Thank you very much,first. The first code runs well:
function f=malibu(m1,m2,m3,k,l,t)
f=(m1.*k+(m1+m2).*l).*exp(-m3.*t);
end
function loglik= modelmalibu(p)
global k l t x n m2 m3;
f =malibu(p,m2,m3,k,l,t);
if f==0;
loglik0=0;
else
loglik0=(x.*log(f)+(n-x).*log(1-f));%minus likelihood
end
loglik=sum(-loglik0);
end
clear all;
global n t x k l m2 m3;
m1=0.1;
m2=0.2;
m3=0.3;
t=[1 3 6 9 12 18]';
k=[1 1 2 3 3 4]';
l=[0 0 1 3 4 5]';
y=meltem(m1,m2,m3,k,l,t);
n=100;%trial
x=y.*n;%correct replies
pstart=0.3;
[p1,modelvalue]=fminsearch(#modelmalibu,pstart);
But the similar code for more variable gives error out.
function w=anemon(m1,m2,m3,X,Y,k,l)
w=(m1.*k+(m1+m2).*l)+X.*exp(-m3.*Y);
end
function loglik= modelanemon(p)
global n x m2 m3 X Y k l ;
f =anemon(p,m2,m3,X,Y,k,l);
if f==0;
loglik0=0;
else
loglik0=(x*log(f)+(n-x)*log(1-f));%minus likelihood
end
loglik=sum(-loglik0);
end
clear;
global n x Ydata kdata ldata m1 m2 m3;
%parameters
m1=0.002;
m2=0.0001;
m3=7;
%given data
Xdata=[1 3 6 9 10 12]';
Ydata=[11 13 41 81 121 181]';
kdata=[1 1 2 4 5 4]';
ldata=[1 1 3 3 4 5]';
y=anemon(m1,m2,m3,Xdata,Ydata,kdata,ldata);
n=10;
x=y.*n;
pstart=2;
[pbest,modelvalue]=fminsearch(#modelanemon,pstart);
I've actually tried to use your advises, but if I would write an inequality instead of f==0, the first code fall down as well.
I think you are mixing up two concepts.
vectorization: does refer in how to write your code so it can use some of the acceleration functions in your CPU. it has nothing to do with fminsearch. See: http://en.wikipedia.org/wiki/Vectorization_(parallel_computing)
I suppose you want to write your function such that it accepts a vector as input. Easiest way is to just use a function handle like this:
fh = #(x) my_complicated_function(const1, const2, x(1), x(2), x(3) )
In this case my_complicated_function has 5 inputs, and you take the first 2 constant and input a 3 dim vector for the other 3. Fminsearch will work with that.
You would call
x_opt = fminsearch(fh, [1,2,3])
Besides some tips for the code:
Don't use == for comparison of numbers - go for eg. abs(x1-x2)<0.1 instead
interpanemon looks very strange - it doesn't do what one would call interpolation, and if you look - in every iteration p is recalculated so only the last calculation takes effect. As it looks it should output a constant value - no use optimizing that.
The use of a precalculated Z is probably not what you want. It already determines your optimum. If you use linear optimization the optimum must be already in Z - no need doing interpolation.
The invocation in your case might look like:
min_p = fminsearch(#(x) interpanemon(5,Ydata,x,m2,m3,Z,X,Y,kdata,ldata) ,1)
Overall it looks very unusual - it could really help if you would explain the ideas WHY you choose to do it like this. Also it might help if you restate the problem in a simpler form.
Related
It's like a reverse version of Pascal's triangle.
I want to create a vector-input function named
y = lastnum(vect) on Octave, that evaluate the sum of each pair of numbers in any vectors to output the single number from the evaluation loops like this
0 1 2 3 4
1 3 5 7
4 8 12
12 20
32
And the input and output would be like this,
lastnum([0 1 2 3 4])
ans = 32
I mean... is there any progresses that I can do??? You may not understand but, the reverse triangle above can guide you about my question.
I also tagged MATLAB since it has similar language. MATLAB pros may help my problem.
Notice that the number of times each element gets added to produce the final result comes from Pascals triangle itself, so, e.g., for the vector [a b c d] the result will be a+3b+3c+d. So create a vector of entries in Pascals triangle and multiply and add with the original vector v.
I only have access to Matlab, Octave may not have all these functions.
This is a one-liner diag(fliplr(pascal(numel(v)))).'*v(:).
Or a looping version
s = 0;
for i = 0:numel(v)-1
s = s+nchoosek(numel(v)-1,i)*v(i+1);
end
s
Simplest thing I can think of is:
while length(x) > 0
disp(x)
x = x(1:end-1) + x(2:end);
end
or did I misunderstand the question?
Here is the version for both MATLAB and Octave:
function y = lastnum(v)
while 1
if length(v) == 2
y = sum(v)
break;
end
# disp(v); # if you want to print the progress
vt = [];
for k = 1:(length(v)-1)
vt(end+1) = sum(v(k:(k+1)));
end
v = vt;
end
end
I wrote a syntax that calculate values of a function in different values.
For example
x1=[1 2 10 11];
x2=[10 11 12 14];
C= arrayfun (#(t1,t2) myfunction(A,B,t1,t2),x1,x2,'UniformOutput',0);
% A and B are matrixs
In this example the function will do an operation on A(x1,x2) and B(x1,x2) . The problem is that arrayfun will work on each couple (x1(1),x2(1)), x1(2),x2(2)),etc. But I want it to work on all the values of x1 and x2 (16 couples of values so that it can be applied also to for example (x1(1),x2(3))).
Is there any way to do that without a loop?
using ndgrid:
[X Y] = ndgrid(x1,x2);
C= arrayfun (#(x1,x2) myfunction(A,B,t1,t2),X(:),Y(:),'UniformOutput',0);
So I'm going to offer the "dumb" way of doing it without you having to modify your call to your function:
x1=[1 2 10 11];
x2=[10 11 12 14];
v = combvec(x1,x2);
X1 = v(1,:);
X2 = v(2,:);
C= arrayfun (#(X1,X2) myfunction(A,B,t1,t2),X1,X2,'UniformOutput',0);
This effectively calculated all combinations of x1 and x2, then you input it thru your function in pairs just like you were doing it before.
I am trying to resolve why the following Matlab syntax does not work.
I have an array
A = [2 3 4 5 8 9...]
I wish to create an indexed cumulative, for example
s(1) = 2; s(2)=5, s(3)=9; ... and so on
Can someone please explain why the following does not work
x = 1:10
s(x) = sum(A(1:x))
The logic is that if a vector is created for s using x, why would not the sum function behave the same way? The above returns just the first element (2) for all x.
For calculating the cumulative sum, you should be using cumsum:
>> A = [2 3 4 5 8 9]
A =
2 3 4 5 8 9
>> cumsum(A)
ans =
2 5 9 14 22 31
The issue is that 1:x is 1 and that sum reduces linear arrays. To do this properly, you need a 2d array and then sum the rows:
s(x)=sum(triu(repmat(A,[prod(size(A)) 1])'))
You are asking two questions, really. One is - how do I compute the cumulative sum. #SouldEc's answer already shows how the cumsum function does that. Your other question is
Can someone please explain why the following does not work
x = 1:10
s(x) = sum(A(1:x))
It is reasonable - you think that the vector expansion should turn
1:x
into
1:1
1:2
1:3
1:4
etc. But in fact the arguments on either side of the colon operator must be scalars - they cannot be vectors themselves. I'm surprised that you say Matlab isn't throwing an error with your two lines of code - I would have expected that it would (I just tested this on Freemat, and it complained...)
So the more interesting question is - how would you create those vectors (if you didn't know about / want to use cumsum)?
Here, we could use arrayfun. It evaluates a function with an array as input element-by-element; this can be useful for a situation like this. So if we write
x = 1:10;
s = arrayfun(#(n)sum(A(1:n)), x);
This will loop over all values of x, substitute them into the function sum(A(1:n)), and voila - your problem is solved.
But really - the right answer is "use cumsum()"...
Actually what you are doing is
s(1:10)= sum(A(1:[1,2,3...10]))
what you should do is
for i=1:10
s(i)=sum(A(1:i))
end
hope it will help you
I believe most functions in MATLAB should be able to receive matrix input and return the output in the form of matrix.
For example sqrt([1 4 9]) would return [1 2 3].
However, when I tried this recurring factorial function:
function k = fact(z)
if z ~= 0
k = z * fact(z-1);
else
k = 1;
end
end
It works perfectly when a number is input into fact. However, when a matrix is input into fact, it returns the matrix itself, without performing the factorial function.
E.g.
fact(3) returns 6
fact([1 2 3]) returns [1 2 3] instead of [1 2 6].
Any help is appreciated. Thank you very much!
Since MATLAB is not known to be good with recursive functions, how about a vectorized approach? Try this for a vector input -
mat1 = repmat([1:max(z)],[numel(z) 1])
mat1(bsxfun(#gt,1:max(z),z'))=1
output1 = prod(mat1,2)
Sample run -
z =
1 2 7
output1 =
1
2
5040
For the sake of answering your original question, here's the annoying loopy code for a vector or 2D matrix as input -
function k1 = fact1(z1)
k1 = zeros(size(z1));
for ii = 1:size(z1,1)
for jj = 1:size(z1,2)
z = z1(ii,jj);
if z ~= 0
k1(ii,jj) = z .* fact1(z-1);
else
k1(ii,jj) = 1;
end
end
end
return
Sample run -
>> fact1([1 2 7;3 2 1])
ans =
1 2 5040
6 2 1
You can use the gamma function to compute the factorial without recursion:
function k = fact(z)
k = gamma(z+1);
Example:
>> fact([1 2 3 4])
ans =
1 2 6 24
Not sure if all of you know, but there is an actual factorial function defined in MATLAB that can take in arrays / matrices of any size, and computes the factorial element-wise. For example:
k = factorial([1 2 3 4; 5 6 7 8])
k =
1 2 6 24
120 720 5040 40320
Even though this post is looking for a recursive implementation, and Divakar has provided a solution, I'd still like to put my two cents in and suggest an alternative. Also, let's say that we don't have access to factorial, and we want to compute this from first principles. What I would personally do is create a cell array that's the same size as the input matrix, but each element in this cell array would be a linear index array from 1 up to the number defined for each location in the original matrix. You would then apply prod to each cell element to compute the factorial. A precondition is that no number is less than 1, and that all elements are integers. As such:
z1 = ... ; %// Define input matrix here
z1_matr = arrayfun(#(x) 1:x, z1, 'uni', 0);
out = cellfun(#prod, z1_matr);
If z1 = [1 2 3 4; 5 6 7 8];, from my previous example, we get the same output with the above code:
out =
1 2 6 24
120 720 5040 40320
This will obviously be slower as there is an arrayfun then cellfun call immediately after, but I figured I'd add another method for the sake of just adding in another method :) Not sure how constructive this is, but I figured I'd add my own method and join Divakar and Luis Mendo :)
I am trying to resolve why the following Matlab syntax does not work.
I have an array
A = [2 3 4 5 8 9...]
I wish to create an indexed cumulative, for example
s(1) = 2; s(2)=5, s(3)=9; ... and so on
Can someone please explain why the following does not work
x = 1:10
s(x) = sum(A(1:x))
The logic is that if a vector is created for s using x, why would not the sum function behave the same way? The above returns just the first element (2) for all x.
For calculating the cumulative sum, you should be using cumsum:
>> A = [2 3 4 5 8 9]
A =
2 3 4 5 8 9
>> cumsum(A)
ans =
2 5 9 14 22 31
The issue is that 1:x is 1 and that sum reduces linear arrays. To do this properly, you need a 2d array and then sum the rows:
s(x)=sum(triu(repmat(A,[prod(size(A)) 1])'))
You are asking two questions, really. One is - how do I compute the cumulative sum. #SouldEc's answer already shows how the cumsum function does that. Your other question is
Can someone please explain why the following does not work
x = 1:10
s(x) = sum(A(1:x))
It is reasonable - you think that the vector expansion should turn
1:x
into
1:1
1:2
1:3
1:4
etc. But in fact the arguments on either side of the colon operator must be scalars - they cannot be vectors themselves. I'm surprised that you say Matlab isn't throwing an error with your two lines of code - I would have expected that it would (I just tested this on Freemat, and it complained...)
So the more interesting question is - how would you create those vectors (if you didn't know about / want to use cumsum)?
Here, we could use arrayfun. It evaluates a function with an array as input element-by-element; this can be useful for a situation like this. So if we write
x = 1:10;
s = arrayfun(#(n)sum(A(1:n)), x);
This will loop over all values of x, substitute them into the function sum(A(1:n)), and voila - your problem is solved.
But really - the right answer is "use cumsum()"...
Actually what you are doing is
s(1:10)= sum(A(1:[1,2,3...10]))
what you should do is
for i=1:10
s(i)=sum(A(1:i))
end
hope it will help you