I have a to-report that divide a range into 11 steps, like this:
to-report random11steps
let step random 11 + 1
let step-value (minLM / 11)
let rnd 0
repeat step [
set rnd (rnd + step-value)
]
report rnd
end
where minLM is a global variable set to 0.005. When I run the procedure minLM changes to be '0.
If I substitute minLM by 0.005 the function works but I do not know why minLM changes to 0.
Besides, is there any primitive to make the code of this procedure a bit shorter?
Regards
This function is not changing minLM; look elsewhere.
Don't write functions that depend on global variables unless absolutely necessary; pass minLM as an argument.
It is unclear why you want to use repeated addition instead of multiplication. That is, you could just return (step * step-value).
Related
I am new to Julia and trying to use the Julia package DifferentialEquations to simultaneously solve for several conditions of the same set of coupled ODEs. My system is a model of an experiment and in one of the conditions, I increase the amount of one of the dependent variables at mid-way through the process.
I would like to be able to adjust the condition of this single trajectory, however so far I am only able to adjust all the trajectories at once. Is it possible to access a single one using callbacks? If not, is there a better way to do this?
Here is a simplified example using the lorentz equations for what I want to be doing:
#Differential Equations setup
function lorentz!(du,u,p,t)
a,r,b=p
du[1]= a*(u[2]-u[1])
du[2]=u[1]*(r-u[3])-u[2]
du[3]=u[1]*u[2]-b*u[3];
end
#function to cycle through inital conditions
function prob_func(prob,i,repeat)
remake(prob; u0 = u0_arr[i]);
end
#inputs
t_span=[(0.0,100.0),(0.0,100.0)];
u01=[0.0;1.0;0.0];
u02=[0.0;1.0;0.0];
u0_arr = [u01,u02];
p=[10.,28.,8/3];
#initialising the Ensemble Problem
prob = ODEProblem(lorentz!,u0_arr[1],t_span[1],p);
CombinedProblem = EnsembleProblem(prob,
prob_func = prob_func, #-> (prob),#repeat is a count for how many times the trajectories had been repeated
safetycopy = true # determines whether a safetly deepcopy is called on the prob before the prob_func (sounds best to leave as true for user-given prob_func)
);
#introducing callback
function condition(u,t,repeat)
return 50 .-t
end
function affect!(repeat)
repeat.u[1]=repeat.u[1] +50
end
callback = DifferentialEquations.ContinuousCallback(condition, affect!)
#solving
sim=solve(CombinedProblem,Rosenbrock23(),EnsembleSerial(),trajectories=2,callback=callback);
# Plotting for ease of understanding example
plot(sim[1].t,sim[1][1,:])
plot!(sim[2].t,sim[2][1,:])
I want to produce something like this:
Example_desired_outcome
But this code produces:
Example_current_outcome
Thank you for your help!
You can make that callback dependent on a parameter and make the parameter different between problems. For example:
function f(du,u,p,t)
if p == 0
du[1] = 2u[1]
else
du[1] = -2u[1]
end
du[2] = -u[2]
end
condition(t,u,integrator) = u[2] - 0.5
affect!(integrator) = integrator.prob.p = 1
For more information, check out the FAQ on this topic: https://diffeq.sciml.ai/stable/basics/faq/#Switching-ODE-functions-in-the-middle-of-integration
I am usually programming in Python, but for an assignment, I am using Simulink. I am wondering why the above elseif ladder does not generate an incremental increase of the variable [IP3] over time. What I would think it should do is return 0.01 until t = 500, then 0.03 until t = 1000, then 0.1 until 1500, 1 until 2000, and 10 from then on. Apologies for the older image btw, I updated the variables in the mean time.
In the Simulink model that you showed, elseif parts will never execute since:
if u1>0 is satisfied, none of the other conditions will be checked and thus it will always be returning 0.01 for all u1>0.
And when u1<=0, all the conditions will be checked but none of them
will be satisfied. (u1 may never be less than zero in your case as u1 is time).
This behavior is same in every programming language.
Fixing your If-elseif Statements:
You need to add this in the If block:
Under If expression (e.g. u1 ~= 0), write this:
u1>0 & u1<=500
Under Elseif expressions (comma-separated list, e.g. u2 ~= 0, u3(2) < u2):, write this:
u1>500 & u1<=1000, u1>1000 & u1<=1500, u1>1500 & u1<=2000, u1>2000
Since u1 is time in your case which cannot be negative, you may also want to use the else part. So instead of the last step you can also do this:
Under Elseif expressions (comma-separated list, e.g. u2 ~= 0, u3(2) < u2):, write this:
u1>500 & u1<=1000, u1>1000 & u1<=1500, u1>1500 & u1<=2000
and connect the output of the else part which was connected with the output of u1>2000 before.
options = optimset('Display','iter','MaxIter',3,'OutputFcn',#outfun);
[x,fval,~,output] = fminsearch(#(param) esm6(param,identi),result(k,1:end-1),options);
This code will find the local Minimum of my esm6 function and due to the 'Display' Option it will Output strings like this
Iteration Func-count min f(x) Procedure
0 1 36.9193
1 5 35.9815 initial simplex
2 7 35.4924 contract inside
3 9 35.4924 contract inside
4 11 33.0085 expand
So in the command window, i get the function Count for each Iteration step. The structure output, which is created by fminsearch has only the total amount of func-count in it. Is there a way to receive all the Information, that is outputed in the command window also in the Output-structure?
EDIT:
I think i'm pretty Close to the solution. I wrote this outputfunction:
function stop = outfun(x,optimvalues,state);
stop = false;
if state == 'iter'
history = evalin('base','history');
history = [history; optimvalues.iteration optimvalues.funcCount];
assignin('base','history',history);
end
end
due to http://de.mathworks.com/help/matlab/math/output-functions.html this should work, but in fact, matlab tells me,
??? Reference to non-existent field 'funcCount'.
any idea, why this happens?
I have the following codes which I wish to have an output matrix Rpp of (10201,3). I run this code (which takes a bit long) then I check the matrix size of Rpp and I see (1,3), I tried so many things I couldn't find any proper way. The logic of the codes is to take the 6 values (contain 4 constant values and 2 variable values (chosen from 101 values)) and make the calculation for 3 different i1 and store every output vector of 3 in a matrix with (101*101 (pairs of those 2 variable values)) rows and 3 (for each i1) columns.
I appreciate your help
Vp1=linspace(3000,3500,101);
Vp2=3850;
rho1=2390;
rho2=2510;
Vs1=linspace(1250,1750,101);
Vs2=2000;
i1=[10 25 40];
Rpp = zeros(length(Vp1)*length(Vs1),length (i1));
for n=1:length(Vp1)*length(Vs1)
for m=1:length (i1)
for l=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(l);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(l)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(l)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp=((b.*(cos(i1)/Vp1(l))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(l))).*(cos(j2)/Vs2)).*H.*p.^2)./D
end
end
end
end
Try this. You 2 outer loops didn't do anything. You never used m or n so I killed those 2 loops. Also you just kept overwriting Rpp on every loop so your initialization of Rpp didn't do anything. I added an index var to assign the results to the equation to what I think is the correct part of Rpp.
Vp1=linspace(3000,3500,101);
Vp2=3850;
rho1=2390;
rho2=2510;
Vs1=linspace(1250,1750,101);
Vs2=2000;
i1=[10 25 40];
Rpp = zeros(length(Vp1)*length(Vs1),length (i1));
index = 1;
for l=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(l);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(l)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(l)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp(index,:)=((b.*(cos(i1)/Vp1(l))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(l))).*(cos(j2)/Vs2)).*H.*p.^2)./D;
index = index+1;
end
end
Results:
>> size(Rpp)
ans =
10201 3
The way you use the for loop is wrong. You're running the calculation for length(Vp1)*length(Vs1) * length (i1) * length(Vp1) * length(Vs1) times. Here's the correct way. I changed l into lll just so I won't confuse it with the number 1. In each iteration of the first for loop, you're running length(Vs1) times, and you need to assign the result (a 1X3 array) to the Rpp by using a row number specified by k+(lll-1)*length(Vp1).
for lll=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(lll);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(lll)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(lll)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp(k+(lll-1)*length(Vp1),:)=((b.*(cos(i1)/Vp1(lll))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(lll))).*(cos(j2)/Vs2)).*H.*p.^2)./D;
end
end
I'm new to CoffeeScript and have been reading the book, The Little Book on CoffeeScript. Here are a few lines from the book's Chapter 2 which confused me while reading :
The only low-level loop that CoffeeScript exposes is the while loop. This has similar behavior to the while loop in pure JavaScript, but has the added advantage that it returns an array of results, i.e. like the Array.prototype.map() function.
num = 6
minstrel = while num -= 1
num + " Brave Sir Robin ran away"
Though it may look good for a CoffeeScript programmer, being a newbie, I'm unable to understand what the code does. Moreover, the words returns an array of results doesn't seem to go together with the fact that while is a loop construct, not a function. So the notion of it returning something seems confusing. Furthermore, the variable num with the string "Brave Sir Robin ran away" in every iteration of the loop seems to be awkward, as the value num is being used as the loop counter.
I would be thankful if you could explain the behavior of the code and perhaps illustrate what the author is trying to convey with simpler examples.
Wow! I didn't know that but it absolutely makes sense if you remember that Coffeescript always returns the last expression of a "block".
So in your case it returns (not via the "return" statement if that is what confuses you) the expression
num + " Brave Sir Robin ran away"
from the block associated with the while condition and as you will return multiple such expressions it pushes them on an array.
Have a look on the generated JavaScript and it might be clearer as the generated code is pretty much procedural
var minstrel, num;
num = 6;
minstrel = (function() {
var _results;
_results = [];
while (num -= 1) {
_results.push(num + " Brave Sir Robin ran away");
}
return _results;
})();
I hope that makes sense to you.
Beware, that function call can be very inefficient!
Below is a prime factors generator
'use strict'
exports.generate = (number) ->
return [] if number < 2
primes = []
candidate = 1
while number > 1
candidate++
while number % candidate is 0
primes.push candidate
number /= candidate
candidate = number - 1 if Math.sqrt(number) < candidate
primes
This is the version using while as expression
'use strict'
exports.generate = (number) ->
return [] if number < 2
candidate = 1
while number > 1
candidate++
primes = while number % candidate is 0
number /= candidate
candidate
candidate = number - 1 if Math.sqrt(number) < candidate
primes
First version ran my tests in 4 milliseconds, the last one takes 18 milliseconds. I believe the reason is the generated closure which returns the primes.