How in matlab I can interactively append matrix with rows?
For example lets say I have empty matrix:
m = [];
and when I run the for loop, I get rows that I need to insert into matrix.
For example:
for i=1:5
row = v - x; % for example getting 1 2 3
% m.append(row)?
end
so after inserting it should look something like:
m = [
1 2 3
3 2 1
1 2 3
4 3 2
1 1 1
]
In most programming languages you can simply append rows into array/matrix. But I find it hard to do it in matlab.
m = [m ; new_row]; in your loop. If you know the total row number already, define m=zeros(row_num,column_num);, then in your loop m(i,:) = new_row;
Just use
m = [m; row];
Take into account that extending a matrix is slow, as it involves memory reallocation. It's better to preallocate the matrix to its full size,
m = NaN(numRows,numCols);
and then fill the row values at each iteration:
m(ii,:) = row;
Also, it's better not to use i as a variable name, because by default it represents the imaginary unit (that's why I'm using ii here as iteration index).
To create and add a value into the matrix you can do this and can make a complete matrix like yours.
Here row = 5 and then column = 3 and for hence two for loop.
Put the value in M(i, j) location and it will insert the value in the matrix
for i=1:5
for j=1:3
M(i, j) = input('Enter a value = ')
end
fprintf('Row %d inserted successfully\n', i)
end
disp('Full Matrix is = ')
disp(M)
Provably if you enter the same values given, the output will be like yours,
Full Matrix is =
1 2 3
3 2 1
1 2 3
4 3 2
1 1 1
Related
I have an NxR random matrix which has in each row shuffled 1:R.
I need another matrix with different sorts of shuffling in each row to have the same distribution of 1 to R for each column.
In other words, is there any way to shuffle rows of a matrix while keeping the frequency of each column the same?
I attached a screenshot of matrix B which I generated manually based on Matrix A.
SS of matrix A and B
Both have 1 to 4 in each row and the distribution of columns of A matches with columns of B. For example, in the first column of both matrices, there are two 1s, four 2s, one 3, and three 4s.
Is there any way to write an algorithm to generate the matrix B for larger dimensions?
The following is my code in which I tried my luck hoping I could get the solution from randomness, but I was unable to.
clear
n=20;
r= 8;
A= cell2mat(arrayfun(#randperm, repmat(r, 1, n), 'UniformOutput', false)'); % a random mtx
%Frequency of each column
for i=1:r
F_A(:,i) = histcounts(A(:,i), r);
end
k=0;
while true
B= A ;
for i=1:n
idx = randperm(r);
B(i, idx) = A(i,:) ;
end
for i=1:r
F_B(:,i) = histcounts(B(:,i), r); %Frequency of each column
end
k=k+1;
if sum(F_A == F_B, 'all') == r^2 || k >100000 % stop if frequency of A and B are similar
break
end
end
sum(F_A == F_B, 'all')
sum(A==B, 'all')
Here is an O(N^2*R/2) time algorithm to solve this problem. It might sacrifice some randomness due to the nature of the constraints, but you can apply multiple iterations of the algorithm if you want higher degree of randomness. The run time of the algorithm is much better than the exponential time of random algorithm that has no guarantee of convergence.
The steps are simple,
Initialize B matrix as a row-shuffled version of A.
Walk throw all elements of B scanning row by row:
pick a random element (i,k) in the same row as the current (i,j) element
if the same (i,j) element is picked, skip to the next (i,j+1) element
search down in the jth and kth columns for the reversed pair (l,k) and (l,j)
now, we can interchange (i,j),(i,k) and (l,j),(l,k) without unbalancing the column frequency
Repeat step 2 if more randomness is required.
So, the idea is to randomly interchange two elements in the same row if there exists an exact reversed pair in another row below. This introduces an unbalance in two columns and is fixed by the reversed change in the other row below; the net change in column frequency will be zero.
clc, clear
rng(123) % for reproducible outputs
N = 10; R = 4;
A = zeros(N,R);
for i = 1:N
A(i,:) = randperm(R);
end
% Initialize B as shuffled rows of A
B = A(randperm(N),:);
for i = 1:N
for j = 1:R
k = randi(R); % pick random element in the row
if j == k, continue, end % if same element, skip
for l = i+1:N % search down for the 2 reverse elements
if B(l,[j k]) == B(i,[k j]) % if found
B(l,[j k]) = B(l,[k j]); % replace the reverse elements
B(i,[j k]) = B(i,[k j]); % replace the k, j elements
break
end
end
end
end
A sample output of 1 iteration:
A =
3 2 4 1
2 4 1 3
3 2 1 4
2 3 1 4
2 1 3 4
4 1 3 2
4 2 3 1
3 1 4 2
4 3 1 2
1 4 2 3
B =
2 3 4 1
1 4 2 3
3 2 1 4
4 1 3 2
4 2 3 1
2 3 1 4
4 1 3 2
3 2 1 4
2 1 4 3
3 4 1 2
There are a matlab function shuffle, or you can take a look this question.
E.g., using the arrayfun( ) function to randperm( ) each column of A into a cell array result, then converting the cell array result into a single matrix:
[m,n] = size(A)
B = cell2mat(arrayfun(#(j)A(randperm(m),j),1:n,'uni',false))
A sample run:
Let say we have the vector v=[1,2,3] and we want to build the matrix of all the combinations of the numbers contained in v, i.e.
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
Since I'm not good in recursion, firstly I tried to write the code to build such a matrix by using for loops
makeLoop([1,2,3])
function A = makeLoop(v)
loops=length(v);
for i = 1:loops
dummy=v;
m=factorial(loops)/loops;
A((1+m*(i-1)):m*i,1)=v(i);
v(i)=[];
loops2=length(v);
for j = 1:loops2
dummy2=v;
m2=factorial(loops2)/loops2;
A(((1+m2*(j-1))+m*(i-1)):(m2*j+m*(i-1)),2)=v(j);
v(j)=[];
loops3=length(v);
for k = 1:loops3
m3=factorial(loops3)/loops3;
A(((1+m2*(j-1))+m*(i-1)):(m2*j+m*(i-1)),3)=v(k);
end
v=dummy2;
end
v=dummy;
end
end
it seems like it work, but obviously write it all for a bigger v would be like hell. Anyway I don't understand how to properly write the recursion, I think the recursive structure will be something like this
function A = makeLoop(v)
if length(v)==1
"do the last for loop"
else
"do a regular loop and call makeLoop(v) (v shrink at each loop)"
end
but I don't get which parts should I remove from the original code, and which to keep.
You were very close! The overall structure that you proposed is sound and your loopy-code can be inserted into it with practically no changes:
function A = makeLoop(v)
% number of (remaining) elements in the vector
loops = length(v);
if loops==1 %"do the last for loop"
A = v; %Obviously, if you input only a single number, the output has to be that number
else %"do a regular loop and call makeLoop(v) (v shrink at each loop)"
%preallocate matrix to store results
A = zeros(factorial(loops),loops);
%number of results per vector element
m = factorial(loops)/loops;
for i = 1:loops
%For each element of the vector, call the function again with that element missing.
dummy = v;
dummy(i) = [];
AOut = makeLoop(dummy);
%Then add that element back to the beginning of the output and store it.
A((1+m*(i-1)):m*i,:) = [bsxfun(#times,v(i),ones(m,1)) AOut];
end
end
Explanation bsxfun() line:
First, read the bsxfun documentation, it explains how it works way better than I could. But long story short, with bsxfun() we can replicate a scalar easily by multiplying it with a column vector of ones. E.g. bsxfun(#times,5,[1;1;1]) will result in the vector [5;5;5]. Note that since Matlab 2016b, bsxfun(#times,5,[1;1;1]) can written shorter as 5.*[1;1;1]
To the task at hand, we want to add v(i) in front (as the first column) of all permutations that may occur after it. Therefore we need to replicate the v(i) into the 1. dimension to match the number of rows of AOut, which is done with bsxfun(#times,v(i),ones(m,1)). Then we just horizontally concatenate this with AOut.
You can simply use the perms function to achieve this:
v = [1 2 3];
perms(v)
ans =
3 2 1
3 1 2
2 3 1
2 1 3
1 3 2
1 2 3
If you want them sorted using the same criterion you applied in the desired output, use the following code (refer to this page for an official documentation of the sortrows functon):
v = [1 2 3];
p = perms(v);
p = sortrows(p)
p =
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
I am working with a n x 1 matrix, A, that has repeating values inside it:
A = [0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4]
which correspond to an n x 1 matrix of B values:
B = [2;4;6;8;10; 3;5;7;9;11; 4;6;8;10;12; 5;7;9;11;13]
I am attempting to produce a generalised code to place each repetition into a separate column and store it into Aa and Bb, e.g.:
Aa = [0 0 0 0 Bb = [2 3 4 5
1 1 1 1 4 5 6 7
2 2 2 2 6 7 8 9
3 3 3 3 8 9 10 11
4 4 4 4] 10 11 12 13]
Essentially, each repetition from A and B needs to be copied into the next column and then deleted from the first column
So far I have managed to identify how many repetitions there are and copy the entire column over to the next column and then the next for the amount of repetitions there are but my method doesn't shift the matrix rows to columns as such.
clc;clf;close all
A = [0;1;2;3;4;0;1;2;3;4;0;1;2;3;4;0;1;2;3;4];
B = [2;4;6;8;10;3;5;7;9;11;4;6;8;10;12;5;7;9;11;13];
desiredCol = 1; %next column to go to
destinationCol = 0; %column to start on
n = length(A);
for i = 2:1:n-1
if A == 0;
A = [ A(:, 1:destinationCol)...
A(:, desiredCol+1:destinationCol)...
A(:, desiredCol)...
A(:, destinationCol+1:end) ];
end
end
A = [...] retrieved from Move a set of N-rows to another column in MATLAB
Any hints would be much appreciated. If you need further explanation, let me know!
Thanks!
Given our discussion in the comments, all you need is to use reshape which converts a matrix of known dimensions into an output matrix with specified dimensions provided that the number of elements match. You wish to transform a vector which has a set amount of repeating patterns into a matrix where each column has one of these repeating instances. reshape creates a matrix in column-major order where values are sampled column-wise and the matrix is populated this way. This is perfect for your situation.
Assuming that you already know how many "repeats" you're expecting, we call this An, you simply need to reshape your vector so that it has T = n / An rows where n is the length of the vector. Something like this will work.
n = numel(A); T = n / An;
Aa = reshape(A, T, []);
Bb = reshape(B, T, []);
The third parameter has empty braces and this tells MATLAB to infer how many columns there will be given that there are T rows. Technically, this would simply be An columns but it's nice to show you how flexible MATLAB can be.
If you say you already know the repeated subvector, and the number of times it repeats then it is relatively straight forward:
First make your new A matrix with the repmat function.
Then remap your B vector to the same size as you new A matrix
% Given that you already have the repeated subvector Asub, and the number
% of times it repeats; An:
Asub = [0;1;2;3;4];
An = 4;
lengthAsub = length(Asub);
Anew = repmat(Asub, [1,An]);
% If you can assume that the number of elements in B is equal to the number
% of elements in A:
numberColumns = size(Anew, 2);
newB = zeros(size(Anew));
for i = 1:numberColumns
indexStart = (i-1) * lengthAsub + 1;
indexEnd = indexStart + An;
newB(:,i) = B(indexStart:indexEnd);
end
If you don't know what is in your original A vector, but you do know it is repetitive, if you assume that the pattern has no repeats you can use the find function to find when the first element is repeated:
lengthAsub = find(A(2:end) == A(1), 1);
Asub = A(1:lengthAsub);
An = length(A) / lengthAsub
Hopefully this fits in with your data: the only reason it would not is if your subvector within A is a pattern which does not have unique numbers, such as:
A = [0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0;]
It is worth noting that from the above intuitively you would have lengthAsub = find(A(2:end) == A(1), 1) - 1;, But this is not necessary because you are already effectively taking the one off by only looking in the matrix A(2:end).
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
is there any possibility to assign multiple values for a matrix from an another vector without a loop?
For example:
I have a matrix filled with zeros:
matrix=zeros(2);
matrix =
0 0
0 0
Now i have an another vector where the first two columns are the positions and the third column are the values wich belongs to the corresponding positions.
values=[2 1 4;1 2 2]
values =
Posx PosY Value
2 1 4
1 2 2
The result should look like:
matrix =
0 2 <-- matrix(values(2,1),values(2,2))=values(2,3) ;
4 0 <-- matrix(values(1,1),values(1,2))=values(1,3);
This isn't pretty, but it is a one liner:
matrix(size(matrix,1) * (values(:,2) - 1) + values(:,1)) = values(:,3)
I can make it a bit clearer by splitting it into two lines. The idea is that you transform the first two columns of values into a one dimensional indexing vector which has as many elements as there are values to be assigned, and then assign values:
index = size(matrix,1) * (values(:,2) - 1) + values(:,1)
matrix(index) = values(:,3)
When you index into a matrix with a vector it counts down the columns first, and then across the rows. To make it even more clear, split the first statement up some more:
numRows = size(matrix,1)
rowIndex = values(:,1)
colIndex = values(:,2)
vals = values(:,3)
index = numRows * (colIndex - 1) + rowIndex
matrix(index) = vals
In fact, you don't need to go through all the trouble of building the index vector, as the function sub2ind exists to do that for you:
index = sub2ind(size(matrix), rowIndex, colIndex)
matrix(index) = vals
although I think it's good to see how to get the results with a call to sub2index, for your own education.
I made a function to do that, you can use it, if you want:
function B = ndassign( A , varargin )
%%% copy A to B, and assign values to A at specified nd indexes
%%% B=ndind(A,X,Y,Z,V)
%%% ---> B(X(i),Y(i),Z(i))=V(i)
%%% Example:
%%% ndassign(eye(3),[1 2 3],[3 2 1],[4 5 6])
%%% ans =
%%% 1 0 4
%%% 0 5 0
%%% 6 0 1
B=A;
inds=sub2ind(size(A),varargin{1:end-1});
B(inds)=varargin{end};
end