I'd like to index matrix
x=[1:5;6:10]
x =
1 2 3 4 5
6 7 8 9 10
using array
[1,2,1,2,1]
to get
1 7 3 9 5
I tried this:
x([1,2,1,2,1],:)
ans =
1 2 3 4 5
6 7 8 9 10
1 2 3 4 5
6 7 8 9 10
1 2 3 4 5
but that is not what I want. Please help
I'd use linear indexing with sub2ind:
>> v = x(sub2ind(size(x),a,1:5))
v =
1 7 3 9 5
Let
ind = [1, 2, 1, 2, 1];
offset = [1:size(x, 1):numel(x)] - 1;
then
x(ind + offset)
returns what you want. This assumes that your index vector has an entry for every column of x and uses linear indexing to add a column offset to every in-column index.
Related
I want to take the union of some of the rows of a matrix x. The row numbers of the rows whose union has to be taken are given by vector r. Is there any built-in function in MATLAB that can do it?
x = [1 2 4 0 0;
3 6 5 0 0;
7 8 10 12 9;
2 4 6 7 0;
3 4 5 8 12];
r = [1, 3, 5];
I think this will work for you - first, take the submatrix x(r,:) with the rows you want, and then find all the unique values in it:
unique(x(r,:))
ans =
0
1
2
3
4
5
7
8
9
10
12
You could do it like this
>>> union(union(x(r(1),:),x(r(2),:)),x(r(3),:))
ans =
0 1 2 3 4 5 7 8 9 10 12
or set up a for loop that iterates over the vector r to compute all the unions
Given two parameters:
n %number of repetitions per value
k %max value to repeat
I would like to create a vector of size n*k, which is a concatenation of k vectors of size n, such that the i'th vector contains the value i at each coordinate.
Example:
n = 5;
k = 9;
Desired result:
[1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,9,9,9,9,9];
Is there an elegant way to achieve this?
Thanks!
quite a few ways to do it:
method 1:
A=1:k
repelem(A',n,1)'
method 2:
A=1:k
kron(A', ones(n,1))'
method 3:
A=1:k
B=repmat(A, n, 1)
B(:)'
method 4:
A=1:k
B=ones(n,1)*A
B(:)'
Here is an alternative method
A = reshape(mtimes((1:k).',ones(1,n)).',1,n*k)
A =
Columns 1 through 22
1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5
Columns 23 through 44
5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9
Column 45
9
It multiplies each element by ones n times
>> mtimes((1:k).',ones(1,5)).'
ans =
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
and then reshapes the whole matrix to one vector
Similarly to How to combine vectors of different length in a cell array into matrix in MATLAB I would like to combine matrix having different dimension, stored in a cell array, into a matrix having zeros instead of the empty spaces. Specifically, I have a cell array {1,3} having 3 matrix of size (3,3) (4,3) (4,3):
A={[1 2 3; 4 5 6; 7 8 9] [1 2 3; 4 5 6; 7 8 9; 9 9 9] [1 2 3; 4 5 6; 7 8 9; 4 4 4]}
and I would like to obtain something like:
B =
1 2 3 1 2 3 1 2 3
4 5 6 4 5 6 4 5 6
7 8 9 7 8 9 7 8 9
0 0 0 9 9 9 4 4 4
I tried using cellfun and cell2mat but I do not figure out how to do this. Thanks.
Even if other answers are good, I'd like to submit mine, using cellfun.
l = max(cellfun(#(x) length(x),A))
B = cell2mat(cellfun(#(x) [x;zeros(l-length(x),3)], A, 'UniformOutput', 0));
Using bsxfun's masking capability -
%// Convert A to 1D array
A1d = cellfun(#(x) x(:).',A,'Uni',0) %//'
%// Get dimensions of A cells
nrows = cellfun('size', A, 1)
ncols = cellfun('size', A, 2)
%// Create a mask of valid positions in output numeric array, where each of
%// those numeric values from A would be put
max_nrows = max(nrows)
mask = bsxfun(#le,[1:max_nrows]',repelem(nrows,ncols)) %//'
%// Setup output array and put A values into its masked positions
B = zeros(max_nrows,sum(ncols))
B(mask) = [A1d{:}]
Sample run
Input -
A={[1 2 3 5 6; 7 8 9 3 8] [1 2 3; 4 5 6; 7 8 9; 9 9 9] [1 2 3; 4 5 6; 7 8 9; 4 4 4]}
Output -
B =
1 2 3 5 6 1 2 3 1 2 3
7 8 9 3 8 4 5 6 4 5 6
0 0 0 0 0 7 8 9 7 8 9
0 0 0 0 0 9 9 9 4 4 4
I would be surprised if this is possible in one or a few lines. You will probably have to do some looping yourself. The following achieves what you want in the specific case of incompatible first dimension lengths:
A={[1 2 3; 4 5 6; 7 8 9] [1 2 3; 4 5 6; 7 8 9; 9 9 9] [1 2 3; 4 5 6; 7 8 9; 4 4 4]}
maxsize = max(cellfun(#(x) size(x, 1), A));
B = A;
for k = 1:numel(B)
if size(B{k}, 1) < maxsize
tmp = B{k};
B{k} = zeros(maxsize, size(tmp,1));
B{k}(1:size(tmp,1),1:size(tmp,2)) = tmp;
end
end
B = cat(2, B{:});
Now B is:
B =
1 2 3 1 2 3 1 2 3
4 5 6 4 5 6 4 5 6
7 8 9 7 8 9 7 8 9
0 0 0 9 9 9 4 4 4
I would do it using a good-old for loop, which is quite intuitive I think.
Here is the commented code:
clc;clear var
A={[1 2 3; 4 5 6; 7 8 9] [1 2 3; 4 5 6; 7 8 9; 9 9 9] [1 2 3; 4 5 6; 7 8 9; 4 4 4]};
%// Find the maximum rows and column # to initialize the output array.
MaxRow = max(cell2mat(cellfun(#(x) size(x,1),A,'Uni',0)));
SumCol = sum(cell2mat(cellfun(#(x) size(x,2),A,'Uni',0)));
B = zeros(MaxRow,SumCol);
%// Create a counter to keep track of the current columns to fill
ColumnCounter = 1;
for k = 1:numel(A)
%// Get the # of rows and columns for each cell from A
NumRows = size(A{k},1);
NumCols = size(A{k},2);
%// Fill the array
B(1:NumRows,ColumnCounter:ColumnCounter+NumCols-1) = A{k};
%// Update the counter
ColumnCounter = ColumnCounter+NumCols;
end
disp(B)
Output:
B =
1 2 3 1 2 3 1 2 3
4 5 6 4 5 6 4 5 6
7 8 9 7 8 9 7 8 9
0 0 0 9 9 9 4 4 4
[max_row , max_col] = max( size(A{1}) , size(A{2}) , size(A{3}) );
A{1}(end:max_row , end:max_col)=0;
A{2}(end:max_row , end:max_col)=0;
A{3}(end:max_row , end:max_col)=0;
B=[A{1} A{2} A{3}];
for this specific problem, simply this will do:
B=cat(1,A{:});
or what I often just give a try for 2D cells, and works for your example as well:
B=cell2mat(A');
if you literally don't give a f* what dimension it will be cut in (and you're exceedingly lazy): put the same into a try-catch-block and loop over some dims as below.
function A=cat_any(A)
for dims=1:10% who needs more than 10 dims? ... otherwise replace 10 with: max(cellfun(#ndims,in),[],'all')
try, A=cat(dims,A{:}); end
if ~iscell(A), return A; end
end
disp('Couldn''t cat!') %if we can't cat, tell the user
end
Beware, this might lead to unexpected results ... but in most cases simply just worked for me.
Given the matrix I = [1,2;3,4], I would like to duplicate the elements to create a matrix I2 such that:
I2 = [1 1 1 2 2 2
1 1 1 2 2 2
1 1 1 2 2 2
3 3 3 4 4 4
3 3 3 4 4 4
3 3 3 4 4 4]
Other than using repmat, what other methods or functions are available?
Use kron:
>> N = 3 %// Number of times to replicate a number in each dimension
>> I = [1,2;3,4];
>> kron(I, ones(N))
ans =
1 1 1 2 2 2
1 1 1 2 2 2
1 1 1 2 2 2
3 3 3 4 4 4
3 3 3 4 4 4
3 3 3 4 4 4
This probably deserves some explanation in case you're not aware of what kron does. kron stands for the Kronecker Tensor Product. kron between two matrices A of size m x n and B of size p x q creates an output matrix of size mp x nq such that:
Therefore, for each coefficient in A, we take this value, multiply it with every value in the matrix B and we position these matrices in the same order as we see in A. As such, if we let A = I, and B be the 3 x 3 matrix full of ones, you thus get the above result.
Using indexing:
I = [1, 2; 3, 4]; %// original matrix
n = 3; %// repetition factor
I2 = I(ceil(1/n:1/n:size(I,1)), ceil(1/n:1/n:size(I,2))); %// result
One-liner with bsxfun -
R = 3; %// Number of replications
I2 = reshape(bsxfun(#plus,permute(I,[3 1 4 2]),zeros(R,1,R)),R*size(I,1),[])
Sample run -
I =
3 2 5
9 8 9
I2 =
3 3 3 2 2 2 5 5 5
3 3 3 2 2 2 5 5 5
3 3 3 2 2 2 5 5 5
9 9 9 8 8 8 9 9 9
9 9 9 8 8 8 9 9 9
9 9 9 8 8 8 9 9 9
I have two matrix A and B. Suppose I would like to find in each row of matrix A the smallest number, and for the same cell that this number is in Matrix A, do find the corresponding number of the same cell in matrix B. For example the number in matrix A will be in the position A(1,3), A(2,9)...and I want the corresponding number in B(1,3), B(2,9)... Is it possible to do it, or I am asking something hard for matlab. Hope someone will help me.
What you can do is use min and find the minimum across all of the rows for each column. You would actually use the second output in order to find the location of each column per row that you want to find. Once you locate these, simply use sub2ind to access the corresponding values in B. As such, try something like this:
[~,ind] = min(A,[],2);
val = B(sub2ind(size(A), (1:size(A,1)).', ind));
val would contain the output values in the matrix B which correspond to the same positions as the minimum values of each row in A. This is also assuming that A and B are the same size. As an illustration, here's an example. Let's set A and B to be a random 4 x 4 array of integers each.
rng(123);
A = randi(10, 4, 4)
B = randi(10, 4, 4)
A =
7 8 5 5
3 5 4 1
3 10 4 4
6 7 8 8
B =
2 7 8 3
2 9 4 7
6 8 4 1
6 7 3 5
By running the first line of code, we get this:
[~,ind] = min(A,[],2)
ind =
3
4
1
1
This tells us that the minimum value of the first row is the third column, the minimum value of the next row is the 4th column, and so on and so forth. Once we have these column numbers, let's access what the corresponding values are in B, so we would want row and columns (1,3), (2,4), etc. Therefore, after running the second statement, we get:
val = B(sub2ind(size(A), (1:size(A,1)).', ind))
val =
8
7
6
6
If you quickly double check the accessed positions in B in comparison to A, we have found exactly those spots in B that correspond to A.
A = randi(9,[5 5]);
B = randi(9,[5 5]);
[C,I] = min(A');
B.*(A == repmat(C',1,size(A,2)))
example,
A =
2 1 6 9 1
2 4 4 4 2
5 6 5 5 5
9 3 9 3 6
4 5 6 8 3
B =
3 5 6 8 1
9 2 9 7 1
5 6 6 5 6
4 6 1 4 5
5 3 7 1 9
ans =
0 5 0 0 1
9 0 0 0 1
5 0 6 5 6
0 6 0 4 0
0 0 0 0 9
You can use it like,
B(A == repmat(C',1,5))
ans =
9
5
5
6
6
5
4
1
1
6
9