Let's say, I have an M x N matrix. Now, I want to insert a constant M x 1 column vector (say all 1's) in between each of the N columns. Therefore, my resulting matrix would be of dimension (M x (2*N-1)), with every other column being 1's.
Is there an easy way to do that?
Vertically concatenate a matrix of ones, reshape, and cut off the last column of ones. For a matrix A:
B = reshape([A; ones(size(A))],size(A,1),[]);
B(:,end)=[]
Here is another way to do it, using the possibility of out of bounds indexing in assignments:
M(:,1:2:end*2)=M;
M(:,2:2:end)=1
If you don't mind creating a temporary matrix, one way to do it would be to do the following:
old_matrix = rand(M,N); % Just for example
new_matrix = ones(M,2*N-1);
new_matrix(:,1:2:end) = old_matrix;
Note that for an arbitrary constant matrix, you could replace the second line with the following:
new_matrix = repmat(const_array,1,2*N-1);
Related
Say A is a 3x4x5 array. I am given a vector a, say of dimension 2 and b of dimension 2. If I do A(a,b,:) it will give 5 matrices of dimensions 2x2. I instead want the piecewise vectors (without writing a for loop).
So, I want the two vectors of A which are given by (a's first element and b's first element) and (a's second element and b's second element)
How do I do this without a for loop? If A were two dimensions I could do this using sub2ind. I don't know how to access the entire vectors.
You can use sub2ind to find the linear index to the first element of each output vector: ind = sub2ind(size(A),a,b). To get the whole vectors, you can't do A(ind,:), because the : has to be the 3rd dimension. However, what you can do is reshape A to be 2D, collapsing the first two dimensions into one. We have a linear index to the vectors we want, that will correctly index the first dimension of this reshaped A:
% input:
A = rand(3,4,5);
a = [2,3];
b = [1,2];
% expected:
B = [squeeze(A(a(1),b(1),:)).';squeeze(A(a(2),b(2),:)).']
% solution:
ind = sub2ind(size(A),a,b);
C = reshape(A,[],size(A,3));
C = C(ind,:)
assert(isequal(B,C))
You can change a and b to be 3d arrays just like A and then the sub2ind should be able to index the whole matrix. Like this:
Edit: Someone pointed out a bug. I have changed it so that a correction gets added. The problem was that ind1, which should have had the index number for each desired element of A was only indexing the first "plane" of A. The fix is that for each additional "plane" in the z direction, the total number of elements in A in the previous "planes" must be added to the index.
A=rand(3,4,5);
a=[2,3];
b=[1,2];
a=repmat(a,1,1,size(A,3));
b=repmat(b,1,1,size(A,3));
ind1=sub2ind(size(A),a,b);
correction=(size(A,1)*size(A,2))*(0:size(A,3)-1);
correction=permute(correction,[3 1 2]);
ind1=ind1+repmat(correction,1,2,1);
out=A(ind1)
I have an array A size of 16X16 and I want to add first 3 rows out of 16 in A. What is the most efficient solution in MATLAB?
I tried this code but this is not efficient because I want to extend it for large arrays:
filename = 'n1.txt';
B = importdata(filename);
i = 1;
D = B(i,:)+ B(i+1,:)+ B(i+2,:);
For example, if I want to extend this for an array of size 256x256 and I want to extract 100 rows and add them, how I will do this?
A(1:3,:);%// first three rows.
This uses the standard indices of matrix notation. Check Luis's answer I linked for the full explanation on indices in all forms. For summing things:
B = A(1:100,:);%// first 100 rows
C = sum(B,1);%// sum per column
D = sum(B,2);%// sum per row
E = sum(B(:));%// sum all elements, rows and columns, to a single scalar
I have a two arrays within a <1x2 cell>. I want to permute those arrays. Of course, I could use a loop to permute each one, but is there any way to do that task at once, without using loops?
Example:
>> whos('M')
Name Size Bytes Class Attributes
M 1x2 9624 cell
>> permute(M,p_matrix)
This does not permute the contents of the two arrays within M.
I could use something like:
>> for k=1:size(M,2), M{k} = permute(M{k},p_matrix); end
but I'd prefer not to use loops.
Thanks.
This seems to work -
num_cells = numel(M) %// Number of cells in input cell array
size_cell = size(M{1}) %// Get sizes
%// Get size of the numeric array that will hold all of the data from the
%// input cell array with the second dimension representing the index of
%// each cell from the input cell array
size_num_arr = [size_cell(1) num_cells size_cell(2:end)]
%// Dimensions array for permuting with the numeric array holding all data
perm_dim = [1 3:numel(size_cell)+1 2]
%// Store data from input M into a vertically concatenated numeric array
num_array = vertcat(M{:})
%// Reshape and permute the numeric array such that the index to be used
%// for indexing data from different cells ends up as the final dimension
num_array = permute(reshape(num_array,size_num_arr),perm_dim)
num_array = permute(num_array,[p_matrix numel(size_cell)+1])
%// Save the numeric array as a cell array with each block from
%// thus obtained numeric array from its first to the second last dimension
%// forming each cell
size_num_arr2 = size(num_array)
size_num_arr2c = num2cell(size_num_arr2(1:end-1))
M = squeeze(mat2cell(num_array,size_num_arr2c{:},ones(1,num_cells)))
Some quick tests show that mat2cell would prove to be the bottleneck, so if you don't mind indexing into the intermediate numeric array variable num_array and use it's last dimension for an equivalent indexing into M, then this approach could be useful.
Now, another approach if you would like to preserve the cell format would be with arrayfun, assuming each cell of M to be a 4D numeric array -
M = arrayfun(#(x) num_array(:,:,:,:,x),1:N,'Uniform',0)
This seems to perform much better than with mat2cell in terms of performance.
Please note that arrayfun isn't a vectorized solution as most certainly it uses loops behind-the-scenes and seems like mat2cell is using for loops inside its source code, so please do keep all these issues in mind.
I have three column vectors, then I want to creat matrix from this column victors
A1(:);
A2(:);
A3(:)
each column vectors has 25 element then the new matrix C will be a matrix with 3x25
I want to make A1(:) the first column of matrix c
A2(:) second column
A3(:) third column
Use cat to concatenate along dimension 1 or 2 depending on how you input those three vectors.
Thus, you can use -
C = cat(2,A1(:),A2(:),A3(:)).'
Or
C = cat(1,A1(:).',A2(:).',A3(:).')
Of course, you can skip (:)'s, if you know that all those are column vectors.
The above two approaches assumes that you intend to get an output of size 3 x N, where is N is the number of elements in the column vectors. If you were looking to get an output of size N x 3 , i.e. where each column is formed from the elements of column vectors A1, A2 and so on, just drop the transpose from the first of the two approaches mentioned above. Thus, use this -
C = cat(2,A1(:),A2(:),A3(:))
I'm writing a function that requires some values in a matrix of arbitrary dimansions to be dropped in a specified dimension.
For example, say I have a 3x3 matrix:
a=[1,2,3;4,5,6;7,8,9];
I might want to drop the third element in each row, in which case I could do
a = a(:,1:2)
But what if the dimensions of a are arbitrary, and the dimension to trim is defined as an argument in the function?
Using linear indexing, and some carefully considered maths is an option but I was wondering if there is a neater soltion?
For those interested, this is my current code:
...
% Find length in each dimension
sz = size(dat);
% Get the proportion to trim in each dimension
k = sz(d)*abs(p);
% Get the decimal part and integer parts of k
int_part = fix(k);
dec_part = abs(k - int_part);
% Sort the array
dat = sort(dat,d);
% Trim the array in dimension d
if (int_part ~=0)
switch d
case 1
dat = dat(int_part + 1 : sz(1) - int_part,:);
case 2
dat = dat(:,int_part + 1 : sz(2) - int_part);
end
end
...
It doesn't get any neater than this:
function A = trim(A, n, d)
%// Remove n-th slice of A in dimension d
%// n can be vector of indices. d needs to be scalar
sub = repmat({':'}, 1, ndims(A));
sub{d} = n;
A(sub{:}) = [];
This makes use of the not very well known fact that the string ':' can be used as an index. With due credit to this answer by #AndrewJanke, and to #chappjc for bringing it to my attention.
a = a(:, 1:end-1)
end, used as a matrix index, always refers to the index of the last element of that matrix
if you want to trim different dimensions, the simplest way is using and if/else block - as MatLab only supports 7 dimensions at most, you wont need an infinite number of these to cover all bases
The permute function allows to permute the dimension of an array of any dimension.
You can place the dimension you want to trim in a prescribed position (the first, I guess), trim, and finally restore the original ordering. In this way you can avoid running loops and do what you want compactly.