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.
Related
I'm a total beginner when it comes to MATLAB, so I have a question for this. How am I supposed to write this code out and if by any chance, can someone be kind enough to write out the code because I've been struggling with this. Use loops to create 3 x 5 matrix in which the value of each element is half of its row number plus three times its column number. for instance, the value of element (2,5) is: 1/22+35
The nested loops control the indexing of the matrix. The outer for-loop traverses through the rows of the matrix and the inner for-loop traverses through the columns of the matrix.
The second part of this question requires using a combination of the looping/scanning variables Row and Column to set the value of the matrix:
Matrix Value = (Row ÷ 2) + (3 × Column)
Number_Of_Rows = 3;
Number_Of_Columns = 5;
Matrix = zeros(Number_Of_Rows,Number_Of_Columns);
%Running through the array indices using two loops%
for Row = 1: Number_Of_Rows
for Column = 1: Number_Of_Columns
%Evaluating the value based on the current row and column index%
Matrix(Row,Column) = (Row/2) + (3*Column);
end
end
Matrix
Result:
Looping Methodology:
Variable Matrix Opened In Workspace:
Here's an intuitive way to do this:
% Initialize row num and column num
row = 3;
column = 5;
% H is the matrix of desire
% Initialize it as a 3*5 zero matrix
H = zeros(3,5);
% Outer loop, over column index
% Remember Matlab's index start with 1 not 0
for c = 1:column
% Inner Loop, over row index
for r = 1:row
% The algorithm of each element in the matrix
H(r,c) = 0.5*r+3*c;
end
end
I have a 102-by-102 matrix. I want to select square sub-matrices of orders from 2 up to 8 using random column numbers. Here is what I have done so far.
matt is the the original matrix of size 102-by-102.
ittr = 30
cols = 3;
for i = 1:ittr
rr = randi([2,102], cols,1);
mattsub = matt([rr(1) rr(2) rr(3)], [rr(1) rr(2) rr(3)]);
end
I have to extract matrices of different orders from 2 to 8. Using the above code I would have to change the mattsub line every time I change cols. I believe it is possible to do with another loop inside but cannot figure out how. How can I do this?
There is no need to extract elements of a vector and concatenate them, just use the vector to index a matrix.
Instead of :
mattsub = matt([rr(1) rr(2) rr(3)], [rr(1) rr(2) rr(3)]);
Use this:
mattsub = matt(rr, rr);
Defining a set of random sizes is pretty easy using the randi function. Once this is done, they can be projected along your iterations number N using arrayfun. Within the iterations, the randperm and sort functions can be used in order to build the random indexers to the original matrix M.
Here is the full code:
% Define the starting parameters...
M = rand(102);
N = 30;
% Retrieve the matrix rows and columns...
M_rows = size(M,1);
M_cols = size(M,2);
% Create a vector of random sizes between 2 and 8...
sizes = randi(7,N,1) + 1;
% Generate the random submatrices and insert them into a vector of cells...
subs = arrayfun(#(x)M(sort(randperm(M_rows,x)),sort(randperm(M_cols,x))),sizes,'UniformOutput',false);
This can work on any type of matrix, even non-squared ones.
You don't need another loop, one is enough. If you use randi to get a random integer as size of your submatrix, and then use those to get random column and row indices you can easily get a random submatrix. Do note that the ouput is a cell, as the submatrices won't all be of the same size.
N=102; % Or substitute with some size function
matt = rand(N); % Initial matrix, use your own
itr = 30; % Number of iterations
mattsub = cell(itr,1); % Cell for non-uniform output
for ii = 1:itr
X = randi(7)+1; % Get random integer between 2 and 7
colr = randi(N-X); % Random column
rowr = randi(N-X); % random row
mattsub{ii} = matt(rowr:(rowr+X-1),colr:(colr+X-1));
end
I have an N-by-M-Matrix as input called GR wich consists of the following numbers: -3,0,2,4,7,10,12
And I have to return a vector. If M=1, then it should just return the input.
If M>1 It should remove the lowest number from the matrix and then calculate the mean of the remaining numbers.
However, if one of the numbers in the row is -3, it should return the value -3 in the output.
My thoughts of the problem:
Is it possible to make a for loop?
for i=1:length(GR(:,1))
If length(GR(1,:))==1
GR=GR
end
If length(GR(1,:))>1
x=min(GR(i,:))=[] % for removing the lowest number in the row
GR=sum(x)/length(x(i,:))
I just don't have any Idea of how to detect if any of the numbers in the row is -3 and then return that value instead of calculating the mean and when I tried to delete the lowest number in the matrix using x=min(GR(i,:)) matlab gave me this error massage 'Deletion requires an existing variable.'
I put in a break function. As soon as it detects a -3 value it breaks from the loop. Same goes for the other function.
Note that it is an i,j (M*N) matrix. So you might need to change your loop.
for i=1:length(GR(:,1))
if GR(i,1)==-3
GR=-3
break
end
If length(GR(1,:))==1
GR=GR
break
end
If length(GR(1,:))>1
x=min(GR(i,:))=[] % for removing the lowest number in the row
GR=sum(x)/length(x(i,:))
end
end
you can use Nan's, nanmean, any, and dim argument in these functions:
% generate random matrix
M = randi(3);
N = randi(3);
nums = [-3,0,2,4,7,10,12];
GR = reshape(randsample(nums,N*M,true),[N M]);
% computation:
% find if GR has only one column
if size(GR,2) == 1
res = GR;
else
% find indexes of rows with -3 in them
idxs3 = any(GR == -3,2);
% the (column) index of the min. value in each row
[~,minCol] = min(GR,[],2);
% convert [row,col] index pair into linear index
minInd = sub2ind(size(GR),1:size(GR,1),minCol');
% set minimum value in each row to nan - to ignore it on averaging
GR(minInd) = nan;
% averaging each rows (except for the Nans)
res = nanmean(GR,2);
% set each row with (-3) in it to (-3)
res(idxs3) = -3;
end
disp(res)
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 have a column vector of data in variable vdata and a list of indeces idx. I want to access vdata at the indeces x before and x after each index in idx. One way I would do it in a for loop is:
x = 10;
accessed_data = [];
for (ii = 1:length(idx))
accessed_data = vdata(idx-x:idx+x);
end
Is there a way to do this in a vectorized function? I found a solution to a very similar question here: Addressing multiple ranges via indices in a vector but I don't understand the code :(.
Assuming min(idx)-x>0 and max(idx)+x<=numel(vdata) then you can simply do
iidx = bsxfun(#plus, idx(:), -x:x); % create all indices
accessed_data = vdata( iidx );
One scheme that uses direct indexing instead of a for loop:
xx = (-x:x).'; % Range of indices
idxx = bsxfun(#plus,xx(:,ones(1,numel(idx))),idx(:).'); % Build array
idxx = idxx(:); % Columnize to interleave columns
idxx = idxx(idxx>=1&idxx<=length(vdata)); % Make sure the idx+/-x is valid index
accessed_data = vdata(idxx); % Indices of data
The second line can be replaced with a form of the first line from #Shai's answer. This scheme checks that all of the resultant indices are valid. Because some might have to be removed, you could end up with a ragged array. One way to solve this is to use cell arrays, but here I just make idxx a vector, and thus accessed_data is as well.
This gives the solution in a matrix, with one row for each value in idx. It assumes that all values in idx are greater than or equal to x, and less than or equal to length(vdata)-x.
% Data
x = 10;
idx = [12 20 15];
vdata = 1:100;
ind = repmat(-x:x,length(idx),1) + repmat(idx(:),1,2*x+1);
vdata(ind)