I am aggregating 3x3 matrix D to 2x2 (adding column/row 1 and 2 together). Names for the rows and columns of matrix D are kept in 1x3 vector m.
And I am looking for a way to aggregate vector M by removing one of elements so it has the size 2x1.
Given 3x3 matrix D:
D =
1 2 3
4 5 6
7 8 9
And 2x3 aggregation matrix S:
S =
1 1 0
0 0 1
I aggregate D into 2x2 matrix by:
D = S*D*S'
D =
12 15
9 9
And the names for the columns and rows are stored in 3x1 vector m.
m =
'A'
'B'
'C'
I want to remove element 2 ('B') so the final output looks like this:
m =
'A'
'C'
Any suggestions ?
I assumed that you want to find the top left corners of blocks of a block-diagonal matrix and return their column indexes.
For example for the following matrix:
11100 10000
11100 00000
00010 => 00010
00010 00000
00001 00001
The result will be [1 4 5].
S = [1 1 0
0 0 1];
m = {'A';'B';'C'};
sz = size(S);
h = diff([zeros(sz(1),1),S],1,2)>0;
v = diff([zeros(1,sz(2));S],1,1)>0;
[~,idx]=find(h&v);
result = m(idx)
Related
I think the question is pretty basic, but still keeps me busy since some time.
Lets assume we have a vector containing 4 integers randomly repetetive, like:
v = [ 1 3 3 3 4 2 1 2 3 4 3 2 1 4 3 3 4 2 2]
I am searching for the vector of all positions of each integer, e.g. for 1 it should be a vector like:
position_one = [1 7 13]
Since I want to search every row of a 100x10000 matrix I was not able to deal with linear indeces.
Thanks in advance!
Rows and columns
Since your output for every integer changes, a cell array will fit the whole task. For the whole matrix, you can do something like:
A = randi(4,10,30); % some data
Row = repmat((1:size(A,1)).',1,size(A,2)); % index of the row
Col = repmat((1:size(A,2)),size(A,1),1); % index of the column
pos = #(n) [Row(A==n) Col(A==n)]; % Anonymous function to find the indices of 'n'
than for every n you can write:
>> pos(3)
ans =
1 1
2 1
5 1
6 1
9 1
8 2
3 3
. .
. .
. .
where the first column is the row, and the second is the column for every instance of n in A.
And for all ns you can use an arrayfun:
positions = arrayfun(pos,1:max(A(:)),'UniformOutput',false) % a loop that goes over all n's
or a simple for loop (faster):
positions = cell(1,max(A(:)));
for n = 1:max(A(:))
positions(n) = {pos(n)};
end
The output in both cases would be a cell array:
positions =
[70x2 double] [78x2 double] [76x2 double] [76x2 double]
and for every n you can write positions{n}, to get for example:
>> positions{1}
ans =
10 1
2 3
5 3
3 4
5 4
1 5
4 5
. .
. .
. .
Only rows
If all you want in the column index per a given row and n, you can write this:
A = randi(4,10,30);
row_pos = #(k,n) A(k,:)==n;
positions = false(size(A,1),max(A(:)),size(A,2));
for n = 1:max(A(:))
positions(:,n,:) = row_pos(1:size(A,1),n);
end
now, positions is a logical 3-D array, that every row corresponds to a row in A, every column corresponds to a value of n, and the third dimension is the presence vector for the combination of row and n. this way, we can define R to be the column index:
R = 1:size(A,2);
and then find the relevant positions for a given row and n. For instance, the column indices of n=3 in row 9 is:
>> R(positions(9,3,:))
ans =
2 6 18 19 23 24 26 27
this would be just like calling find(A(9,:)==3), but if you need to perform this many times, the finding all indices and store them in positions (which is logical so it is not so big) would be faster.
Find linear indexes in a matrix: I = find(A == 1).
Find two dimensional indexes in matrix A: [row, col] = find(A == 1).
%Create sample matrix, with elements equal one:
A = zeros(5, 4);
A([2 10 14]) = 1
A =
0 0 0 0
1 0 0 0
0 0 0 0
0 0 1 0
0 1 0 0
Find ones as linear indexes in A:
find(A == 1)
ans =
2
10
14
%This is the same as reshaping A to a vector and find ones in the vector:
B = A(:);
find(B == 1);
B' =
0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
Find two dimensional indexes:
[row, col] = find(A == 1)
row =
2
5
4
col =
1
2
3
You can do that with accumarray using an anonymous function as follows:
positions = accumarray(v(:), 1:numel(v), [], #(x) {sort(x.')});
For
v = [ 1 3 3 3 4 2 1 2 3 4 3 2 1 4 3 3 4 2 2];
this gives
positions{1} =
1 7 13
positions{2} =
6 8 12 18 19
positions{3} =
2 3 4 9 11 15 16
positions{4} =
5 10 14 17
I have matrix lets say MAT of size 2*n
I want to sort the matrix by row num 2
BUT to keep each info from row 1 to its row 2 info
prev
C K A L E Y B
4 2 1 3 6 7 7
and after sort
A K L C E Y B
1 2 3 4 6 7 7
any idea?
you can use the second output argument of sort:
[~, si] = sort(MAT(2,:));
res = MAT(:,si);
Consider an n-by-k matrix M and an p-by-1 vector of indexes V ranging from 1 to n. How can I create the p-by-k matrix C where each row corresponds to the entry of M referred to by the value in each row of V.
Example
M = 1 1
1 2
1 3
1 4
and
V = 2
1
3
What I require is the matrix
C = 1 2
1 1
1 3
To assign the rows V of matrix M to a matrix C, you would write:
C = M(V,:);
I have a 4x5 matrix called A from which I want to select randomly 3 rows, then 4 random columns and then select those elements which coincide in those selected rows and columns so that I have 12 selected elements.Then I want to create a diagonal matrix called B which will have entries either 1 or 0 so that multiplication of that B matrix with reshaped A matrix (20x1) will give me those selected 12 elements of A.
How can I create that B matrix? Here is my code:
A=1:20;
A=reshape(A,4,5);
Mr=4;
Ma=3;
Na=4;
Nr=5;
M=Ma*Mr;
[S1,S2]=size(A);
N=S1*S2;
y2=zeros(size(A));
k1=randperm(S1);
k1=k1(1:Ma);
k2=randperm(S2);
k2=k2(1:Mr);
y2(k1,k2)=A(k1,k2);
It's a little hard to understand what you want and your code isn't much help but I think I've a solution for you.
I create a matrix (vector) of zeros of the same size as A and then use bsxfun to determine the indexes in this vector (which will be the diagonal of B) that should be 1.
>> A = reshape(1:20, 4, 5);
>> R = [1 2 3]; % Random rows
>> C = [2 3 4 5]; % Random columns
>> B = zeros(size(A));
>> B(bsxfun(#plus, C, size(A, 1)*(R-1).')) = 1;
>> B = diag(B(:));
>> V = B*A(:);
>> V = V(V ~= 0)
V =
2
3
4
5
6
7
8
9
10
11
12
13
Note: There is no need for B = diag(B(:)); we could have simply used element by element multiplication in Matlab.
>> V = B(:).*A(:);
>> V = V(V ~= 0)
Note: This may be overly complex or very poorly put together and there is probably a better way of doing it. It's my first real attempt at using bsxfun on my own.
Here is a hack but since you are creating y2 you might as well just use it instead of creating the useless B matrix. The bsxfun answer is much better.
A=1:20;
A=reshape(A,4,5);
Mr=4;
Ma=3;
Na=4;
Nr=5;
M=Ma*Mr;
[S1,S2]=size(A);
N=S1*S2;
y2=zeros(size(A));
k1=randperm(S1);
k1=k1(1:Ma);
k2=randperm(S2);
k2=k2(1:Mr);
y2(k1,k2)=A(k1,k2);
idx = reshape(y2 ~= 0, numel(y2), []);
B = diag(idx);
% "diagonal" matrix 12x20
B = B(all(B==0,2),:) = [];
output = B * A(:)
output =
1
3
4
9
11
12
13
15
16
17
19
20
y2 from example.
y2 =
1 0 9 13 17
0 0 0 0 0
3 0 11 15 19
4 0 12 16 20
I have a picture matrix A, the size of which is 200*3000 double. And I have another picture matrix B, the size of which is 200*1000 double. The 1000 columns of matrix B exactly comes from the columns of matrix A. My question is:
How to get a matrix C with the same size of matrix A, but only keep the original values of columns in matrix B? I mean the size of matrix C is 200*3000 double, but only 1000 columns have the same values as matrix B. The other 2000 columns will be set to another value d, that is my second question, what is the value I should set for d, so that the picture matrix C can distinguish from picture matrix A?
Use ismember with the 'rows' option. Here's an example:
A = [1 2 3 4; 5 6 7 8]; %// example A
B = [3 10 1; 7 20 5]; %// example B
val = NaN; %// example value to indicate no match
C = A; %// initiallize
ind = ismember(A.',B.','rows'); %// matching columns
C(:,~ind) = val; %// set non-matching columns to val
Equivalently, you coud replace ismember by bsxfun, so that line becomes
ind = any(all(bsxfun(#eq, A, permute(B, [1 3 2])), 1), 3);
In this example,
A =
1 2 3 4
5 6 7 8
B =
3 10 1
7 20 5
C =
1 NaN 3 NaN
5 NaN 7 NaN