I am trying to find three parameters (a, b, c) to fit my experimental data using ODE solver and optimization by least squares using Scilab in-built functions.
However, I keep having the message "submatrix incorrectly defined" at line "y_exp(:,1) = [0.135 ..."
When I try another series of data (t, yexp) such as the one used in the original template I get no error messages. The template I use was found here: https://wiki.scilab.org/Non%20linear%20optimization%20for%20parameter%20fitting%20example
function dy = myModel ( t , y , a , b, c )
// The right-hand side of the Ordinary Differential Equation.
dy(1) = -a*y(1) - b*y(1)*y(2)
dy(2) = a*y(1) - b*y(1)*y(2) - c*y(2)
endfunction
function f = myDifferences ( k )
// Returns the difference between the simulated differential
// equation and the experimental data.
global MYDATA
t = MYDATA.t
y_exp = MYDATA.y_exp
a = k(1)
b = k(2)
c = k(3)
y0 = y_exp(1,:)
t0 = 0
y_calc=ode(y0',t0,t,list(myModel,a,b,c))
diffmat = y_calc' - y_exp
// Make a column vector
f = diffmat(:)
MYDATA.funeval = MYDATA.funeval+ 1
endfunction
// Experimental data
t = [0,20,30,45,75,105,135,180,240]';
y_exp(:,1) =
[0.135,0.0924,0.067,0.0527,0.0363,0.02445,0.01668,0.012,0.009]';
y_exp(:,2) =
[0,0.00918,0.0132,0.01835,0.0261,0.03215,0.0366,0.0393,0.0401]';
// Store data for future use
global MYDATA;
MYDATA.t = t;
MYDATA.y_exp = y_exp;
MYDATA.funeval = 0;
function val = L_Squares ( k )
// Computes the sum of squares of the differences.
f = myDifferences ( k )
val = sum(f.^2)
endfunction
// Initial guess
a = 0;
b = 0;
c = 0;
x0 = [a;b;c];
[fopt ,xopt]=leastsq(myDifferences, x0)
Does anyone know how to approach this problem?
Just rewrite lines 28,29 as
y_exp = [0.135,0.0924,0.067,0.0527,0.0363,0.02445,0.01668,0.012,0.009
0,0.00918,0.0132,0.01835,0.0261,0.03215,0.0366,0.0393,0.0401]';
or insert a clear at line 1 (you may have defined y_exp before with a different size).
Related
I've been trying to write a program that can render 4D lines, the specific function doing this gets the lines already rotated, and the function attempts to clip the lines at planes z = p and w = p if needed, and then draw the line to the screen.
I think that I am doing at least most of this properly, however I am unsure, and not having much experience viewing the fourth dimension I cannot tell what might be a visual bug, or what is actually how it should be rendered.
The function first loads a line into two variables, each is one of the two endpoints of the line. If both points are beyond clippl (the clipping plane variable) for z = clippl and w = clippl, it then applies perspective transformation to them, and subsequently renders a line on the screen correspondingly.
If certain logic is met for the points, the function goes through a process of clipping them, and then continues the same as it would outside the clipping planes.
The location of the camera is held in the variables Ox, Oy, Oz, Ow at the beginning of the full program.
I can't tell if I've done this properly, can anyone tell me if this works right as a 4D perspective projection from a first person camera?
EDIT: I've added points to the rendering list that are at the corners of the cube I'm rendering, and it seems to show that there is in fact some problem with the line clipping, as I am fairly certain that the points are rendering properly, and there is not always a line showing up at it. Could the problem have to do with the w = p clip?
Here's the function, the program uses p5.js:
function drawPLines(P){
var lA,lB;
for(var i=0;i<P.length;i++){
lA = [P[i][0],P[i][1],P[i][2],P[i][3]];
lB = [P[i][4],P[i][5],P[i][6],P[i][7]];
//X: ( x*VS+(width*0.5)+(ox*VS) )
//Y: ( y*VS+(height*0.5)+(oy*VS) )
//x: (XV[0]*P[i][0])+(YV[0]*P[i][1])+(ZV[0]*P[i][2])+(WV[0]*P[i][3])
//y: (XV[1]*P[i][0])+(YV[1]*P[i][1])+(ZV[1]*P[i][2])+(WV[1]*P[i][3])
var x0,y0,x1,y1;
//x0 = (XV[0]*lA[0])+(YV[0]*lA[1])+(ZV[0]*lA[2])+(WV[0]*lA[3]);
//y0 = (XV[1]*lA[0])+(YV[1]*lA[1])+(ZV[1]*lA[2])+(WV[1]*lA[3]);
//new rendering pipeline
//old rendering pipeline
if(lA[2]>clippl&&lB[2]>clippl&&lA[3]>clippl&&lB[3]>clippl){
x0 = XV[0]*lA[0];
y0 = YV[1]*lA[1];
x0 = (x0/lA[3])/(lA[2]/lA[3]);
y0 = (y0/lA[3])/(lA[2]/lA[3]);
//console.log(y);
x0 = ( x0*VS+(width*0.5)+(ox*VS) );
y0 = ( y0*VS+(height*0.5)+(oy*VS) );
//x1 = (XV[0]*lB[0])+(YV[0]*lB[1])+(ZV[0]*lB[2])+(WV[0]*lB[3]);
//y1 = (XV[1]*lB[0])+(YV[1]*lB[1])+(ZV[1]*lB[2])+(WV[1]*lB[3]);
x1 = XV[0]*lB[0];
y1 = YV[1]*lB[1];
x1 = (x1/lB[3])/(lB[2]/lB[3]);
y1 = (y1/lB[3])/(lB[2]/lB[3]);
//console.log(y);
x1 = ( x1*VS+(width*0.5)+(ox*VS) );
y1 = ( y1*VS+(height*0.5)+(oy*VS) );
stroke([P[i][8],P[i][9],P[i][10],P[i][11]]);
line(x0,y0,x1,y1);
}else if((lA[2]>clippl||lA[3]>clippl||lB[2]>clippl||lB[3]>clippl)){
var V = 0;
var zV = 0;
var wV = 0;
//var oV = 0;
if(lA[2]>clippl&&lA[3]>clippl){
V++;
}else if(lA[2]>clippl&&lA[3]<=clippl){
zV++;
}else if(lA[2]<=clippl&&lA[3]>clippl){
wV++;
}/*else{
oV++;
}*/
if(lB[2]>clippl&&lB[3]>clippl){
V++;
}else if(lB[2]>clippl&&lB[3]<=clippl){
zV++;
}else if(lB[2]<=clippl&&lB[3]>clippl){
wV++;
}/*else{
oV++;
}*/
if((V==1)||(wV==1&&(V==1||zV==1))||(zV==1&&(V==1||wV==1))){
var lin = lB;
var out = lA;
if(lA[2]<=clippl){
out = lB;
lin = lA;
}
if(lin[2]<=clippl){
lin = [((((lA[0]-lB[0])*clippl)-((lA[0]-lB[0])*lB[2]))/(lA[2]-lB[2]))+lB[0],((((lA[1]-lB[1])*clippl)-((lA[1]-lB[1])*lB[2]))/(lA[2]-lB[2]))+lB[1],clippl,((((lA[3]-lB[3])*clippl)-((lA[3]-lB[3])*lB[2]))/(lA[2]-lB[2]))+lB[3]];
}
if((lA[2]-lB[2])!==0){
lA = lin;
lB = out;
}
lin = lA;
out = lB;
if(lB[3]<=clippl){
out = lA;
lin = lB;
}
if(lin[3]<=clippl){
lin = [((((lA[0]-lB[0])*clippl)-((lA[0]-lB[0])*lB[3]))/(lA[3]-lB[3]))+lB[0],((((lA[1]-lB[1])*clippl)-((lA[1]-lB[1])*lB[3]))/(lA[3]-lB[3]))+lB[1],((((lA[2]-lB[2])*clippl)-((lA[2]-lB[2])*lB[3]))/(lA[3]-lB[3]))+lB[2],clippl];
//alert(lin);
//alert(out);
}
if((lA[3]-lB[3])!==0){
lA = lin;
lB = out;
}
if(lA[2]>clippl||lB[2]>clippl||lA[3]>clippl||lB[3]>clippl){
x0 = XV[0]*lA[0];
y0 = YV[1]*lA[1];
x0 = (x0/lA[3])/(lA[2]/lA[3]);
y0 = (y0/lA[3])/(lA[2]/lA[3]);
//console.log(y);
x0 = ( x0*VS+(width*0.5)+(ox*VS) );
y0 = ( y0*VS+(height*0.5)+(oy*VS) );
//x1 = (XV[0]*lB[0])+(YV[0]*lB[1])+(ZV[0]*lB[2])+(WV[0]*lB[3]);
//y1 = (XV[1]*lB[0])+(YV[1]*lB[1])+(ZV[1]*lB[2])+(WV[1]*lB[3]);
x1 = XV[0]*lB[0];
y1 = YV[1]*lB[1];
x1 = (x1/lB[3])/(lB[2]/lB[3]);
y1 = (y1/lB[3])/(lB[2]/lB[3]);
//console.log(y);
x1 = ( x1*VS+(width*0.5)+(ox*VS) );
y1 = ( y1*VS+(height*0.5)+(oy*VS) );
stroke([P[i][8],P[i][9],P[i][10],P[i][11]]);
line(x0,y0,x1,y1);
}
}
}
}
}
You can see the full program at https://editor.p5js.org/hpestock/sketches/Yfagz4Bz3
First, a picture:
Column A is my source data, 50 points.
Column C and D are the SMA calculated with numpy and mathdotnet.com, respectively, with a window of 15.
Column F is the delta.
As we can see, about halfway, the data becomes identical, but the first half is not. I do not understand why, and, more importantly, do not know what to trust.
So I got from SO an optimized version of the SMA and ran the data through it.
The code is here:
private static NDArray SMA(this NDArray Data, int Period)
{
var Length = Data.len;
// calculate the moving average
var Buffer = new double[Period];
var Output = new double[Length];
var CurrentIndex = 0;
for (var i = 0; i < Length; i++)
{
Buffer[CurrentIndex] = Data.GetDouble(i) / Period;
var MA = 0.0;
for (var j = 0; j < Period; j++)
{
MA += Buffer[j];
}
Output[i] = MA;
CurrentIndex = (CurrentIndex + 1) % Period;
}
var R = new ArraySegment<double>(Output, Period - 1, Length - Period + 1);
return new NDArray(R.ToArray());
}
It is using NumSharp, the .net port of numpy, to hold the source array.
While it is all different code, the C# code and python numpy output the same results (differences happen after the 12th decimal point, so we can consider them identical).
This points out to mathdotnet.com being different; so I guess I can trust the numpy / C# versions more.
Are there different variations of the SMA that could cause this? or something obvious I don't see?
I have put all the data here: https://pastebin.com/WgYJUUJF
Edit:
Here is the numpy code:
import numpy as np
def calcSma(data, smaPeriod):
j = next(i for i, x in enumerate(data) if x is not None)
our_range = range(len(data))[j + smaPeriod - 1:]
empty_list = [None] * (j + smaPeriod - 1)
sub_result = [np.mean(data[i - smaPeriod + 1: i + 1]) for i in our_range]
return np.array(empty_list + sub_result)
def calcSma2(data_set, periods=3):
weights = np.ones(periods) / periods
return np.convolve(data_set, weights, mode='valid')
a = np.array([1.1282553063375, 1.13157696082132, 1.13275406120136, 1.1332879715733, 1.12761933580452, 1.12621836040801, 1.12282485875706, 1.12265572041877, 1.13094386506532, 1.12320520490577, 1.12427293064877, 1.1328332027022, 1.13099445663901, 1.12843355605048, 1.13002750724853, 1.12843355605048, 1.13099445663901, 1.12709476494142, 1.12684879712348, 1.12672349888807, 1.12600933402474, 1.13112070248549, 1.12985951088976, 1.12822416032659, 1.12471789559362, 1.12651004224413, 1.12442669033881, 1.12334638977164, 1.12714333124378, 1.1312233808195, 1.12713229372575, 1.128255040952, 1.12585669781931, 1.12763457442902, 1.12470631424376, 1.12223443223443, 1.12506842815956, 1.12691187181355, 1.12385654130971, 1.13026344596074, 1.12237927400894, 1.1245915922457, 1.13088395780284, 1.13211944646759, 1.12590649028825, 1.12829127560895, 1.11876736364966, 1.12222667492441, 1.12169543369019, 1.12199031071285])
b = calcSma(a, 15)
c = calcSma2(a, 15)
print b
print "----------------------------------"
print c
and here is the mathdotnet one:
var data = Vector<double>.Build.Dense(new[] { 1.1282553063375, 1.13157696082132, 1.13275406120136, 1.1332879715733, 1.12761933580452, 1.12621836040801, 1.12282485875706, 1.12265572041877, 1.13094386506532, 1.12320520490577, 1.12427293064877, 1.1328332027022, 1.13099445663901, 1.12843355605048, 1.13002750724853, 1.12843355605048, 1.13099445663901, 1.12709476494142, 1.12684879712348, 1.12672349888807, 1.12600933402474, 1.13112070248549, 1.12985951088976, 1.12822416032659, 1.12471789559362, 1.12651004224413, 1.12442669033881, 1.12334638977164, 1.12714333124378, 1.1312233808195, 1.12713229372575, 1.128255040952, 1.12585669781931, 1.12763457442902, 1.12470631424376, 1.12223443223443, 1.12506842815956, 1.12691187181355, 1.12385654130971, 1.13026344596074, 1.12237927400894, 1.1245915922457, 1.13088395780284, 1.13211944646759, 1.12590649028825, 1.12829127560895, 1.11876736364966, 1.12222667492441, 1.12169543369019, 1.12199031071285 });
var sma = Vector<double>.Build.Dense(data.MovingAverage(15).Skip(14).ToArray());
var s = sma.Aggregate(string.Empty, (Current, v) => Current + $"{v}, ");
Console.WriteLine(s);
I have found many times a solution for my problems from here, but this time I am totally baffled. I don't know what's wrong at my code.
I made a code to create a box with charged particles inside with Vpython. As I launch the program, I get only a grey screen and the program crash. No error message, nothing.
from visual import *
from random import *
def electronizer(num):
list = []
electron_charge = -1.60217662e-19
electron_mass = 9.10938356e-31
for i in range(num):
another_list = []
e = sphere(pos=(random(), random(),random()), radius=2.818e-15,
color=color.cyan)
e.v = vector(random(), random(), random())
another_list.append(e)
another_list.append(e.v)
another_list.append(electron_charge)
another_list.append(electron_mass)
list.append(another_list)
return list
def protonizer(num):
list = []
proton_charge = 1.60217662e-19
proton_mass = 1.6726219e-27
for i in range(num):
another_list = []
p = sphere(pos=(random(), random(),random()), radius=0.8408739e-15, color=color.red)
p.v = vector(random(), random(), random())
another_list.append(p)
another_list.append(p.v)
another_list.append(proton_charge)
another_list.append(proton_mass)
list.append(another_list)
return list
def cross(a, b):
c = vector(a[1]*b[2] - a[2]*b[1],
a[2]*b[0] - a[0]*b[2],
a[0]*b[1] - a[1]*b[0])
return c
def positioner(work_list):
k = 8.9875517873681764e3 #Nm2/C2
G = 6.674e-11 # Nm2/kg2
vac_perm = 1.2566370614e-6 # H/m
pi = 3.14159265
dt = 0.1e-3
constant = 1
force = vector(0,0,0)
for i in range(len(work_list)):
for j in range(len(work_list)):
if i != j:
r = work_list[i][0].pos - work_list[j][0].pos
r_mag = mag(r)
r_norm = norm(r)
F = k * ((work_list[i][2] * work_list[j][2]) / (r_mag**2)) * r_norm
force += F
B = constant*(vac_perm / 4*pi) * (cross(work_list[j][2] * work_list[j][1], norm(r)))/r_mag**2
F = cross(work_list[i][2] * work_list[i][1], B)
force += F
F = -(G * work_list[i][3] * work_list[j][3]) / r_mag**2 * r_norm
force += F
acceleration = force / work_list[i][3]
difference_in_velocity = acceleration * dt
work_list[i][1] += difference_in_velocity
difference_in_position = work_list[i][1] * dt
work_list[i][0].pos += difference_in_position
if abs(work_list[i][0].pos[0]) > 2.5e-6:
work_list[i][1][0] = -work_list[i][1][0]
elif abs(work_list[i][0][1]) > 2.5e-6:
work_list[i][1][1] = -work_list[i][1][1]
elif abs(work_list[i][0][2]) > 2.5e-6:
work_list[i][1][2] = -work_list[i][1][2]
return work_list
box = box(pos=(0, 0, 0), length = 5e-6, width = 5e-6, height = 5e-6, opacity = 0.5)
protons_num = raw_input("number of protons: ")
electrons_num = raw_input("number of electrons: ")
list_of_electrons = electronizer(int(electrons_num))
list_of_protons = protonizer(int(protons_num))
work_list = list_of_electrons + list_of_protons
while True:
work_list = positioner(work_list)
You should ask your question on the VPython.org forum where the VPython experts hang out and will be able to answer your question. You should mention which operating system you are using and which version of python you are using. From your code I see that you are using classic VPython. There is a newer version of VPython 7 that just came out but the VPython syntax has changed.
I have tried to write a function that takes in notes in MIDI form (C2,A4,Bb6) and returns their respective frequencies in hertz. I'm not sure what the best method of doing this should be. I am torn between two approaches. 1) a list based one where I can switch on an input and return hard-coded frequency values given that I may only have to do this for 88 notes (in the grand piano case). 2) a simple mathematical approach however my math skills are a limitation as well as converting the input string into a numerical value. Ultimately I've been working on this for a while and could use some direction.
You can use a function based on this formula:
The basic formula for the frequencies of the notes of the equal
tempered scale is given by
fn = f0 * (a)n
where
f0 = the frequency of one fixed note which must be defined. A common choice is setting the A above middle C (A4) at f0 = 440 Hz.
n = the number of half steps away from the fixed note you are. If you are at a higher note, n is positive. If you are on a lower note, n is negative.
fn = the frequency of the note n half steps away. a = (2)1/12 = the twelth root of 2 = the number which when multiplied by itself 12 times equals 2 = 1.059463094359...
http://www.phy.mtu.edu/~suits/NoteFreqCalcs.html
In Objective-C, this would be:
+ (double)frequencyForNote:(Note)note withModifier:(Modifier)modifier inOctave:(int)octave {
int halfStepsFromA4 = note - A;
halfStepsFromA4 += 12 * (octave - 4);
halfStepsFromA4 += modifier;
double frequencyOfA4 = 440.0;
double a = 1.059463094359;
return frequencyOfA4 * pow(a, halfStepsFromA4);
}
With the following enums defined:
typedef enum : int {
C = 0,
D = 2,
E = 4,
F = 5,
G = 7,
A = 9,
B = 11,
} Note;
typedef enum : int {
None = 0,
Sharp = 1,
Flat = -1,
} Modifier;
https://gist.github.com/NickEntin/32c37e3d31724b229696
Why don't you use a MIDI pitch?
where f is the frequency, and d the MIDI data.
Till now x has two colomns and there was no problems, but now x have got various num of colomns, and I don't know how to write analog code but with dynamic number of colomns in x?
min_x = min(x);
max_x = max(x);
step = (max_x - min_x)/50;
[X, Y] = ndgrid(min_x(1):step(1):max_x(1), min_x(2):step(2):max_x(2));
You can use cellarrays to generate a comma separated list:
%# sample data
x = rand(10,3); %# you can change the column numbers here
%# calculate step sizes
mn = min(x);
mx = max(x);
step = (mx-mn)/50;
%# vec{i} = mn(i):s(i):mx(i)
vec = arrayfun(#(a,s,b)a:s:b, mn,step,mx, 'UniformOutput',false);
%# [X,Y,...] = ndgrid(vec{1},vec{2},...)
C = cell(1,numel(vec));
[C{:}] = ndgrid( vec{:} );
%# result = [X(:),Y(:),...]
result = cell2mat( cellfun(#(v)v(:), C, 'UniformOutput',false) );