I work on a matrix like this :
A=[1,2,3;5,4,6;9,8,7];
I want to get the index of each line. Here it is : Index = [1;2;3]
But how can I get this without a loop ?
I do this for the moment :
for k=1:length(A)
[~,Index(k)] = min(A(k,:));
end
From the documentation:
M = min(A,[],dim) returns the smallest elements along dimension dim. For example, if A is a matrix, then min(A,[],2) is a column vector containing the minimum value of each row.
Looking at output options, you can see
[M,I] = min(___) finds the indices of the minimum values of A and returns them in output vector I.
You were already using the second part of the above documentation notes, so just combine the two...
A=[1,2,3; 5,4,6; 9,8,7];
[~, idx] = min(A, [], 2);
% result: idx = [1; 2; 3];
Related
I have some binary matrix. I want to remove all first ones from each column, but keep one if this value is alone in column. I have some code, which produces correct result, but it looks ugly- I should iterate through all columns.
Could You give me a piece of advice how to improve my code?
Non-vectorised code:
% Dummy matrix for SE
M = 10^3;
N = 10^2;
ExampleMatrix = (rand(M,N)>0.9);
ExampleMatrix1=ExampleMatrix;
% Iterate columns
for iColumn = 1:size(ExampleMatrix,2)
idx = find(ExampleMatrix(:,iColumn)); % all nonzeroes elements
if numel(idx) > 1
% remove all ones except first
ExampleMatrix(idx(1),iColumn) = 0;
end
end
I think this does what you want:
ind_col = find(sum(ExampleMatrix, 1)>1); % index of relevant columns
[~, ind_row] = max(ExampleMatrix(:,ind_col), [], 1); % index of first max of each column
ExampleMatrix(ind_row + (ind_col-1)*size(ExampleMatrix,1)) = 0; % linear indexing
The code uses:
the fact that the second output of max gives the index of the first maximum value. In this case max is applied along the first dimension, to find the first maximum of each column;
linear indexing.
I'd like to be able to create a column vector whose values in each row correspond to the column in a matrix with the max value in that specific row.
For example,
If I have a matrix such as:
A = [1,5,2;3,1,1;0,1,0];
I'd end up with the matrix:
maxValueColumns = transpose([2,1,2]);
Is there an easy/efficient way to do this?
You're looking for max():
A = [1,5,2;3,1,1;0,1,0];
[~, maxValueColumns] = max(A, [], 2); % 'maxValueColumns' will contain [2; 1; 2]
I have a source matrix A(m,n) for which I used "find" and now I have a list of desired indices [y,x].
I also have a 3D matrix with dimensions B(m,n,3).
I want to extract all the elements in B using the result from find.
So if find yields 4 pairs of results, I would like to have a 4x3 matrix with the contents of the Z dimension of B for the resulting indices.
I tried many things but keep failing:
A = rand(480,640);
[y,x] = find(A < 0.5);
o = B(y,x,:);
Requested 39024x39024x3 (34.0GB) array exceeds maximum array size preference.
I am clearly doing something wrong since B has dimensions (640,640,3).
With the way you are trying to index, matlab tries to index B with every combination of the elements in y and x resulting in a massive matrix. I've implemented a for loop to do what I think you are asking.
I would also note that in order to index across B the first two dimensions need to be the same size as A, Otherwise you will not be able to index B past the maximum row or column index in A.
A = rand(480,640);
B = rand(480,640,3);
[x,y] = find(A < 0.5);
o = zeros(size(x,1),1,3); % x and y are the same length so it doesn't matter
for i = 1:size(x,1)
o(i,1,:)= B(x(i),y(i),:);
end
o = reshape(o,size(x,1),3);
You can reshape B to a 2D matrix of size [m*n , 3] then use logical indexing to extract elements:
C = reshape(B, [], 3);
o = C(A<0.5, :);
I have two matrices A and B. A(:,1) corresponds to an x-coordinate, A(:,2) corresponds to a y-coordinate, and A(:,3) corresponds to a certain radius. All three values in a row describe the same circle. Now let's say...
A =
[1,4,3]
[8,8,7]
[3,6,3]
B =
[1,3,3]
[1, 92,3]
[4,57,8]
[5,62,1]
[3,4,6]
[9,8,7]
What I need is to be able to loop through matrix A and determine if there are any rows in matrix B that are "similar" as in the x value is within a range (-2,2) of the x value of A (Likewise with the y-coordinate and radius).If it satisfies all three of these conditions, it will be added to a new matrix with the values that were in A. So for example I would need the above data to return...
ans =
[1,4,3]
[8,8,7]
Please help and thank you in advance to anyone willing to take the time!
You can use ismembertol.
result = A(ismembertol(A,B,2,'ByRows',1,'DataScale',1),:)
Manual method
A = [1,4,3;
8,8,7;
3,6,3];
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]; % example matrices
t = 2; % desired threshold
m = any(all(abs(bsxfun(#minus, A, permute(B, [3 2 1])))<=t, 2), 3);
result = A(m,:);
The key is using permute to move the first dimension of B to the third dimension. Then bsxfun computes the element-wise differences for all pairs of rows in the original matrices. A row of A should be selected if all the absolute differences with respect to any column of B are less than the desired threshold t. The resulting variable m is a logical index which is used for selecting those rows.
Using pdist2 (Statistics and Machine Learning Toolbox)
m = any(pdist2(A, B, 'chebychev')<=t, 2);
result = A(m,:);
Ths pdist2 function with the chebychev option computes the maximum coordinate difference (Chebychev distance, or L∞ metric) between pairs of rows.
With for loop
It should work:
A = [1,4,3;
8,8,7;
3,6,3]
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]
index = 1;
for i = 1:size(A,1)
C = abs(B - A(i,:));
if any(max(C,[],2)<=2)
out(index,:) = A(i,:);
index = index + 1
end
end
For each row of A, computes the absolute difference between B and that row, then checks if there exists a row in which the maximum is less than 2.
Without for loop
ind = any(max(abs(B - permute(A,[3 2 1])),[],2)<=2);
out = A(ind(:),:);
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.