If I have matrix size(mat)= X*Y*6
let call mat(:,:,1)=A
and mat(:,:,2)=B and so on
how do i rezise mat to X*Y*12
where
mat(:,:,1)=mat(:,:,2)= A
mat(:,:,3)=mat(:,:,4)=B
and so on
You can use the following syntax:
%defines input matrix (in your case it is already defined)
m = 500;
n = 400;
z = 6;
mat = rand(m,n,z);
%initialize output matrix
newMat = zeros(m,n,z*2);
%assign old matrix values into the new matrix
newMat(:,:,1:2:end) = mat;
newMat(:,:,2:2:end) = mat;
If you have Matlab 2015a or newer you can use repelem:
N = 2; %// number of times to repeat
result = repelem(mat, 1, 1, N); %// repeat N times along 3rd dim
For older Matlab versions you can do it manually as follows:
N = 2; %// number of times to repeat
ind = ceil(1/N:1/N:size(mat,3)); %// build index with repetitions
result = mat(:,:,ind); %// apply index along desired dim
Example:
>> %// Data
>> mat = randi(9,2,4,2)
mat(:,:,1) =
5 8 9 2
7 3 1 5
mat(:,:,2) =
5 7 1 1
1 8 8 2
>> %// First approach
>> N = 2;
>> result = repelem(mat, 1, 1, N)
result(:,:,1) =
5 8 9 2
7 3 1 5
result(:,:,2) =
5 8 9 2
7 3 1 5
result(:,:,3) =
5 7 1 1
1 8 8 2
result(:,:,4) =
5 7 1 1
1 8 8 2
>> %// Second approach
>> N = 2;
>> ind = ceil(1/N:1/N:size(mat,3));
>> result = mat(:,:,ind)
result(:,:,1) =
5 8 9 2
7 3 1 5
result(:,:,2) =
5 8 9 2
7 3 1 5
result(:,:,3) =
5 7 1 1
1 8 8 2
result(:,:,4) =
5 7 1 1
1 8 8 2
Related
If I have this vector:
x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6]
I would like to get the position of each unique number according to itself.
y = [1 2 3 4 5 1 2 3 1 1 2 1 2 3 4]
At the moment I'm using:
y = sum(triu(x==x.')) % MATLAB 2016b and above
It's compact but obviously not memory efficient.
For the pure beauty of MATLAB programming I would avoid using a loop. Do you have a better simple implementation ?
Context:
My final goal is to sort the vector x but with the constraint that a number that appear N times has the priority over another number that has appeared more than N times:
[~,ind] = sort(y);
x_relative_sort = x(ind);
% x_relative_sort = 1 2 3 4 6 1 2 4 6 1 2 6 1 6 1
Assuming x is sorted, here's one vectorized alternative using unique, diff, and cumsum:
[~, index] = unique(x);
y = ones(size(x));
y(index(2:end)) = y(index(2:end))-diff(index).';
y = cumsum(y);
And now you can apply your final sorting:
>> [~, ind] = sort(y);
>> x_relative_sort = x(ind)
x_relative_sort =
1 2 3 4 6 1 2 4 6 1 2 6 1 6 1
If you have positive integers you can use sparse matrix:
[y ,~] = find(sort(sparse(1:numel(x), x, true), 1, 'descend'));
Likewise x_relative_sort can directly be computed:
[x_relative_sort ,~] = find(sort(sparse(x ,1:numel(x),true), 2, 'descend'));
Just for variety, here's a solution based on accumarray. It works for x sorted and containing positive integers, as in the question:
y = cell2mat(accumarray(x(:), x(:), [], #(t){1:numel(t)}).');
You can be more memory efficient by only comparing to unique(x), so you don't have a large N*N matrix but rather N*M, where N=numel(x), M=numel(unique(x)).
I've used an anonymous function syntax to avoid declaring an intermediate matrix variable, needed as it's used twice - this can probably be improved.
f = #(X) sum(cumsum(X,2).*X); y = f(unique(x).'==x);
Here's my solution that doesn't require sorting:
x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
y = cell2mat( splitapply(#(v){cumsum(v)},x,cumsum(logical([1 diff(x)]))) ) ./ x;
Explanation:
% Turn each group new into a unique number:
t1 = cumsum(logical([1 diff(x)]));
% x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
% t1 = [1 1 1 1 1 2 2 2 3 4 4 5 5 5 5 6 6 6];
% Apply cumsum separately to each group:
t2 = cell2mat( splitapply(#(v){cumsum(v)},x,t1) );
% t1 = [1 1 1 1 1 2 2 2 3 4 4 5 5 5 5 6 6 6];
% t2 = [1 2 3 4 5 2 4 6 3 4 8 6 12 18 24 1 2 3];
% Finally, divide by x to get the increasing values:
y = t2 ./ x;
% x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
% t2 = [1 2 3 4 5 2 4 6 3 4 8 6 12 18 24 1 2 3];
I have a matrix which contains values and I wish to find the index of the top n minimum values.
I use the following code for finding the minimum most value:
[r,c]=find(Result==min(min(Result)));
I cant find any other questions on stack overflow which answer the question, please help
Maybe you could do something like this:
sorted = sort(Result(:));
topten = sorted(1:10);
[~,ia,~] = intersect(Result(:),topten(:)); % // Get the indices of the top ten values
[r,c]=ind2sub(size(Result),ia); % // Convert the indices to rows and columns
Or without Intersect in the other answer
[sorted,I] = sort(Result(:));
[r,c] = ind2sub(size(Result),I(1:10)); %//Change 10 to any other required value
Use prctile (Statistics Toolbox) to find the appropriate threshold, and then use indexing to select the elements above that threshold:
x = magic(4); %// example
n = 5; %// we want the top n elements
M = numel(x);
p = prctile(x(:), (M-n)/M*100);
indices = find(x>p); %// result in the form linear indices
[row, col] = find(x>p); %// result in the form of row and column indices
In this example:
>> x
x =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
>> indices.'
ans =
1 8 12 13 15
>> row.'
ans =
1 4 4 1 3
>> col.'
ans =
1 2 3 4 4
>> x(indices).'
ans =
16 14 15 13 12
Example with repeated elements:
>> x = [1 1 2 5; 3 4 3 5];
>> n = 5;
gives
>> indices.'
ans =
2 4 6 7 8
>> row.'
ans =
2 2 2 1 2
>> col.'
ans =
1 2 3 4 4
>> x(indices).'
ans =
3 4 3 5 5
I have a matrix having rows with repeated numbers. I want to find those rows and replace them with a dummy row so as to keep the number of rows of the matrix constant.
Dummy_row = [1 2 3]
(5x3) Matrix A
A = [2 3 6;
4 7 4;
8 7 2;
1 3 1;
7 8 2]
(5x3) Matrix new_A
new_A = [2 3 6;
1 2 3;
8 7 2;
1 2 3;
7 8 2]
I tried the following which deleted the rows having repeated numbers.
y = [1 2 3]
w = sort(A,2)
v = all(diff(t,1,2)~=0|w(:,1:2)==0,2) % When v is zero, the row has repeated numbers
z = A(w,:)
Can you please help?
bsxfun based solution -
%// Create a row mask of the elements that are to be edited
mask = any(sum(bsxfun(#eq,A,permute(A,[1 3 2])),2)>1,3);
%// Setup output variable and set to-be-edited rows as copies of [1 2 3]
new_A = A;
new_A(mask,:) = repmat(Dummy_row,sum(mask),1)
Code run -
A =
2 3 6
4 7 4
8 7 2
1 3 1
7 8 2
new_A =
2 3 6
1 2 3
8 7 2
1 2 3
7 8 2
You could use the following:
hasRepeatingNums = any(diff(sort(A, 2), 1, 2)==0, 2);
A(hasRepeatingNums,:) = repmat(Dummy_row, nnz(hasRepeatingNums), 1);
See if this works for you,
A= [ 2 3 6;
4 7 4;
8 7 2;
5 5 5;
1 8 8;
1 3 1;
7 8 2 ];
Dummy_row = [1 2 3];
b = diff(sort(A,2),1,2);
b = sum(b == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
gives,
newA =
2 3 6
1 2 3
8 7 2
1 2 3
1 2 3
1 2 3
7 8 2
Edit
Not much change is needed, try this,
Dummy_row = [1 2 3];
b = sum(A == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
i want to control the creation of random numbers in this matrix :
Mp = floor(1+(10*rand(2,20)));
mp1 = sort(Mp,2);
i want to modify this code in order to have an output like this :
1 1 2 2 3 3 3 4 5 5 6 7 7 8 9 9 10 10 10 10
1 2 3 3 3 3 3 3 4 5 6 6 6 6 7 8 9 9 9 10
i have to fill each row with all the numbers going from 1 to 10 in an increasing order and the second matrix that counts the occurences of each number should be like this :
1 2 1 2 1 2 3 1 1 2 1 1 2 1 1 2 1 2 3 4
1 1 1 2 3 4 5 6 1 1 1 2 3 4 1 1 1 2 3 1
and the most tricky matrix that i'v been looking for since the last week is the third matrix that should skim through each row of the first matrix and returns the numbers of occurences of each number and the position of the last occcurence.here is an example of how the code should work. this example show the intended result after running through the first row of the first matrix.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (positions)
1 2
2 2
3 3
4 1
5 2
6 1
7 2
8 1
9 2
10 4
(numbers)
this example show the intended result after running through the second row of the first matrix.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (positions)
1 1 2
2 1 2
3 3 6
4 1 1
5 3
6 1 4
7 2 1
8 1 1
9 2 3
10 4
(numbers)
so the wanted matrix must be filled up with zeros from the beginning and each time after running through each row of the first matrix, we add the new result to the previous one...
I believe the following code does everything you asked for. If I didn't understand, you need to get a lot clearer in how you pose your question...
Note - I hard coded some values / sizes. In "real code" you would never do that, obviously.
% the bit of code that generates and sorts the initial matrix:
Mp = floor(1+(10*rand(2,20)));
mp1 = sort(Mp, 2);
clc
disp(mp1)
occCount = zeros(size(mp1));
for ii = 1:size(mp1,1)
for jj = 1:size(mp1,2)
if (jj == 1)
occCount(ii,jj) = 1;
else
if (mp1(ii,jj) == mp1(ii,jj-1))
occCount(ii,jj) = occCount(ii, jj-1) + 1;
else
occCount(ii,jj) = 1;
end
end
end
end
% this is the second matrix you asked for
disp(occCount)
% now the third:
big = zeros(10, 20);
for ii = 1:size(mp1,1)
for jj = 1:10
f = find(mp1(ii,:) == jj); % index of all of them
if numel(f) > 0
last = f(end);
n = numel(f);
big(jj, last) = big(jj, last) + n;
end
end
end
disp(big)
Please see if this is indeed what you had in mind.
The following code solves both the second and third matrix generation problems with a single loop. For clarity, the second matrix M2 is the 2-by-20 array in the example containing the cumulative occurrence count. The third matrix M3 is the sparse matrix of size 10-by-20 in the example that encodes the number and position of the last occurrence of each unique value. The code only loops over the rows, using accumarray to do most of the work. It is generalized to any size and content of mp1, as long as the rows are sorted first.
% data
mp1 = [1 1 2 2 3 3 3 4 5 5 6 7 7 8 9 9 10 10 10 10;
1 2 3 3 3 3 3 3 4 5 6 6 6 6 7 8 9 9 9 10]; % the example first matrix
nuniq = max(mp1(:));
% accumulate
M2 = zeros(size(mp1));
M3 = zeros(nuniq,size(mp1,2));
for ir=1:size(mp1,1),
cumSums = accumarray(mp1(ir,:)',1:size(mp1,2),[],#numel,[],true)';
segments = arrayfun(#(x)1:x,nonzeros(cumSums),'uni',false);
M2(ir,:) = [segments{:}];
countCoords = accumarray(mp1(ir,:)',1:size(mp1,2),[],#max,[],true);
[ii,jj] = find(countCoords);
nzinds = sub2ind(size(M3),ii,nonzeros(countCoords));
M3(nzinds) = M3(nzinds) + nonzeros(cumSums);
end
I won't print the outputs because they are a bit big for the answer, and the code is runnable as is.
NOTE: For new test data, I suggest using the commands Mp = randi(10,[2,20]); mp1 = sort(Mp,2);. Or based on your request to user2875617 and his response, ensure all numbers with mp1 = sort([repmat(1:10,2,1) randi(10,[2,10])],2); but that isn't really random...
EDIT: Error in code fixed.
I am editing the previous answer to check if it is fast when mp1 is large, and apparently it is:
N = 20000; M = 200; P = 100;
mp1 = sort([repmat(1:P, M, 1), ceil(P*rand(M,N-P))], 2);
tic
% Initialise output matrices
out1 = zeros(M, N); out2 = zeros(P, N);
for gg = 1:M
% Frequencies of each row
freqs(:, 1) = mp1(gg, [find(diff(mp1(gg, :))), end]);
freqs(:, 2) = histc(mp1(gg, :), freqs(:, 1));
cumfreqs = cumsum(freqs(:, 2));
k = 1;
for hh = 1:numel(freqs(:, 1))
out1(gg, k:cumfreqs(hh)) = 1:freqs(hh, 2);
out2(freqs(hh, 1), cumfreqs(hh)) = out2(freqs(hh, 1), cumfreqs(hh)) + freqs(hh, 2);
k = cumfreqs(hh) + 1;
end
end
toc
My input is the following:
X = [1 1; 1 2; 1 3; 1 4; 2 5; 1 6; 2 7; 1 8];
X =
1 1
1 2
1 3
1 4
2 5
1 6
2 7
1 8
I am looking to output a new matrix based on the value of the first column. If the value is equal to 1 -- the output will remain the same, when the value is equal to 2 then I would like to output two of the values contained in the second row. Like this:
Y =
1
2
3
4
5
5
6
7
7
8
Where 5 is output two times because the value in the first column is 2 and the same for 7
Here it is (vectorized):
C = cumsum(X(:,1))
A(C) = X(:,2)
D = hankel(A)
D(D==0) = inf
Y = min(D)
Edit:
Had a small bug, now it works.
% untested code:
Y = []; % would be better to pre-allocate
for ii = 1:size(X,1)
Y = [Y; X(ii,2)*ones(X(ii,1),1)];
end