I'm trying to do a solver for equations. When I run the code the X variable appears to be undefined, but it prints out perfectly. What am I missing?
I should give the program some numbers, than operations as Macros and it should create an outer product matrix of the operations applied.
function msu()
print("Insert how many values: ")
quantity = parse(Int64, readline())
values = []
for i in 1:quantity
println("x$i")
num1 = parse(Float64, readline())
push!(values, num1)
end
println(values)
print("How many operations? ")
quantity = parse(Int64, readline())
ops = []
for i in 1:quantity
push!(ops, Meta.parse(readline()))
end
mat = zeros((quantity, quantity))
for i in 1:length(mat)
sum = 0
for j in 1:length(values)
# here begins problems, the following prints are for debugging purpose
print(length(values))
func = Meta.parse("$(ops[convert(Int64, ceil(j / quantity))]) * $(ops[convert(Int64, j % quantity)])")
print(func)
x = values[j]
println(x)
sum += eval(func)
end
mat[i] = sum
end
println(mat)
end
msu()
The original code was in Spanish, if you find any typo it's probably because I skipped a translation.
I'm struggling to amend the Julia-specific tutorial on NLopt to meet my needs and would be grateful if someone could explain what I'm doing wrong or failing to understand.
I wish to:
Minimise the value of some objective function myfunc(x); where
x must lie in the unit hypercube (just 2 dimensions in the example below); and
the sum of the elements of x must be one.
Below I make myfunc very simple - the square of the distance from x to [2.0, 0.0] so that the obvious correct solution to the problem is x = [1.0,0.0] for which myfunc(x) = 1.0. I have also added println statements so that I can see what the solver is doing.
testNLopt = function()
origin = [2.0,0.0]
n = length(origin)
#Returns square of the distance between x and "origin", and amends grad in-place
myfunc = function(x::Vector{Float64}, grad::Vector{Float64})
if length(grad) > 0
grad = 2 .* (x .- origin)
end
xOut = sum((x .- origin).^2)
println("myfunc: x = $x; myfunc(x) = $xOut; ∂myfunc/∂x = $grad")
return(xOut)
end
#Constrain the sums of the x's to be 1...
sumconstraint =function(x::Vector{Float64}, grad::Vector{Float64})
if length(grad) > 0
grad = ones(length(x))
end
xOut = sum(x) - 1
println("sumconstraint: x = $x; constraint = $xOut; ∂constraint/∂x = $grad")
return(xOut)
end
opt = Opt(:LD_SLSQP,n)
lower_bounds!(opt, zeros(n))
upper_bounds!(opt,ones(n))
equality_constraint!(opt,sumconstraint,0)
#xtol_rel!(opt,1e-4)
xtol_abs!(opt,1e-8)
min_objective!(opt, myfunc)
maxeval!(opt,20)#to ensure code always terminates, remove this line when code working correctly?
optimize(opt,ones(n)./n)
end
I have read this similar question and documentation here and here, but still can't figure out what's wrong. Worryingly, each time I run testNLopt I see different behaviour, as in this screenshot including occasions when the solver uselessly evaluates myfunc([NaN,NaN]) many times.
You aren't actually writing to the grad parameters in-place, as you write in the comments;
grad = 2 .* (x .- origin)
just overrides the local variable, not the array contents -- and I guess that's why you see these df/dx = [NaN, NaN] everywhere. The simplest way to fix that would be with broadcasting assignment (note the dot):
grad .= 2 .* (x .- origin)
and so on. You can read about that behaviour here and here.
How can you get the length of the curve down below between 0 and 4*pi? The commands you should use are inttrap and diff. Here is what I have now:
t=linspace(0,4*%pi)
x=(4+sin(a*t)).*cos(3*t)
y=(4+sin(a*t)).*sin(3*t)
z=cos(3*t)
xx=diff(x)
yy=diff(y)
zz=diff(z)
aid=sqrt(xx^2+yy^2+zz^2)
length=inttrap([t],aid)
Getting error message, the last step is not right.
The reason for error message is that t and aid have different sizes. And that is because diff returns a vector with 1 entry fewer than the input. You can see how it works on an example: diff([3 1 5]) is [-2 4].
To fix this, use t(1:$-1), which omits the last entry of t. That is,
len = inttrap(t(1:$-1), aid)
(Please don't use length, which is a function name in Scilab.)
Another problem you have is that diff is just differences, not a derivative. To get the derivative, you need to divide by the step size, which in your case is t(2)-t(1).
Also, the syntax xx^2 is deprecated for elementwise power; use xx.^2 instead
t = linspace(0,4*%pi)
a = 1
x = (4+sin(a*t)).*cos(3*t)
y = (4+sin(a*t)).*sin(3*t)
z = cos(3*t)
step = t(2)-t(1)
xx = diff(x)/step
xy = diff(y)/step
xz = diff(z)/step
aid = sqrt(xx.^2+yy.^2+zz.^2)
len = inttrap(t(1:$-1), aid)
I have the image and the vector
a = imread('Lena.tiff');
v = [0,2,5,8,10,12,15,20,25];
and this M-file
function y = Funks(I, gama, c)
[m n] = size(I);
for i=1:m
for j=1:n
J(i, j) = (I(i, j) ^ gama) * c;
end
end
y = J;
imshow(y);
when I'm trying to do this:
f = Funks(a,v,2)
I am getting this error:
??? Error using ==> mpower
Integers can only be combined with integers of the same class, or scalar doubles.
Error in ==> Funks at 5
J(i, j) = (I(i, j) ^ gama) * c;
Can anybody help me, with this please?
The error is caused because you're trying to raise a number to a vector power. Translated (i.e. replacing formal arguments with actual arguments in the function call), it would be something like:
J(i, j) = (a(i, j) ^ [0,2,5,8,10,12,15,20,25]) * 2
Element-wise power .^ won't work either, because you'll try to "stuck" a vector into a scalar container.
Later edit: If you want to apply each gamma to your image, maybe this loop is more intuitive (though not the most efficient):
a = imread('Lena.tiff'); % Pics or GTFO
v = [0,2,5,8,10,12,15,20,25]; % Gamma (ar)ray -- this will burn any picture
f = cell(1, numel(v)); % Prepare container for your results
for k=1:numel(v)
f{k} = Funks(a, v(k), 2); % Save result from your function
end;
% (Afterwards you use cell array f for further processing)
Or you may take a look at the other (more efficient if maybe not clearer) solutions posted here.
Later(er?) edit: If your tiff file is CYMK, then the result of imread is a MxNx4 color matrix, which must be handled differently than usual (because it 3-dimensional).
There are two ways I would follow:
1) arrayfun
results = arrayfun(#(i) I(:).^gama(i)*c,1:numel(gama),'UniformOutput',false);
J = cellfun(#(x) reshape(x,size(I)),results,'UniformOutput',false);
2) bsxfun
results = bsxfun(#power,I(:),gama)*c;
results = num2cell(results,1);
J = cellfun(#(x) reshape(x,size(I)),results,'UniformOutput',false);
What you're trying to do makes no sense mathematically. You're trying to assign a vector to a number. Your problem is not the MATLAB programming, it's in the definition of what you're trying to do.
If you're trying to produce several images J, each of which corresponds to a certain gamma applied to the image, you should do it as follows:
function J = Funks(I, gama, c)
[m n] = size(I);
% get the number of images to produce
k = length(gama);
% Pre-allocate the output
J = zeros(m,n,k);
for i=1:m
for j=1:n
J(i, j, :) = (I(i, j) .^ gama) * c;
end
end
In the end you will get images J(:,:,1), J(:,:,2), etc.
If this is not what you want to do, then figure out your equations first.
I'm having some problems with my code. Here it is:
lambdaz = 1.2;
n = 24;
mu = 0.00055e9;
lambda = sym('lambda','clear');
W = (((2.*mu)./n.^2)).*((lambda.^n)+(lambdaz.^n)+((lambda.^-n).*(lambdaz.^-n))-3);
dW_dlambda = diff(W, lambda);
W2=(((2.*mu)./n.^2).*(lambda.^n))+(((2.*mu)./n.^2).*(lambdaz.^n))+(((2.*mu)./n.^2).*((lambda.^-n).*(lambdaz.^-n)))-(3.*((2.*mu)./n.^2))
dW2_dlambda=diff(W2,lambda)
x=((((lambda.^2).*(lambdaz))-1).^-1).*(dW_dlambda);
x2=((((lambda.^2).*(lambdaz))-1).^-1).*(dW2_dlambda)
P2 = int(x2,lambda)
P=int(x,lambda);
P=(0:1000:26700)
plot(lambda,P)
When I try to plot lambda against P I get the "conversion to double from sym is not possible" error message. I'm not particularly fantastic at Matlab so any help would be gratefully received!
The plot function only works for numeric inputs. Both lambda and P are symbolic expressions (at least before overwrote P by setting it equal to a vector after the integration) that cannot be directly converted to to floating point. You get the same error if try to something like double(sym('exp(x)')). You have two options. The first is the ezplot function in the Symbolic Toolbox:
...
P = int(x,lambda);
ezplot(P,[-5 5]); % Plot's P from lambda = -5 to lambda = 5
Or you can use the subs function:
...
P = int(x,lambda);
lambda = -5:0.01:5;
plot(lambda,real(subs(P,'lambda',lambda)))
axis([lambda(1) lambda(end) -1e15 1e15])
I used real to suppress a warning for negative values of lambda.