with {x,y} in the line referenced by the error but it says nothing or it raises an error about the function subs not being declared. I have also tried moving the pkg and syms lines above the function header
Given a 2D circle with 2 angles in the range -PI -> PI around a coordinate, what is the value of the smallest angle between them?
Taking into account that the difference between PI and -PI is not 2 PI but zero.
An Example:
Imagine a circle, with 2 lines coming out from the center, there are 2 angles between those lines, the angle they make on the inside aka the smaller angle, and the angle they make on the outside, aka the bigger angle.
Both angles when added up make a full circle. Given that each angle can fit within a certain range, what is the smaller angles value, taking into account the rollover
This gives a signed angle for any angles:
a = targetA - sourceA
a = (a + 180) % 360 - 180
Beware in many languages the modulo operation returns a value with the same sign as the dividend (like C, C++, C#, JavaScript, full list here). This requires a custom mod function like so:
mod = (a, n) -> a - floor(a/n) * n
Or so:
mod = (a, n) -> (a % n + n) % n
If angles are within [-180, 180] this also works:
a = targetA - sourceA
a += (a>180) ? -360 : (a<-180) ? 360 : 0
In a more verbose way:
a = targetA - sourceA
a -= 360 if a > 180
a += 360 if a < -180
x is the target angle. y is the source or starting angle:
atan2(sin(x-y), cos(x-y))
It returns the signed delta angle. Note that depending on your API the order of the parameters for the atan2() function might be different.
If your two angles are x and y, then one of the angles between them is abs(x - y). The other angle is (2 * PI) - abs(x - y). So the value of the smallest of the 2 angles is:
min((2 * PI) - abs(x - y), abs(x - y))
This gives you the absolute value of the angle, and it assumes the inputs are normalized (ie: within the range [0, 2π)).
If you want to preserve the sign (ie: direction) of the angle and also accept angles outside the range [0, 2π) you can generalize the above. Here's Python code for the generalized version:
PI = math.pi
TAU = 2*PI
def smallestSignedAngleBetween(x, y):
a = (x - y) % TAU
b = (y - x) % TAU
return -a if a < b else b
Note that the % operator does not behave the same in all languages, particularly when negative values are involved, so if porting some sign adjustments may be necessary.
An efficient code in C++ that works for any angle and in both: radians and degrees is:
inline double getAbsoluteDiff2Angles(const double x, const double y, const double c)
{
// c can be PI (for radians) or 180.0 (for degrees);
return c - fabs(fmod(fabs(x - y), 2*c) - c);
}
See it working here:
https://www.desmos.com/calculator/sbgxyfchjr
For signed angle:
return fmod(fabs(x - y) + c, 2*c) - c;
In some other programming languages where mod of negative numbers are positive, the inner abs can be eliminated.
I rise to the challenge of providing the signed answer:
def f(x,y):
import math
return min(y-x, y-x+2*math.pi, y-x-2*math.pi, key=abs)
For UnityEngine users, the easy way is just to use Mathf.DeltaAngle.
Arithmetical (as opposed to algorithmic) solution:
angle = Pi - abs(abs(a1 - a2) - Pi);
I absolutely love Peter B's answer above, but if you need a dead simple approach that produces the same results, here it is:
function absAngle(a) {
// this yields correct counter-clock-wise numbers, like 350deg for -370
return (360 + (a % 360)) % 360;
}
function angleDelta(a, b) {
// https://gamedev.stackexchange.com/a/4472
let delta = Math.abs(absAngle(a) - absAngle(b));
let sign = absAngle(a) > absAngle(b) || delta >= 180 ? -1 : 1;
return (180 - Math.abs(delta - 180)) * sign;
}
// sample output
for (let angle = -370; angle <= 370; angle+=20) {
let testAngle = 10;
console.log(testAngle, "->", angle, "=", angleDelta(testAngle, angle));
}
One thing to note is that I deliberately flipped the sign: counter-clockwise deltas are negative, and clockwise ones are positive
There is no need to compute trigonometric functions. The simple code in C language is:
#include <math.h>
#define PIV2 M_PI+M_PI
#define C360 360.0000000000000000000
double difangrad(double x, double y)
{
double arg;
arg = fmod(y-x, PIV2);
if (arg < 0 ) arg = arg + PIV2;
if (arg > M_PI) arg = arg - PIV2;
return (-arg);
}
double difangdeg(double x, double y)
{
double arg;
arg = fmod(y-x, C360);
if (arg < 0 ) arg = arg + C360;
if (arg > 180) arg = arg - C360;
return (-arg);
}
let dif = a - b , in radians
dif = difangrad(a,b);
let dif = a - b , in degrees
dif = difangdeg(a,b);
difangdeg(180.000000 , -180.000000) = 0.000000
difangdeg(-180.000000 , 180.000000) = -0.000000
difangdeg(359.000000 , 1.000000) = -2.000000
difangdeg(1.000000 , 359.000000) = 2.000000
No sin, no cos, no tan,.... only geometry!!!!
A simple method, which I use in C++ is:
double deltaOrientation = angle1 - angle2;
double delta = remainder(deltaOrientation, 2*M_PI);