please help me to understand this code:
x(:,i) = mean( (y(:,((i-1)*j+1):i*j)), 2 )';
i can't find it in my book. thanks.
The code you posted can be made more readable using temporary variables:
a = (i-1)*j+1;
b = i*j;
val = y(:,a:b);
x(:,i) = mean( val, 2 )'; %# =mean( val' )
What exactly you do not understand? For meaning of mean , : and ' consult matlab help.
It would help if you said exactly what you don't understand, but here are a few tips:
if you have something like a(r,c), that means matrix a, row r, column c (always in this order). In other words, you should have two elements inside the brackets separated by a comma where the first represents the row, the second the column.
If you have : by itself in one of the sides of the comma, that means "all". Thus, if you had a(r,:), then you would have matrix a, row r, all columns.
If : is not alone in one of the sides of the comma, then it will mean "to". So if you have a(r, z:y), that means matrix a, row r, columns z to y.
Mean = average. The format of the function in Matlab is M = mean(A,dim). A will be the matrix you take the average (or mean) of, M will be the place where the results are going to go. If dim = 1, you will get a row vector with each element being the average of a column. If dim = 2 (as it is in your case), then you should get a column vector, with each element being the average of a row. Be careful, though, because at the end of your code you have ', which means transpose. That means that your column vector will be transformed into a row vector.
OK, so your code:
x(:,i) = mean( (y(:,((i-1)*j+1):i*j)), 2 )';
Start with the bit inside, that is
y(:,((i-1)*j+1):i*j)
So that is saying
matrix y(r,c)
where
r (row) is :, that is, all rows
c (column) is ((i-1)j+1):ij, that is, columns going from (i-1)j+1 until ij
Your code will then get the matrix resulting from that, which I called y(r,c), and will do the following:
mean( (y(r,c), 2 )
so get the result from above and take the mean (average) of each row. As your code has the ' afterwards, that is, you have:
mean( (y(r,c), 2 )'
then it will get the column vector and transform into a row vector. Each element of this row will be the average of a row of y(r,c).
Finally:
x(:,i) = mean( (y(r,c), 2 )';
means that the result of the above will be put in column i of matrix x.
Shouldn't this be x(i,:) instead?
The i-th column of the array x is the average of the i-th group of j columns of the array y.
For example, if i is 1 and j is 3, the 1st column of x is the average of the first three columns of y.
Related
So I want to concatenate an m x n matrix to obtain a 1 x mn matrix. The matrix I want to concatenate are generated from a while loop. Although the number of columns will always be 3, I however cannot tell how many rows there will be for each iteration. Also, the row sizes for each iteration may not always be the same.
The code runs in cases where the row sizes were all equal to 6, but in cases where they aren't equal I get an error:
Error using vertcat Dimensions of matrices being concatenated are not consistent.
parts of the code are as follows:
A = [];
B = [];
searchArea = 2;
for ii = 1: numel(velocity)
Do ....
while area(ii,:) < searchArea
Do ....
% COLLATE vectors for A
A = [A; [Ax(ii), Ay(ii), Az(ii)]];
Do ...
end
%# Copy the A into new variable (B) and Reshape into row vector so as to associate each row to its corresponding velocity
B = [B; reshape(A.',1,[])];
A = [];
end
Could someone please advice me on what I am doing wrong here. I would clarify further if there be need. Thanks guys!
If it's your intent that B ends up being a row vector, then you need to change this:
B = [B; reshape(A.',1,[])]; % Does vertical concatenation
to this:
B = [B reshape(A.',1,[])]; % Does horizontal concatenation (note there's no semicolon)
so that each row vector gotten from reshaping A gets added to the end of the row instead of as a new row (as the semicolon indicates).
I have sum of 3 cell arrays
A=72x1
B=72x720
C=72x90
resultant=A+B+C
size of resultant=72x64800
now when I find the minimum value with row and column indices I can locate the row element easily but how can I locate the column element in variables?
for example
after dong calculations for A,B,C I added them all and got a resultant in from of <72x(720x90)> or can say a matrix of integers of size <72x64800> then I found the minimum value of resultant with row and column index using the code below.
[minimumValue,ind]=min(resultant(:));
[row,col]=find(result== minimumValue);
then row got 14 and column got 6840 value..
now I can trace row 14 of all A,B,C variables easily but how can I know that the resultant column 6480 belongs to which combination of A,B,C?
Instead of using find, use the ind output from the min function. This is the linear index for minimumValue. To do that you can use ind2sub:
[r,c] = ind2sub(size(resultant),ind);
It is not quite clear what do you mean by resultant = A+B+C since you clearly don't sum them if you get a bigger array (72x64800), on the other hand, this is not a simple concatenation ([A B C]) since this would result in a 72x811 array.
However, assuming this is a concatenation you can do the following:
% get the 2nd dimension size of all matrices:
cols = cellfun(#(x) size(x,2),{A,B,C})
% create a vector with reapiting matrices names for all their columns:
mats = repelem(['A' 'B' 'C'],cols);
% get the relevant matrix for the c column:
mats(c)
so mats(c) will be the matrix with the minimum value.
EDIT:
From your comment I understand that your code looks something like this:
% arbitrary data:
A = rand(72,1);
B = rand(72,720);
C = rand(72,90);
% initializing:
K = size(B,2);
N = size(C,2);
counter = 1;
resultant = zeros(72,K*N);
% summing:
for k = 1:K
for n = 1:N
resultant(:,counter) = A + B(:,k) + C(:,n);
counter = counter+1;
end
end
% finding the minimum value:
[minimumValue,ind] = min(resultant(:))
and from the start of the answer you know that you can do this:
[r,c] = ind2sub(size(resultant),ind)
to get the row and column of minimumValue in resultant. So, in the same way you can do:
[Ccol,Bcol] = ind2sub([N,K],c)
where Bcol and Ccol is the column in B and C, respectively, so that:
minimumValue == A(r) + B(r,Bcol) + C(r,Ccol)
To see how it's working imagine that the loop above fills a matrix M with the value of counter, and M has a size of N-by-K. Because we fill M with a linear index, it will be filled in a column-major way, so the row will correspond to the n iterator, and the column will correspond to the k iterator. Now c corresponds to the counter where we got the minimum value, and the row and column of counter in M tells us the columns in B and C, so we can use ind2sub again to get the subscripts of the position of counter. Off course, we don't really need to create M, because the values within it are just the linear indices themselves.
I need to create a function that has the input argument n, a integer , n>1 , and an output argument v, which is a column vector of length n containing all the positive integers smaller than or equal to n, arranged in such a way that no element of the vector equals its own index.
I know how to define the function
This is what I tried so far but it doesn't work
function[v]=int_col(n)
[1,n] = size(n);
k=1:n;
v=n(1:n);
v=k'
end
Let's take a look at what you have:
[1,n] = size(n);
This line doesn't make a lot of sense: n is an integer, which means that size(n) will give you [1,1], you don't need that. (Also an expression like [1,n] can't be on the left hand side of an assignment.) Drop that line. It's useless.
k=1:n;
That line is pretty good, k is now a row vector of size n containing the integers from 1 to n.
v=n(1:n);
Doesn't make sense. n isn't a vector (or you can say it's a 1x1 vector) either way, indexing into it (that's what the parentheses do) doesn't make sense. Drop that line too.
v=k'
That's also a nice line. It makes a column vector v out of your row vector k. The only thing that this doesn't satisfy is the "arranged in such a way that no element of the vector equals its own index" part, since right now every element equals its own index. So now you need to find a way to either shift those elements or shuffle them around in some way that satisfies this condition and you'd be done.
Let's give a working solution. You should really look into it and see how this thing works. It's important to solve the problem in smaller steps and to know what the code is doing.
function [v] = int_col(n)
if n <= 1
error('argument must be >1')
end
v = 1:n; % generate a row-vector of 1 to n
v = v'; % make it a column vector
v = circshift(v,1); % shift all elements by 1
end
This is the result:
>> int_col(5)
ans =
5
1
2
3
4
Instead of using circshift you can do the following as well:
v = [v(end);v(1:end-1)];
So I ran into this bug today
A_TEST(dest,:)=A_TEST(source,:)+A_TEST(dest,:);
A_TEST(:,dest)=A_TEST(:,source)+A_TEST(:,dest);
If dest is non-unique, this fails (which makes sense). So my quick fix is to do a for loop over dest
for (k=1:numel(dest))
A(dest(k),:)=A(source(k),:)+A(dest(k),:);
A(:,dest(k))=A(:,source(k))+A(:,dest(k));
end
And matlab is bad at such for loops. How would one vectorize this call?
With the following, I show how to do it with rows.
To do it with columns, it's a similar approach but different code, I'll explain why.
To summarize, you have a matrix A, with n rows and p columns.
You have a list of integers in the range [1,n], src, and idem for dst.
I'm assuming that they both have the same size, and might contain more than n elements (so that there are repetitions in both potentially).
Grouping the srcs by dsts, it's clear that the operation you're talking about is equivalent to a linear recombination of rows. This is equivalent to a pre-multiplication by a n x n matrix in which element (i,j) = k means "the recombination corresponding to the destination row i contains row j with multiplicity k".
This is what the following code does:
function A = madScience( A, src, dst )
n = size(A,1);
M = eye(n);
ix = sub2ind( [n,n], dst, src );
[ux,~,mlt] = unique( ix );
nux = length(ux);
mlt = accumarray( mlt(:), 1, [nux,1] );
M(ux) = M(ux) + mlt';
A = M*A;
end
Note 1: The two codes that you give in your post are NOT equivalent; you would need two separate for loops to make them equivalent.
Note 2: The same operation on columns is equivalent to a post multiplication by a matrix in which element (i,j) = k means "the recombination corresponding to column j contains column i with multiplicity k".
Note 3: If A is square, then both operations can be performed with the same matrix M as (M*A) * M' (the parenthesis are important).
There are two matrices; the first one is my input matrix
and the second one ("renaming matrix") is used to replace the values of the first one
That is, looking at the renaming matrix; 701 must be replaced by 1,...,717 must be replaced by 10,etc.. such that the input matrix becomes as such
The ? values are defined but i didn't put them. The second column of the input matrix is already sorted(ascending order from top down) but the values are not consecutive(no "710": see first pic).
The question is how to get the output matrix(last pic) from the first two.
Looks to me like it's screaming for a sparse matrix solution. In matlab you can create a sparse matrix with the following command:
SM = sparse( ri, ci, val );
where ri is the row index of the non-zero elements, ci is the corresponding column index, and val is the values.
Let's call your input matrix IM and your lookup matrix LUM, then we construct the sparse matrix:
nr = size(LUM, 1);
SM = sparse( ones(nr, 1), LUM(:, 1), LUM(:, 2) );
Now we can get your result in a single line:
newMatrix = reshape(SM(1, IM), size(IM));
almost magic.
I didn't have a chance to check this tonight - but if it doesn't work exactly as described, it should be really really close...
If the values in the first column all appear in the second column, and if all you want is replace the values in the second column by 1..n and change the values in the first column accordingly, you can do all of this with a simple call to ismember:
%# define "inputMatrix" here as the first array in your post
[~,newFirstColumn] = ismember(inputMatrix(:,1),inputMatrix(:,2));
To create your output, you'd then write
outputMatrix = [newFirstColumn,(1:length(newFirstColumn))'];
If M is the original matrix and R is the renaming matrix, here's how you do it
N = M;
for n = 1:size(M,1)
N(find(M==R(n,1))) = R(n,2);
end
Note that in this case you're creating a new matrix N with the renamed values. You don't have to do that if you like.