I would like your help to make more efficient (maybe, by vectorising) the Matlab code below. The code below does essentially the following: take a row vector A; consider the maximum elements of such a row vector and let, for example, be i and j their positions; construct two columns vectors, the first with all zeros but a 1 positioned at i, the second with all zeros but a 1 positioned at j.
This is my attempt with loops, but it looks more complicated than needed.
clear
rng default
A=[3 2 3];
max_idx=ismember(A,max(A));
vertex=cell(size(A,2),1);
for j=1:size(max_idx,2)
if max_idx(j)>0
position=find(max_idx(j));
vertex_temp=zeros(size(A,2),1);
vertex_temp(position)=1;
vertex{j}=vertex_temp;
else
vertex{j}=[];
end
end
vertex=vertex(~cellfun('isempty',vertex));
Still using a for loop, but more readable:
A = [3 2 3];
% find max indices
max_idx = find(A == max(A));
vertex = cell(numel(max_idx),1);
for k = 1:numel(max_idx)
vertex{k} = zeros(size(A,2),1); % init zeros
vertex{k}(max_idx(k)) = 1; % set value in vector to 1
end
If you really wanted to avoid a for loop, you could probably also use something like this:
A=[3 2 3];
max_idx = find(A==max(A));
outMat = zeros(numel(A), numel(max_idx));
outMat((0:(numel(max_idx)-1)) * numel(A) + max_idx) = 1;
then optionally, if you want them in separate cells rather than columns of a matrix:
outCell = mat2cell(outMat, numel(A), ones(1,numel(max_idx)))';
However, I think this might be less simple and readable than the existing answers.
Is there a specific reason you want a cell array rather than a matrix?
If you can have it all in one vector:
A = [3 2 3]
B_rowvec = A == max(A)
B_colvec = B_rowvec'
If you need them separated into separate vectors:
A = [3 2 3]
Nmaxval = sum(A==max(A))
outmat = zeros(length(A),Nmaxval)
for i = 1:Nmaxval
outmat(find(A==max(A),i),i)=1;
end
outvec1 = outmat(:,1)
outvec2 = outmat(:,2)
Basically, the second input for find will specify which satisfactory instance of the first input you want.
so therefore
example = [ 1 2 3 1 2 3 1 2 3 ]
first = find(example == 1, 1) % returns 1
second = find(example == 1, 2) % returns 4
third = find(example == 1, 3) % returns 7
Related
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).
Okay, so I have a script that will produce my vector of repeated integers of a certain interval, but now theres a particular instance where I need to make sure that once it is shuffled, the numbers do not repeat. So for example, I produced a vector of repeating 1-5, 36 times, shuffled. How do I ensure that there are no repeated numbers after shuffling? And to make things even more complex, I need to produce two such vectors that do not ever have the same value at the same index. For example, lets say 1:5 was repeated twice for these vectors, so then this would be what I'm looking for:
v1 v2
4 2
2 4
3 2
5 3
4 5
1 4
5 1
1 5
3 1
2 3
I made that right now by taking an example of 1 vector and just shifting it off by 1 to create another vector that will satisfy the requirements, but in my situation, that wont actually work because I can't have them be systematically dependent like that.
So I tried a recursive technique to make the script start over if the vectors did not make the cut and as expected, that did not go over so well. I hit my maximum recursive iterations and I've realized this is clearly not the way to go. Is there some other alternative?
EDIT:
So I found a way to satisfy some of the conditions I needed above in the following code:
a = nchoosek(1:5,2);
b = horzcat(a(:,2),a(:,1));
c = vertcat(a,b);
cols = repmat(c,9,1);
cols = cols(randperm(180),:);
I just need to find a way to shuffle cols that will also enforce no repeating numbers in columns, such that cols(i,1) ~= cols(i+1,1) and cols(i,2) ~= cols(i+1,2)
This works, but it probably is not very efficient for a large array:
a = nchoosek(1:5, 2);
while (any(a(1: end - 1, 1) == a(2: end, 1)) ...
|| any(a(1: end - 1, 2) == a(2: end, 2)))
random_indices = randperm(size(a, 1));
a = a(random_indices, :);
end
a
If you want something faster, the trick is to logically insert each row in a place where your conditions are satisfied, rather than randomly re-shuffling. For example:
n1 = 5;
n2 = 9;
a = nchoosek(1:n1, 2);
b = horzcat(a(:,2), a(:,1));
c = vertcat(a, b);
d = repmat(c, n2, 1);
d = d(randperm(n1 * n2), :);
% Perform an "insertion shuffle"
for k = 2: n1 * n2
% Grab row k from array d. Walk down the rows until a position is
% found where row k does not repeat with its upstairs or downstairs
% neighbors.
m = 1;
while (any(d(k,:) == d(m,:)) || any(d(k,:) == d(m+1,:)))
m = m + 1;
end
% Insert row k in the proper position.
if (m < k)
ind = [ 1: m k m+1: k-1 k+1: n1 * n2 ];
else
ind = [ 1: k-1 k+1: m k m+1: n1 * n2 ];
end
d = d(ind,:);
end
d
One way to solve this problem is to think both vectors as being created as follows:
For every row of arrays v1 and v2
Shuffle the array [1 2 3 4 5]
Set the values of v1 and v2 at the current row with the first and second value of the shuffle. Both values will always be different.
Code:
s = [1 2 3 4 5];
Nrows = 36;
solution = zeros(Nrows,2);
for k=1:Nrows
% obtain indexes j for shuffling array s
[x,j] = sort(rand(1,5));
%row k takes the first two values of shuffled array s
solution(k,1:2) = s(j(1:2));
end
v1 = solution(:,1);
v2 = solution(:,2);
Main edit: random => rand,
With this method there is no time wasted in re-rolling repeated numbers because the first and second value of shuffling [1 2 3 4 5] will always be different.
Should you need more than two arrays with different numbers the changes are simple.
I have a vector whose elements identify the indices (per column) that I need to set in a different matrix. Specifically, I have:
A = 7
1
2
and I need to create a matrix B with some number of rows of zeros, except for the elements identified by A. In other words, I want B:
B = zeros(10, 3); % number of rows is known; num columns = size(A)
B(A(1), 1) = 1
B(A(2), 2) = 1
B(A(3), 3) = 1
I would like to do this without having to write a loop.
Any pointers would be appreciated.
Thanks.
Use linear indexing:
B = zeros(10, 3);
B(A(:).'+ (0:numel(A)-1)*size(B,1)) = 1;
The second line can be written equivalently with sub2ind (may be a little slower):
B(sub2ind(size(B), A(:).', 1:numel(A))) = 1;
Here is part of a function that counts the Nyquist stability criterion. Function gives an error when vectors re or im do not have five elements:
Attempted to access im(5); index out of bounds because numel(im)=4.
Error in stability (line 27)
tran(w) = re(1) + re(2)*w.^1 + re(3)*w.^2 + re(4)*w.^3 +
re(5)w.^4 +1i(...
I want to assign 0 to the missing elements.
function answear=stability(re,im)
%% Function check stability of system
%re is real part of transmitation
%im is imagine part of transmitation
%% data
for n=1:length(re)
if(re(n) == [ ])
re(n) = 0;
end
end
for n=1:length(im)
if(im (n)== [ ])
im(n) = 0;
end
end
if( length(re) > length(im))
root = length(re);
else
root = length(im);
end
for w=1:root
tran(w) = re(1) + re(2)*w.^1 + re(3)*w.^2 + re(4)*w.^3 + re(5)*w.^4 +1i*(...
im(1) + im(2)*w.^1 + im(3)*w.^2 + im(4)*w.^3 +im(5)*w.^4);
end
Here are two ways to expand a vector to ensure it has at least 5 elements. Each working from a different thought but doing the same.
Most compact:
x = [1 2 3];
x(end+1:5) = 0;
Assigns zero to all elements behind the matrix till the fifth element.
However, if you already know the amount of elements, this is more compact:
x = [1 2 3];
if numel(x)<5
x(5) = 0;
end
Note that this will automatically expand with zeros untill it has 5 elements. The advantage of these methods it that it works, regardless of the orientation of your vector.
If you want to pad your vector with zeros to make certain it has at least 5 elements I suggest the following:
Make a vecotr of 5 zeros
Fill the first X elements of pad with your vector (lets call it A)
For examples with A begin less than 5 and A being more than 5 elements:
A= [1 2 3]
n=5
pad = zeros(1,n)
pad(1:length(A)) = A
clear
A = [5 6 7 8 9 4 3]
n=5
pad = zeros(1,n)
pad(1:length(A)) = A
Given a vector
A = [1,2,3,...,100]
I want to extract all elements, except every n-th. So, for n=5, my output should be
B = [1,2,3,4,6,7,8,9,11,...]
I know that you can access every n-th element by
A(5:5:end)
but I need something like the inverse command.
If this doesn't exist I would iterate over the elements and skip every n-th entry, but that would be the dirty way.
You can eliminate elements like this:
A = 1:100;
removalList = 1:5:100;
A(removalList) = [];
Use a mask. Let's say you have
A = 1 : 100;
Then
m = mod(0 : length(A) - 1, 5);
will be a vector of the same length as A containing the repeated sequence 0 1 2 3 4.
You want everything from A except the elements where m == 4, i.e.
B = A(m ~= 4);
will result in
B == [1 2 3 4 6 7 8 9 11 12 13 14 16 ...]
Or you can use logical indexing:
n = 5; % remove the fifth
idx = logical(zeroes(size(A))); % creates a blank mask
idx(n) = 1; % makes the nth element 1
A(idx) = []; % ta-da!
About the "inversion" command you cited, it is possible to achieve that behavior using logical indexing. You can negate the vector to transform every 1 in 0, and vice-versa.
So, this code will remove any BUT the fifth element:
negatedIdx = ~idx;
A(negatedIdx) = [];
why not use it like this?
say A is your vector
A = 1:100
n = 5
B = A([1:n-1,n+1:end])
then
B=[1 2 3 4 6 7 8 9 10 ...]
One possible solution for your problem is the function setdiff().
In your specific case, the solution would be:
lenA = length(A);
index = setdiff(1:lenA,n:n:lenA);
B = A(index)
If you do it all at once, you can avoid both extra variables:
B = A( setdiff(1:end,n:n:end) )
However, Logical Indexing is a faster option, as tested:
lenA = length(A);
index = true(1, lenA);
index(n:n:lenA) = false;
B = A(index)
All these codes assume that you have specified the variable n, and can adapt to a different value.
For the shortest amount of code, you were nearly there all ready. If you want to adjust your existing array use:
A(n:n:end)=[];
Or if you want a new array called B:
B=A;
B(n:n:end)=[];