I have 6 different data sets from sensor and I want to find out the Root mean Square error but of the limited signal with limits means RMSE of sig_diff_lim.
I have tried to apply loop but it’s not working can anybody tell me where I am at fault.
Thanks in anticipation.
clc
drv(1)=load('a.mat');
drv(2)=load('b.mat');
drv(3)=load('c.mat');
drv(4)=load('d.mat');
drv(5)=load('e.mat');
drv(6)=load('h.mat');
for i= 1:numel(drv)
t=drv(i).T;
ref=drv(i).P;
lws=drv(i).SWA;
sig_diff(i,:) =lws(i)-ref(i);
swvel_thres=10;
vehvel=30;
SAmax=90;
sig_diff_lim(i,:)=sig_diff((lws(i)<SAmax)&(lws(i)>-SAmax)&(swav(i)<swvel_thres)&(vel(i)>vehvel));
square_error(i,:) = (sig_diff_lim(i)).^2;
mse(i,:)= mean(square_error(i));
rmse(i,:) = sqrt(mse(i));
end
rmse
mse
It seems to me that while assigning to sig_diff_lim(i,:) requires a row vector of size 1*n, the matrix that is returned by
vec=sig_diff((lws_7(i)<SAmax)&(lws_7(i)>-SAmax)&(swav(i)<swvel_thres)&(vel(i)>vehvel))
is of a different size.
Edit
First, are you sure that sig_diff((lws(i)... shouldn't be sig_diff(i,:)((lws(i).... on the assignemnt line?
Now, try element-wise multiplication (operator .* ) between sig_diff (or sig_diff(i,:)), depending on your answer to my previous point) with your logical indexing, like this:
sig_diff.*((lws(i)<... %%%% instead of sig_diff((lws(i)<...
%%% or
sig_diff(i,:).*(lws(i)<...
I think that should do the trick.
Regardless of whether or not this worked, try this example to understand what is going on with that sort of logical indexing:
>> t=[1 2 3 5 7;1 5 6 8 10];
>> t(1,:)(l<7 & l>-7)
ans =
1 2 3 5
>> t(1,:)(l<7 & l>-2)
ans =
1 2 5
Of course, in this example, t is fixed while the thresholds are changing , while in your code the matrix itself is changing. But it shows you that your indexing will return matrices of variable length.
But, with the solution I proposed above,
>> t(1,:).*(l<7 & l>-2)
>> t(1,:).*(l<7 & l>-7)
are both of size 5*1.
Related
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
Imagine the following matrix is defined in MATLAB:
>> matrix=10:18
matrix =
10 11 12 13 14 15 16 17 18
Now I want to use another matrix to index the first one
>> index=[1,2;3,4]
index =
1 2
3 4
>> matrix(index)
ans =
10 11
12 13
So far so good, the size of the answer matches that of the matrix 'index'. If I use a row vector as the indexing matrix, the output is a row vector too. But the problem appears when I use a column vector as the indexing matrix:
>> index=(1:3)'
index =
1
2
3
>> matrix(index)
ans =
10 11 12
As you can see, here the size of the answer does not agree with the size of the matrix 'index'. This inconsistency in the sizes of the indexing matrix and the ans matrix prevents me from writing a piece of code accepting an indexing matrix of an arbitrary size.
I wonder if someone else has come across this problem before and has found some sort of solution to it; in other words, how can I force MATLAB to give me an ans matrix of the same size as the arbitrarily sized indexing matrix?
Cheers
Solution
#Dev-iL has nicely explained here why this is the way Matlab behaves, and #Dan has presented a general solution here. However, there was a simpler ad-hoc workaround for my code which I've explained here.
The reason comes from the function subsref.m, which is called when using parentheses:
%SUBSREF Subscripted reference.
% A(I) is an array formed from the elements of A specified by the
% subscript vector I. The resulting array is the same size as I except
% for the special case where A and I are both vectors. In this case,
% A(I) has the same number of elements as I but has the orientation of A.
As you can see, in the special case of vectors, the shape of the result will be the same as that of matrix in your example.
As for a workaround, IMHO Dan has the cleanest solution.
Preallocate the output matrix with the size of index matrix.
Then assign corresponding indices the corresponding values
out = zeros(size(index));
out(index) = matrix(index);
Example
index = (1:3)'
index =
1
2
3
>> out = zeros(size(index))
out =
0
0
0
>> in = magic(3)
in =
8 1 6
3 5 7
4 9 2
>> out(index) = in(index)
out =
8
3
4
>>
Dev-iL has explained why it is like that. Here is a potential workaround:
reshape(matrix(index),size(index))
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
suppose that we are determine peaks in vector as follow:
we have real values one dimensional vector with length m,or
x(1),x(2),.....x(m)
if x(1)>x(2) then clearly for first point peak(1)=x(1);else we are then comparing x(3) to x(2),if x(3)
[ indexes,peaks]=function(x,m);
c=[];
b=[];
if x(1)>x(2)
peaks(1)=x(1);
else
for i=2:m-1
if x(i+1)< x(i) & x(i)>x(i-1)
peak(i)=x(i);
end;
end
end
end
peaks are determined also using following picture:
sorry for the second picture,maybe it is not triangle,just A and C are on straight line,but here peak is B,so i can't continue my code for writing algorithm to find peak values in my vector.please help me to continue it
updated.numercial example given
x=[2 1 3 5 4 7 6 8 9]
here because first point is more then second,so it means that peak(1)=2,then we are comparing 1 to 3,because 3 is more then 1,we now want to compare 5 to 3,it is also more,compare 5 to 4,because 5 is more then 4,then it means that peak(2)=5,,so if we continue next peak is 7,and final peak would be 9
in case of first element is less then second,then we are comparing second element to third one,if second is more then third and first elements at the same time,then peak is second,and so on
You could try something like this:
function [peaks,peak_indices] = find_peaks(row_vector)
A = [min(row_vector)-1 row_vector min(row_vector)-1];
j = 1;
for i=1:length(A)-2
temp=A(i:i+2);
if(max(temp)==temp(2))
peaks(j) = row_vector(i);
peak_indices(j) = i;
j = j+1;
end
end
end
Save it as find_peaks.m
Now, you can use it as:
>> A = [2 1 3 5 4 7 6 8 9];
>> [peaks, peak_indices] = find_peaks(A)
peaks =
2 5 7 9
peak_indices =
1 4 6 9
This would however give you "plateaus" as well (adjacent and equal "peaks").
You can use diff to do the comparison and add two points in the beginning and end to cover the border cases:
B=[1 diff(A) -1];
peak_indices = find(B(1:end-1)>=0 & B(2:end)<=0);
peaks = A(peak_indices);
It returns
peak_indices =
1 4 6 9
peaks =
2 5 7 9
for your example.
findpeaks does it if you have a recent matlab version, but it's also a bit slow.
This proposed solution would be quite slow due to the for loop, and you also have a risk of rounding error due to the fact that you compare the maximal value to the central one instead of comparing the position of the maximum, which is better for your purpose.
You can stack the data so as to have three columns : the first one for the preceeding value, the second is the data and the third one is the next value, do a max, and your local maxima are the points for which the position of the max along columns is 2.
I've coded this as a subroutine of my own peak detection function, that adds a further level of iterative peak detection
http://www.mathworks.com/matlabcentral/fileexchange/42927-find-peaks-using-scale-space-approach
In MATLAB, is there a more concise way to handle discrete conditional indexing by column than using a for loop? Here's my code:
x=[1 2 3;4 5 6;7 8 9];
w=[5 3 2];
q=zeros(3,1);
for i = 1:3
q(i)=mean(x(x(:,i)>w(i),i));
end
q
My goal is to take the mean of the top x% of a set of values for each column. The above code works, but I'm just wondering if there is a more concise way to do it?
You mentioned that you were using the function PRCTILE, which would indicate that you have access to the Statistics Toolbox. This gives you yet another option for how you could solve your problem, using the function NANMEAN. In the following code, all the entries in x less than or equal to the threshold w for a column are set to NaN using BSXFUN, then the mean of each column is computed with NANMEAN:
x(bsxfun(#le,x,w)) = nan;
q = nanmean(x);
I don't know of any way to index the columns the way you want. This may be faster than a for loop, but it also creates a matrix y that is the size of x.
x=[1 2 3;4 5 6;7 8 9];
w=[5 3 2];
y = x > repmat(w,size(x,1),1);
q = sum(x.*y) ./ sum(y)
I don't claim this is more concise.
Here's a way to solve your original problem: You have an array, and you want to know the mean of the top x% of each column.
%# make up some data
data = magic(5);
%# find out how many rows the top 40% are
nRows = floor(size(data,1)*0.4);
%# sort the data in descending order
data = sort(data,1,'descend');
%# take the mean of the top 20% of values in each column
topMean = mean(data(1:nRows,:),1);