I am using Matlab to do one of my projects. I am stuck at one basic thing.
I have 2 matricies - A and B and a vector V. What I want to do is this:
A(i, j) = V(B(i,j)) for all i, j.
I tried doing this in the most obvious way - nested loops. For some reason, A is not getting populated. Am I missing something? Is there a more efficient (in-built function) way of doing this.
Thanks,
Anil.
If all entries in B are integers larger than zero, and if the maximum of B is not larger than the number of elements in V, then you can simply write
A = V(B);
Related
I am writing some CUDA code for finding the 3 parameters of a circle (centre X,Y & radius) from many (m) measurements of positions around the perimeter.
As m > 3 I am (successfully) using Singular Value Decomposition (SVD) for this purpose (using the cuSolver library). Effectively I am solving m simulaneous equations with 3 unknowns.
However, not all of my perimeter positions are valid (say q of them), and so I have to go through my initial set of m measurements and remove the q invalid ones. This involves moving the size m data array from the card to the host, processing linearly to remove the q invalid entries and then re loading the smaller (m-q) array back onto the card...
My question is; if I were to set all terms on both sides of the q invalid equations to zero, could I just run the m equations (including the zeros) through my SVD analysis (without the data transfer etc) or would this cause other problems?
My instinct tells me that this is a bit like applying weights to the data but instinct and SVD are not terms that sit well together in my experience...
I am hesitant just to try this as I don't know if it will work in some cases and not in others...
I have tested the idea by inserting rows of zeros into my matrix. The solution that I am getting is not significantly affected by this.
So I am answering my own question with a non-rigorous Yes it is OK do do this.
If anybody has a more rigorous or more considered answer I would very much like to hear it.
I need to use a for-loop in a function in order to find spring constants of all possible combinations of springs in series and parallel. I have 5 springs with data therefore I found the spring constant (K) of each in a new matrix by using polyfit to find the slope (using F=Kx).
I have created a function that does so, however it returns data not in a matrix, but as individual outputs. So instead of KP (Parallel)= [1 2 3 4 5] it says KP=1, KP=2, KP=3, etc. Because of this, only the final output is stored in my workspace. Here is the code I have for the function. Keep in mind that the reason I need to use the +2 in the for loop for b is because my original matrix K with all spring constants is ten columns, with every odd number being a 0. Ex: K=[1 0 2 0 3 0 4 0 5] --- This is because my original dataset to find K (slope) was ten columns wide.
function[KP,KS]=function_name1(K)
L=length(K);
c=1;
for a=1:2:L
for b=a+2:2:L
KP=K(a)+K(b)
KS=1/((1/K(a))+(1/K(b)))
end
end
c=c+1;
and then a program calling that function
[KP,KS]=function_name1(K);
What I tried: - Suppressing and unsuppressing lines of code (unsuccessful)
Any help would be greatly appreciated.
hmmm...
your code seems workable, but you aren't dealing with things in the most practical manner
I'd start be redimensioning K so that it makes sense, that is that it's 5 spaces wide instead of your current 10 - you'll see why in a minute.
Then I'd adjust KP and KS to the size that you want (I'm going to do a 5X5 as that will give all the permutations - right now it looks like you are doing some triangular thing, I wouldn't worry too much about space unless you were to do this for say 50,000 spring constants or so)
So my code would look like this
function[KP,KS]=function_name1(K)
L=length(K);
KP = zeros(L);
KS = zeros(l);
c=1;
for a=1:L
for b=1:L
KP(a,b)=K(a)+K(b)
KS(a,b)=1/((1/K(a))+(1/K(b)))
end
end
c=c+1;
then when you want the parallel combination of springs 1 and 4 KP(1,4) or KP(4,1) will do the trick
I am unsure how to use the Distributive property on the following function:
F = B'D + A'D + BD
I understand that F = xy + x'z would become (xy + x')(xy + z) but I'm not sure how to do this with three terms with two variables.
Also another small question:
I was wondering how to know what number a minterm is without having to consult (or memorise) the table of minterms.
For example how can I tell that xy'z' is m4?
When you're trying to use the distributive property there, what you're doing is converting minterms to maxterms. This is actually very related to your second question.
To tell that xy'z' is m4, think of function as binary where false is 0 and true is 1. xy'z' then becomes 100, binary for the decimal 4. That's really what a k-map/minterm table is doing for you to give a number.
Now an important extension of this: the number of possible combinations is 2^number of different variables. If you have 3 variables, there are 2^3 or 8 different combinations. That means you have min/maxterm possible numbers from 0-7. Here's the cool part: anything that isn't a minterm is a maxterm, and vice versa.
So, if you have variables x and y, and you have the expression xy', you can see that as 10, or m2. Because the numbers go from 0-3 with 2 variables, m2 implies M0, M1, and M3. Therefore, xy'=(x+y)(x+y')(x'+y').
In other words, the easiest way to do the distributive property in either direction is to note what minterm or maxterm you're dealing with, and just switch it to the other.
For more info/different wording.
I need to write my own function which has the form f(x,y)=Integrate(g(x,y,z),z from 0 to inf). so the code I used was:
function y=f(x,y)
g=#(z)exp(-z.^2)./(z.^x).*(z.^2+y.^2).^(x/2);% as a function of x,y and z
y=quadgk(g,0,inf)
and if I call it for a single value like f(x0,y0), it works but if I try to calculate something like f([1:10],y0), then the error message says that there is something wrong with the times and dimension. In principle I can use for loops but then my code slows down and takes forever. Is there any help I can get from you guys? or references?
I'm trying to avoid the for loop since in matlab it's much faster to use matrix computation than to use for loop. I wonder if there is any trick that I can take advantage of this feature.
Thanks for any help in advance,
Lynn
Perhaps you can try to transpose the interval, creating row based values instead of column based f([1:10]',y0). Otherwise something in your function might be wrong, for example to get x^y to work with lists as input, you have to prefix with a dot x.^y. The same for mulitply and division I think..
If loop is no problem for you, you should do something like:
function y2=f(x,y)
y2=zeros(size(x));
for n=1:numel(x)
g=#(z)exp(-z.^2)./(z.^x(n)).*(z.^2+y.^2).^(x(n)/2);% as a function of x,y and z
y2(n)=quadgk(g,0,inf)
end
The problem here is that quadk itself uses vectors as argument for g. Then you have in g somethink like z.^x, which is the power of two vectors that is only defined if z and x have the same dimension. But this is not what you want.
I assume that you want to evaluate the function for all arguments in x and that the output vector has the same dimension as x. But this does not seem to be possible since even this simple example
g=#(x)[x;x.^2]
quad(g,0,1)
does not work:
Error using quad (line 79)
The integrand function must return an output vector of the same length as the
input vector.
A similar error shows when using quadgk. The documentation also says that this routine works only for scalar functions and this is not surprising since an adaptive quadrature rule would in general use different points for each function to evaluate the integral.
You have to use quadvinstead, which can integrate vector valued functions. But this gives wrong results since your function is integrated in the interval [0,\infty).
Started learning octave recently. How do I generate a matrix from another matrix by applying a function to each element?
eg:
Apply 2x+1 or 2x/(x^2+1) or 1/x+3 to a 3x5 matrix A.
The result should be a 3x5 matrix with the values now 2x+1
if A(1,1)=1 then after the operation with output matrix B then
B(1,1) = 2.1+1 = 3
My main concern is a function that uses the value of x like that of finding the inverse or something as indicated above.
regards.
You can try
B = A.*2 + 1
The operator . means application of the following operation * to each element of the matrix.
You will find a lot of documentation for Octave in the distribution package and on the Web. Even better, you can usually also use the extensive documentation on Matlab.
ADDED. For more complex operations you can use arrayfun(), e.g.
B = arrayfun(#(x) 2*x/(x^2+1), A)