Suppose I have a matrix of dimension [4x4], and a vector of [16x1], I need to multiply every 4 element in the vector in one element in the matrix, (instead of multiplying element in row by element in vector), how can I do that using loop ?
For example here below, the results of the first four elements in the resulted vector as shown in the below example, then the same thing for the second, third and fourth rows in the matrix. :
So the results in that example is supposed to be [16x1]
Thank you
Using kron you can use this one-liner:
%A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
%v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
sum(kron(A,ones(4,4)).'.*v).'/4
I use the kronecker tensor product to "replicate" 4x4 time the A matrice. After that it's pure algebra.
This is just matrix multiplication in disguise... If your tall vector was a matrix of the same size as the matrix shown, where each highlighted block is a row, it's matrix multiplication. We can set this up, then reshape back into a vector.
You can use indexing to turn this into simple matrix multiplication. A question I answered already today details how the below indexing works using bsxfun, then we just reshape at the end:
% Setup
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
% Matrix mutliplication
r = numel(v)/size(A,1);
b = A * v( bsxfun( #plus, (1:r:numel(v)).', 0:r-1 ) );
% Make result a column vector
b = reshape( b.', [], 1 );
See if this is what you want:
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
r = reshape(sum(bsxfun(#times, permute(A, [3 2 1]), permute(reshape(v, 1, [], size(A,2)), [2 3 1])), 2), [], 1);
which gives
r =
17
17
17
17
41
41
41
41
65
65
65
65
89
89
89
89
There are details that I assumed, but this shoudl do the trick:
A=reshape(1:16,4,4).';
b=repelem([2,0,1,3],1,4).';
c=[];
for row=1:size(A,1)
c=[ c; sum(reshape(repelem(A(row,:),4).*b.',4,[]),2)];
end
I am assuming here that your demo for the vector is just a bad example and that you wont have repeated values, otherwise an easier version can be achieved by just not doing 3/4ths of the multiplications.
If you do not have access to repelem, have a look at alterative codes that do the same thing:Element-wise array replication in Matlab
Related
I have two matrix M x N, For simplicity, we take 4x4:
Matrix A:
1 4 2 5
4 5 8 2
3 4 5 6
2 3 5 8
Matrix B:
10 11 12 13
56 11 23 45
34 44 33 25
25 63 35 78
If an element of matrix A is greater then 5, then we change it from matrix B.
At the end we must get a matrix C:
1 4 2 5
4 5 23 2
3 4 5 25
2 3 5 78
How can I make it, should I use something like logical indexing..
Yes, you should use logical indexing:
C = A;
C(C>5) = B(C>5);
This means that every element in C that is >5 is set to the corresponding value in B.
or
C = A.*(A<=5) + B.*(A>5);
The comparisons in the parentheses create arrays with 0 and 1, so the first multiplication sets all elements of A to zero that should be taken from B, and second multiplication sets all elements of B to zero that should be taken from A.
In MATLAB, say I have a set of square matrices, say A, with trace(A)=0 as follows:
For example,
A = [0 1 2; 3 0 4; 5 6 0]
How can I remove the zeros and then vertically collapse the matrix to become as follow:
A_reduced = [1 2; 3 4; 5 6]
More generally, what if the zeroes can appear anywhere in the column (i.e., not necessarily at the long diagonal)? Assuming, of course, that the total number of zeros for all columns are the same.
The matrix can be quite big (hundreds x hundreds in dimension). So, an efficient way will be appreciated.
To compress the matrix vertically (assuming every column has the same number of zeros):
A_reduced_v = reshape(nonzeros(A), nnz(A(:,1)), []);
To compress the matrix horizontally (assuming every row has the same number of zeros):
A_reduced_h = reshape(nonzeros(A.'), nnz(A(1,:)), []).';
Case #1
Assuming that A has equal number of zeros across all rows, you can compress it horizontally (i.e. per row) with this -
At = A' %//'# transpose input array
out = reshape(At(At~=0),size(A,2)-sum(A(1,:)==0),[]).' %//'# final output
Sample code run -
>> A
A =
0 3 0 2
3 0 0 1
7 0 6 0
1 0 6 0
0 16 0 9
>> out
out =
3 2
3 1
7 6
1 6
16 9
Case #2
If A has equal number of zeros across all columns, you can compress it vertically (i.e. per column) with something like this -
out = reshape(A(A~=0),size(A,1)-sum(A(:,1)==0),[]) %//'# final output
Sample code run -
>> A
A =
0 3 7 1 0
3 0 0 0 16
0 0 6 6 0
2 1 0 0 9
>> out
out =
3 3 7 1 16
2 1 6 6 9
This seems to work, quite fiddly to get the behaviour right with transposing:
>> B = A';
>> C = B(:);
>> reshape(C(~C==0), size(A) - [1, 0])'
ans =
1 2
3 4
5 6
As your zeros are always in the main diagonal you can do the following:
l = tril(A, -1);
u = triu(A, 1);
out = l(:, 1:end-1) + u(:, 2:end)
A correct and very simple way to do what you want is:
A = [0 1 2; 3 0 4; 5 6 0]
A =
0 1 2
3 0 4
5 6 0
A = sort((A(find(A))))
A =
1
2
3
4
5
6
A = reshape(A, 2, 3)
A =
1 3 5
2 4 6
I came up with almost the same solution as Mr E's though with another reshape command. This solution is more universal, as it uses the number of rows in A to create the final matrix, instead of counting the number of zeros or assuming a fixed number of zeros..
B = A.';
B = B(:);
C = reshape(B(B~=0),[],size(A,1)).'
I am considering an easy algorithm to rank my 2D array and mark their rank in the same size of the 2D array.
For example, I have a matrix in below:
[0 2 15 34;
0 15 21 24;
0 3 5 8;
1 14 23 29]
The output should be as follow:
[1 5 10 16;
1 10 12 14;
1 6 7 8;
4 9 13 15]
I am kind of new to matlab, I not sure if the matlab have the functionality to directly do it. Or it would be even better if you could provide some ideas for implementing the algorithm. Thank you very much!
If I understand correctly, you want to replace each element by its rank. I offer three ways to do it; the third seems to be what you want.
Let your example data be defined as
data = [0 2 15 34;
0 15 21 24;
0 3 5 8;
1 14 23 29];
This assigns equal ranks to equal data values (as in your example), but doesn't skip ranks in that case (your example seems to do so):
[~, ~, vv] = unique(data(:));
result = reshape(vv, size(data));
With your example data, this gives
result =
1 3 8 13
1 8 9 11
1 4 5 6
2 7 10 12
This assigns different ranks to equal data values (so skipping ranks is out of the question):
[~, vv] = sort(data(:));
[~, vv] = sort(vv);
result = reshape(vv, size(data));
With your example data,
result =
1 5 11 16
2 10 12 14
3 6 7 8
4 9 13 15
This assigns equal ranks to equal data values, and in that case it skips ranks:
[~, vv] = sort(data(:));
[~, vv] = sort(vv);
[~, jj, kk] = unique(data(:), 'first');
result = reshape(vv(jj(kk)), size(data));
With your example data,
result =
1 5 10 16
1 10 12 14
1 6 7 8
4 9 13 15
Another approach, single-line: for each entry, find how many other entries are smaller, and add 1:
result = reshape(sum(bsxfun(#lt,data(:),data(:).'))+1, size(data));
everyone
I have problem when make iteration for two variable but, it's combine just in one vector or array
At first i write my input for iteration w(0) as w
w=[1 50];
for number 1, I use array
e=0:1:(n-1);
f=0:2:(2*n-2); %for 50 in column 2.
I try to use this code
w=[1 50];
ww=kron(ones((n),1),w)
e=0:1:(n-1);
f=0:2:(2*n-2);
r=[e',f']
x=ww+r
and the output is
ww =
1 50
1 50
1 50
1 50
1 50
1 50
r =
0 0
1 2
2 4
3 6
4 8
5 10
x =
1 50
2 52
3 54
4 56
5 58
6 60
I want x is output just in one array, in example
x =
1
50
2
52
3
54
4
56
5
58
6
60
where w=[1 50] can be use difference addition for iteration
Apply this to your x matrix:
x = reshape(x.',[],1);
See reshape doc for details.
Here is a simple method to create your vector from scratch:
x = [1:6;50:2:60];
x(:)
Or with your variables:
x = [e; f];
x(:)
There are multiple occurrence of same combination of values in different rows of matlab, for example 1 1 in first and second row. I want to remove all those duplicates but adding the values in third column. In case of 1 1 it will be 7. Finally I want to create a similarity matrix as shown below in Answer. I don't mind 2*values in diagonals because I will not be considering diagonal elements in further work. The code below does this but it is not vectorized. Can this be vectorized somehow. Example is given below.
datain = [ 1 1 3;
1 1 4;
1 2 5;
1 2 4;
1 2 3;
1 3 8;
1 3 7;
1 3 12;
2 2 22;
2 2 77;
2 3 111;
2 3 113;
3 3 456;
3 3 568];
cmp1=unique(datain(:,1));
cmp1sz=size(cmp1,1);
cmp2=unique(datain(:,2));
cmp2sz=size(cmp2,1);
thetotal=zeros(cmp1sz,cmp2sz);
for i=1:size(datain,1)
for j=1:cmp1sz
for k=1:cmp2sz
if datain(i,1)==cmp1(j,1) && datain(i,2)== cmp2(k,1)
thetotal(j,k)=thetotal(j,k)+datain(i,3);
thetotal(k,j)=thetotal(k,j)+datain(i,3);
end
end
end
end
The answer is
14 12 27
12 198 224
27 224 2048
This is a poster case for using ACCUMARRAY.
thetotal = accumarray(datain(:,1:2),datain(:,3),[],#sum,0);
%# to make the array symmetric, you simply add its transpose
thetotal = thetotal + thetotal'
thetotal =
14 12 27
12 198 224
27 224 2048
EDIT
So what if datain does not contain only integer values? In this case, you can still construct a table, but e.g. thetotal(1,1) will not correspond to datain(1,1:2) == [1 1], but to the smallest entry in the first two columns of datain.
[uniqueVals,~,tmp] = unique(reshape(datain(:,1:2),[],1));
correspondingIndices = reshape(tmp,size(datain(:,1:2)));
thetotal = accumarray(correspondingIndices,datain(:,3),[],#sum,0);
The value at [1 1] now corresponds to the row [uniqueVals(1) uniqueVals(1)] in the first two cols of datain.