I am trying to randomly disperse different numbers in MATLAB array:
I have two 3's, four 2's and I want to randomly populate ones vector (size 10,1).
End result look something like this:
A = [1;3;1;2;3;2;2;1;1;2;1;1]
Then I want to fix the values in A but add more random elements but I can only replace with higher numbers:
For example, to the matrix above I will randomly add two more 2's and two more 3's giving something like this
A= [3;3;2;2;3;2;2;2;1;2;1;3]
M = [3;3;2;2;2;2];
M(end+1:end+4) = 1;
M=M(randperm(10))
The second half of your question needs a lot of clarification.
First part
You can use randsample for that:
A = ones(1,12); %// original values
v = [3 3 2 2 2 2]; %// values to "disperse" in A
ind_replace = randsample(1:numel(A), numel(v)); %// index of entries to be replaced
A(ind_replace) = v;
If you don't have randsample (which is part of the Statistics Toolbox), use randperm and select the first few elements:
ind_replace = randperm(numel(A));
ind_replace = ind_replace(1:numel(v));
A(ind) = v;
Second part
To only replace entries which equal 1:
v = [2 2 3 3]; %// values to "disperse" among the 1 values in A
ind_ones = find(A==1); %// index of entries which equal one
ind_replace = randsample(1:numel(ind_ones), numel(v)); %// index within the above
%// Or: ind_replace = randperm(numel(ind_ones));
%// ind_replace = ind_replace(1:numel(v));
A(ind_ones(ind_replace)) = v;
Note this generalizes the first part, that is, it can also be used when all entries of A equal 1.
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).
I have an array with n dimensions, and I have a sequence along one dimension at a certain location on all other dimensions. How do I find the location of this sequence? Preferably without loops.
I use matlab. I know what dimension it should be in, but the sequence isnt necessarily there. Find and == dont work. I could make an nd find function using crosscorrelation but Im guessing this is already implemented and I just dont know what function to call.
example:
ND = rand(10,10,10,10);
V = ND(randi(10),randi(10),randi(10),:);
[I1, I2, I3] = find(ND==V);
Edit: The sequence to be found spans the entire dimension it is on, I did not mention this in my original formulation of the problem. Knedlsepp`s solution solves exactly the problem I had, but Luis' solution solves a more general problem for when the sequence doesn't necessarily span the entire dimension.
As there are multiple ways to interpret your question, I will clarify: This approach assumes a 1D sequence of size: numel(V) == size(ND, dimToSearch). So, for V = [1,2] and ND = [1,2,1,2] it is not applicable. If you want this functionality go with Luis Mendo's answer, if not this will likely be faster.
This will be a perfect opportunity to use bsxfun:
We start with some example data:
ND = rand(10,10,10,10);
V = ND(3,2,:,3);
If you don't have the vector V given in the correct dimension (in this case [1,1,10,1]) you can reshape it in the following way:
dimToSearch = 3;
Vdims = ones(1, ndims(ND));
Vdims(dimToSearch) = numel(V);
V = reshape(V, Vdims);
Now we generate a cell that will hold the indices of the matches:
I = cell(1, ndims(ND));
At this point we compute the size of ND if it were collapsed along the dimension dimToSearch (we compute dimToSearch according to V, as at this point it will have the correct dimensions):
dimToSearch = find(size(V)>1);
collapsedDims = size(ND);
collapsedDims(dimToSearch) = 1;
Finally the part where we actually look for the pattern:
[I{:}] = ind2sub(collapsedDims, find(all(bsxfun(#eq, ND, V), dimToSearch)));
This is done in the following way: bsxfun(#eq, ND, V) will implicitly repmat the array V so it has the same dimensions as ND and do an equality comparison. After this we do a check with all to see if all the entries in the dimension dimToSearch are equal. The calls to find and ind2sub will then generate the correct indices to your data.
Let d be the dimension along which to search. I'm assuming that the sought sequence V may be shorter than size(ND,d). So the sequence may appear once, more than once, or never along each dimension-d- "thread".
The following code uses num2cell to reshape ND into a cell array such that each dimension-d-thread is in a different cell. Then strfind is applied to each cell to determine matches with V, and the result is a cell array with the same dimensions as ND, but where the dimension d is a singleton. The contents of each cell tell the d-dimension-positions of the matches, if any.
Credit goes to #knedlsepp for his suggestion to use num2cell, which greatly simplified the code.
ND = cat(3, [1 2 1 2; 3 4 5 6],[2 1 0 5; 0 0 1 2] ); %// example. 2x4x2
V = 1:2; %// sought pattern. It doesn't matter if it's a row, or a column, or...
d = 2; %// dimension along which to search for pattern V
result = cellfun(#(x) strfind(x(:).', V(:).'), num2cell(ND,d), 'UniformOutput', 0);
This gives
ND(:,:,1) =
1 2 1 2
3 4 5 6
ND(:,:,2) =
2 1 0 5
0 0 1 2
V =
1 2
result{1,1,1} =
1 3 %// V appears twice (at cols 1 and 3) in 1st row, 1st slice
result{2,1,1} =
[] %// V doesn't appear in 2nd row, 1st slice
result{1,1,2} =
[] %// V appears appear in 1st row, 2nd slice
result{2,1,2} =
3 %// V appears once (at col 3) in 2nd row, 2nd slice
One not very optimal way of doing it:
dims = size(ND);
Vrep = repmat(V, [dims(1), dims(2), dims(3), 1]);
ND_V_dist = sqrt(sum(abs(ND.^2-Vrep.^2), 4));
iI = find(ND_V_dist==0);
[I1, I2, I3] = ind2sub([dims(1), dims(2), dims(3)], iI);
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;
I would like to generate three sets of random variables and I dont want them to be the same
for example if I want to generate the following sets ranged from 1 to 10
Set1= [ 1 4 ]
set2= [ 3 5 ]
Set3= [ 7 9]
You can create an array of unique random numbers using the function "randperm" and then divide the array into sets.
x = randperm(10,6);
Set1 = x(1:2);
Set2 = x(3:4);
Set3 = x(5:6);
After choosing two elements for each set, it's essential to randomize the remaining elements to choose from for the next iteration . The code below is trying to achieve it -
set123 = zeros(3,2); %// Pre-allocate for the output
array1 = 1:10; %// Array of numbers to choose from for the first iteration
for k = 1:3
%// Create many possible combinations of such two numbers from the sets.
%// The sets would have 10 elements to choose from at the start and
%// then 8 and then 6.
combs = nchoosek(array1,2);
%// At each stage all the possible combinations of the pairs must be
%// juggled and one of them must be selected randomly and then indexed
%// to get the set for this iteration
choosen_rowind = randperm(size(combs,1),1);
set123(k,:) = combs(choosen_rowind,:);
%// Setup the array of numbers left to choose from the next iteration
array1 = setdiff(array1,set123);
end
%// set123 is the desired output in a single array
If I understand correctly, you want to take r-times-c random variables (in your case r=3, c=2) from the set {m,m+1,...,n} (in your case n=1 and m=10), without repetition. I assume you want a (jointly) uniform distribution.
For that you use randsample:
result = reshape(randsample(m:n, r*c), r, c);
Is there any way that I can sum up columns values for each group of three rows in a matrix?
I can sum three rows up in a manual way.
For example
% matrix is the one I wanna store the new data.
% data is the original dataset.
matrix(1,1:end) = sum(data(1:3, 1:end))
matrix(2,1:end) = sum(data(4:6, 1:end))
...
But if the dataset is huge, this wouldn't work.
Is there any way to do this automatically without loops?
Here are four other ways:
The obligatory for-loop:
% for-loop over each three rows
matrix = zeros(size(data,1)/3, size(data,2));
counter = 1;
for i=1:3:size(data,1)
matrix(counter,:) = sum(data(i:i+3-1,:));
counter = counter + 1;
end
Using mat2cell for tiling:
% divide each three rows into a cell
matrix = mat2cell(data, ones(1,size(data,1)/3)*3);
% compute the sum of rows in each cell
matrix = cell2mat(cellfun(#sum, matrix, 'UniformOutput',false));
Using third dimension (based on this):
% put each three row into a separate 3rd dimension slice
matrix = permute(reshape(data', [], 3, size(data,1)/3), [2 1 3]);
% sum rows, and put back together
matrix = permute(sum(matrix), [3 2 1]);
Using accumarray:
% build array of group indices [1,1,1,2,2,2,3,3,3,...]
idx = floor(((1:size(data,1))' - 1)/3) + 1;
% use it to accumulate rows (appliead to each column separately)
matrix = cell2mat(arrayfun(#(i)accumarray(idx,data(:,i)), 1:size(data,2), ...
'UniformOutput',false));
Of course all the solution so far assume that the number of rows is evenly divisble by 3.
This one-liner reshapes so that all the values needed for a particular cell are in a column, does the sum, and then reshapes the back to the expected shape.
reshape(sum(reshape(data, 3, [])), [], size(data, 2))
The naked 3 could be changed if you want to sum a different number of rows together. It's on you to make sure the number of rows in each group divides evenly.
Slice the matrix into three pieces and add them together:
matrix = data(1:3:end, :) + data(2:3:end, :) + data(3:3:end, :);
This will give an error if size(data,1) is not a multiple of three, since the three pieces wouldn't be the same size. If appropriate to your data, you might work around that by truncating data, or appending some zeros to the end.
You could also do something fancy with reshape and 3D arrays. But I would prefer the above (unless you need to replace 3 with a variable...)
Prashant answered nicely before but I would have a simple amendment:
fl = filterLength;
A = yourVector (where mod(A,fl)==0)
sum(reshape(A,fl,[]),1).'/fl;
There is the ",1" that makes the line run even when fl==1 (original values).
I discovered this while running it in a for loop like so:
... read A ...
% Plot data
hold on;
averageFactors = [1 3 10 30 100 300 1000];
colors = hsv(length(averageFactors));
clear legendTxt;
for i=1:length(averageFactors)
% ------ FILTERING ----------
clear Atrunc;
clear ttrunc;
clear B;
fl = averageFactors(i); % filter length
Atrunc = A(1:L-mod(L,fl),:);
ttrunc = t(1:L-mod(L,fl),:);
B = sum(reshape(Atrunc,fl,[]),1).'/fl;
tB = sum(reshape(ttrunc,fl,[]),1).'/fl;
length(B)
plot(tB,B,'color',colors(i,:) )
%kbhit ()
endfor