Consider a set of points (just an example)
x = [0 1 2 5 4 8 5 6];
y = [5 8 4 2 5 6 4 5];
and another reference point:
xc=1;
yc=1;
In which I use to represent these points as vectors:
vec=[x-xc y-yc];
I wish to obtain a matrix with all the angles between all vectors which is obtained by the calculation (for single vectors)
angle = acosd(dot(v,u)/norm(u)*norm(v));
How can I obtain this calculation in a few lines without going vector by vector in a loop? In my calculation the number of points is very very large.
I think you mean vec = [x-xc; y-yc];. To calucate the dotproduct between all rows, you can use
vec.'*vec
The norm (Euclidean) of each vector can be determined as
no = sqrt(sum(vec.*vec,1))
The product of the different norms can be calculated the same as for vec:
no.'*no
The angles can thus be found as
no = sqrt(sum(vec.*vec,1));
angles = acosd(vec.'*vec./(no.'*no));
Related
I have two columns with data, x and y. Now I want to connect these data points in the order they appear in the columns. Say I have x=[1 2 3 4 3 2] and y=[3 4 2 1 3 3]. Now if I use spline to create a smooth curve, it 'sorts' the columns in an increasing order. I would like it to just take the data points, thus firstly x(1),y(1) and connect these to x(2), y(2) and so on.
Is this possible?
spline generates a function from the reals to the reals. This means a more general curve cannot be expressed as y = f(x) but we need to parametrize it as (x(t), y(t)):
x=[1 2 3 4 3 2];
y=[3 4 2 1 3 3];
plot(x,y,'o-');
% cannot be represented as function y=f(x)
% because x=2 and 3 have two different y values
% -> parametrize x and y:
t = 1:numel(x);
tt = linspace(min(t), max(t), 1000);;
tx = spline(t,x,tt);
ty = spline(t,y,tt);
hold on
plot(tx,ty,'-');
I am trying to compute the variance of elements which are organised in matrices (in MATLAB). As an example, let's be A and B two matrices 2x2.
My goal is to find the matrix V (2x2 as well), being the variance of each element of A and each element of B, that is:
Can somebody help me on this?
This is a very simple use case of the var function:
A = [1 2;
3 4];
B = [5 6;
7 8];
V0 = var(cat(3,A,B),0,3);
V1 = var(cat(3,A,B),1,3);
This results in:
V0 =
8 8
8 8
V1 =
4 4
4 4
What happens is that you concatenate your matrices along some unused dimension and then compute the variance along that dimensions.
NOTE: The example of 2 matrices is not very meaningful, but I'm assuming your actual dataset is larger, in which case you could use this method.
Let
input = [0 0 0 5 5 7 8 8];
I now want to transform this vector into the form
output = [3 3 3 3 5 5 6 8];
Which basically is a stairs plot.
Explanation
The input vector is used to plot data points along the x-axis. The y-axis is thereby provided by 1:length(input). So the resulting plot shows the cumulative number of datapoints along the y-axis and the time of occurrence along the x-axis.
I now want to fit a model against my dataset. Therefor I need a vector that provides the correct value for a certain time (x-value).
The desired output vector basically is the result of a stairs plot. I am looking for an efficient way to generate the desired vector in matlab. The result of
[x, y] = stairs(input, 1:length(input));
did not bring me any closer.
It can be done with bsfxun as follows:
x = [0 0 0 5 5 7 8 8];
y = sum(bsxfun(#le, x(:), min(x):max(x)), 1);
This counts, for each element in 1:numel(x), how many elements of x are less than or equal to that.
I have two vectors of the same size. The first one can have any different numbers with any order, the second one is decreasing (but can have the same elements) and consists of only positive integers. For example:
a = [7 8 13 6];
b = [5 2 2 1];
I would like to plot them in the following way: on the x axis I have points from a vector and on the y axis I have the sum of elements from vector b before this points divided by the sum(b). Therefore I will have points:
(7; 0.5) - 0.5 = 5/(5+2+2+1)
(8; 0.7) - 0.7 = (5+2)/(5+2+2+1)
(13; 0.9) ...
(6; 1) ...
I assume that this explanation might not help, so I included the image
Because this looks to me as a cumulative distribution function, I tried to find luck with cdfplot but with no success.
I have another option is to draw the image by plotting each line segment separately, but I hope that there is a better way of doing this.
I find the values on the x axis a little confusing. Leaving that aside for the moment, I think this does what you want:
b = [5 2 2 1];
stairs(cumsum(b)/sum(b));
set(gca,'Ylim',[0 1])
And if you really need those values on the x axis, simply rename the ticks of that axis:
a = [7 8 13 6];
set(gca,'xtick',1:length(b),'xticklabel',a)
Also grid on will add grid to the plot
I am working on some fourier transform code in matlab, and have come across the following:
xx = meshgrid(1:N);
% Center on DC
xx = xx - dcN;
% normalize dynamic range from -1 to 1
xx = xx./max(abs(xx(:)));
% form y coordinate from negative transpose of x coordinate (maintains symmetry about DC)
yy = -xx';
% compute the related radius of the x/y coordinates centered on DC
rr = sqrt(xx.^2 + yy.^2);
How can I generalize this for non-square matrices? This code is assuming my matrix is square, so dcN is the center of the square matrix (in other words, with 11x11, dcN = 6).
The math doesnt work out for that yy variable when the transpose is taken for a non-square matrix.
I have tried to figure out if I can make a meshgrid going from "top to bottom" instead of left to right - but I havent been able to figure taht out either.
Thanks
I have tried to figure out if I can
make a meshgrid going from "top to
bottom" instead of left to right - but
I havent been able to figure taht out
either.
>> N=5
N =
5
>> rot90(meshgrid(N:-1:1))
ans =
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
From your question I guess that you want to find rr, i.e. the distance of any element in the matrix from the center.
If you want this for a M-by-N array, you'd do the following
%# note that using meshgrid instead of ndgrid will swap xx and yy
[xx,yy] = ndgrid(-(M-1)/2:(M-1)/2,-(N-1)/2:(N-1)/2);
%# normalize to the max of xx,yy
nrm = max((M-1)/2,(N-1)/2);
xx = xx./nrm;
yy = yy./nrm;
rr = sqrt(xx.^2+yy.^2)