I am trying to generate random numbers between 1 and 6 using Matlab's randperm and calling randperm = 6.
Each time this gives me a different array let's say for example:
x = randperm(6)
x = [3 2 4 1 5 6]
I was wondering if it was possible to create pairs of random numbers such that you end up with x like:
x = [3 4 1 2 5 6]
I need the vector to be arranged such that 1 and 2 are always next to each other, 3 and 4 next to each other and 5 and 6 next to each other. As I'm doing something in Psychtoolbox and this order is important.
Is it possible to have "blocks" of random order? I can't figure out how to do it.
Thanks
x=1:block:t ; %Numbers
req = bsxfun(#plus, x(randperm(t/block)),(0:block-1).'); %generating random blocks of #
%or req=x(randperm(t/block))+(0:block-1).' ; if you have MATLAB R2016b or later
req=req(:); %reshape
where,
t = total numbers
block = numbers in one block
%Sample run with t=12 and block=3
>> req.'
ans =
10 11 12 4 5 6 1 2 3 7 8 9
Edit:
If you also want the numbers within each block in random order, add the following 3 lines before the last line of above code:
[~, idx] = sort(rand(block,t/block)); %generating indices for shuffling
idx=bsxfun(#plus,idx,0:block:(t/block-1)*block); %shuffled linear indices
req=req(idx); %shuffled matrix
%Sample run with t=12 and block=3
req.'
ans =
9 8 7 2 3 1 12 10 11 5 6 4
I can see a simple 3 step process to get your desired output:
Produce 2*randperm(3)
Double up the values
Add randperm(2)-2 (randomly ordered pair of (-1,0)) to each pair.
In code:
x = randperm(3)
y = 2*x([1 1 2 2 3 3])
z = y + ([randperm(2),randperm(2),randperm(2)]-2)
with result
x = 3 1 2
y = 6 6 2 2 4 4
z = 6 5 2 1 3 4
Related
I am trying to generate random numbers between 1 and 5 using Matlab's randperm and calling randperm = 5.
Each time this gives me a different array let's say for example:
x = randperm(5)
x = [3 2 4 1 5]
I need the vector to be arranged such that 4 and 5 are always next to each other and 2 is always between 1 and 3... so for e.g. [3 2 1 4 5] or [4 5 1 2 3].
So essentially I have two "blocks" of unequal length - 1 2 3 and 4 5. The order of the blocks is not so important, just that 4 & 5 end up together and 2 in between 1 and 3.
I can basically only have 4 possible combinations:
[1 2 3 4 5]
[3 2 1 4 5]
[4 5 1 2 3]
[4 5 3 2 1]
Does anyone know how I can do this?
Thanks
I'm not sure if you want a solution that would somehow generalize to a larger problem, but based on how you've described your problem above it looks like you are only going to have 8 possible combinations that satisfy your constraints:
possible = [1 2 3 4 5; ...
1 2 3 5 4; ...
3 2 1 4 5; ...
3 2 1 5 4; ...
4 5 1 2 3; ...
5 4 1 2 3; ...
4 5 3 2 1; ...
5 4 3 2 1];
You can now randomly select one or more of these rows using randi, and can even create an anonymous function to do it for you:
randPattern = #(n) possible(randi(size(possible, 1), [1 n]), :)
This allows you to select, for example, 5 patterns at random (one per row):
>> patternMat = randPattern(5)
patternMat =
4 5 3 2 1
3 2 1 4 5
4 5 3 2 1
1 2 3 5 4
5 4 3 2 1
You can generate each block and shuffle each one then and set them as members of a cell array and shuffle the cell array and finally convert the cell array to a vector.
b45=[4 5]; % block 1
b13=[1 3]; % block 2
r45 = randperm(2); % indices for shuffling block 1
r13 = randperm(2); % indices for shuffling block 2
r15 = randperm(2); % indices for shuffling the cell
blocks = {b45(r45) [b13(r13(1)) 2 b13(r13(2))]}; % shuffle each block and set them a members of a cell array
result = [blocks{r15}] % shuffle the cell and convert to a vector
I will try to explain what I need through an example.
Suppose you have a matrix x as follows:
1 2 3
4 5 6
And another matrix y as follows:
1 4 5
7 4 8
What I need is (without looping over the rows) to perform an intersection between each 2 corresponding rows in x & y. So I wish to get a matrix z as follows:
1
4
The 1st rows in x and y only have 1 as the common value. The 2nd rows have 4 as the common value.
EDIT:
I forgot to add that in my case, it is guaranteed that the intersection results will have the same length and the length is always 1 actually.
I am thinking bsxfun -
y(squeeze(any(bsxfun(#eq,x,permute(y,[1 3 2])),2)))
Sample runs -
Run #1:
>> x
x =
1 2 3
4 5 6
>> y
y =
1 4 5
7 4 8
>> y(squeeze(any(bsxfun(#eq,x,permute(y,[1 3 2])),2)))
ans =
1
4
Run #2:
>> x
x =
3 5 7 9
2 7 9 0
>> y
y =
6 4 3
6 0 2
>> y(squeeze(any(bsxfun(#eq,x,permute(y,[1 3 2])),2)))
ans =
0
3
2
The idea is to put the matrices together and to look for duplicates in the rows. One idea to find duplicated numeric values is to diff them; the duplicates will be marked by the value 0 in result.
Which leads to:
%'Initial data'
A = [1 2 3; 8 5 6];
B = [1 4 5; 7 4 8];
%'Look in merged data'
V = sort([A,B],2); %'Sort matrix values in rows'
R = V(diff(V,1,2)==0); %'Find duplicates in rows'
This should work with any number of matrices that can be concatenated horizontally. It will detect all the duplicates, but it will return a column the same size as the number of rows only if there is one and only one duplicate per row in the matrices.
I'm trying to create a script to solve my problem, but I got stuck in one place.
So I have imported .txt file with 4x1 sized matrix (simplified to give an example in my case it might be 1209x1 matrix) which contains some coordinate X. And it's look like this:
0
1
2
3
That's coordinates for one period, and I need to get one column for different number of periods N . Each period is the same and lenght=L
So you can do it manually by this script, for example for N=3 periods:
X=[X; X+L; X+2*L];
so for example if L=3
then i will get
0
1
2
3
3
4
5
6
6
7
8
9
it works well but it's not efficient in case if I need to work with number of periods let's say N=1000 or if I need to change their number quickly. Any solution to parameterize this operation so I can just put number for N and get X for N periods?
Thanks and Regards
I don't have MATLAB on this machine so I can't test, but the most straightforward implementation would be something like:
n = 1000;
L = 3;
nvalues = length(X); % Assuming X is your initial vector
newx = zeros(n*nvalues, 1); % Preallocate new array
for ii = 0:(n-1)
startidx = (nvalues*ii) + 1;
endidx = nvalues*(ii+1);
newx(startidx:endidx) = X + ii*L
end
You can use bsxfun to create X, X+L, X+2*L, ... and then reshape it to a vector
>> F=bsxfun(#plus, X, (0:(N-1))*L)
F =
0 3 6
1 4 7
2 5 8
3 6 9
>> X=F(:)
X =
0
1
2
3
3
4
5
6
6
7
8
9
or in a more concise form:
>> X=reshape(bsxfun(#plus, X, (0:(N-1))*L), [], 1)
X =
0
1
2
3
3
4
5
6
6
7
8
9
I have a variable like this that is all one row:
1 2 3 4 5 6 7 8 9 2 4 5 6 5
I want to write a for loop that will find where a number is less than the previous one and put the rest of the numbers in a new row, like this
1 2 3 4 5 6 7 8 9
2 4 5 6
5
I have tried this:
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
m = zeros(size(test));
for i=1:numel(test)-1;
for rows=1:size(m,1)
if test(i) > test(i+1);
m(i+1, rows+1) = test(i+1:end)
end % for rows
end % for
But it's clearly not right and just hangs.
Let x be your data vector. What you want can be done quite simply as follows:
ind = [find(diff(x)<0) numel(x)]; %// find ends of increasing subsequences
ind(2:end) = diff(ind); %// compute lengths of those subsequences
y = mat2cell(x, 1, ind); %// split data vector according to those lenghts
This produces the desired result in cell array y. A cell array is used so that each "row" can have a different number of columns.
Example:
x = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
gives
y{1} =
1 2 3 4 5 6 7 8 9
y{2} =
2 4 5 6
y{3} =
5
If you are looking for a numeric array output, you would need to fill the "gaps" with something and filling with zeros seem like a good option as you seem to be doing in your code as well.
So, here's a bsxfun based approach to achieve the same -
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5] %// Input
idx = [find(diff(test)<0) numel(test)] %// positions of row shifts
lens = [idx(1) diff(idx)] %// lengths of each row in the proposed output
m = zeros(max(lens),numel(lens)) %// setup output matrix
m(bsxfun(#le,[1:max(lens)]',lens)) = test; %//'# put values from input array
m = m.' %//'# Output that is a transposed version after putting the values
Output -
m =
1 2 3 4 5 6 7 8 9
2 4 5 6 0 0 0 0 0
5 0 0 0 0 0 0 0 0
I have an array A, and want to reshape the last four elements of each column into a 2x2 matrix. I would like the results to be stored in a cell array B.
For example, given:
A = [1:6; 3:8; 5:10]';
I would like B to contain three 2x2 arrays, such that:
B{1} = [3, 5; 4, 6];
B{2} = [5, 7; 6, 8];
B{3} = [7, 9; 8, 10];
I can obviously do this in a for loop using something like reshape(A(end-3:end, ii), 2, 2) and looping over ii. Can anyone propose a vectorized method, perhaps using something similar to cellfun that can apply an operation repeatedly to columns of an array?
The way I do this is to look at the desired indices and then figure out a way to generate them, usually using some form of repmat. For example, if you want the last 4 items in each column, the (absolute) indices into A are going to be 3,4,5,6, then add the number of rows to that to move to the next column to get 9,10,11,12 and so on. So the problem becomes generating that matrix in terms of your number of rows, number of columns, and the number of elements you want from each column (I'll call it n, in your case n=4).
octave:1> A = [1:6; 3:8; 5:10]'
A =
1 3 5
2 4 6
3 5 7
4 6 8
5 7 9
6 8 10
octave:2> dim=size(A)
dim =
6 3
octave:3> n=4
n = 4
octave:4> x=repmat((dim(1)-n+1):dim(1),[dim(2),1])'
x =
3 3 3
4 4 4
5 5 5
6 6 6
octave:5> y=repmat((0:(dim(2)-1)),[n,1])
y =
0 1 2
0 1 2
0 1 2
0 1 2
octave:6> ii=x+dim(1)*y
ii =
3 9 15
4 10 16
5 11 17
6 12 18
octave:7> A(ii)
ans =
3 5 7
4 6 8
5 7 9
6 8 10
octave:8> B=reshape(A(ii),sqrt(n),sqrt(n),dim(2))
B =
ans(:,:,1) =
3 5
4 6
ans(:,:,2) =
5 7
6 8
ans(:,:,3) =
7 9
8 10
Depending on how you generate x and y, you can even do away with the multiplication, but I'll leave that to you. :D
IMO you don't need a cell array to store them either, a 3D matrix works just as well and you index into it the same way (but don't forget to squeeze it before you use it).
I gave a similar answer in this question.