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.
Related
Suppose I initialized two vectors,
x=[1 2 3 4 5]';
y=[6 7 8 9 10]';
both representing a column matrix, or vector. Now,
z=[x;y];
The z vector will be combination of the two, in a similar column format. y vector will be in continuation to the x vector by this method.
But what should be the approach if I wanted y to be in continuation from a certain given position, leaving the remaining values in between as blank. For example, I want the continuation of y from 8th position to get an output of z as:-
1
2
3
4
5
NaN
NaN
6
7
8
9
10
Just count, how many blanks (NaN) do you need using the desired position and the number of elements of x, and assemble your output z:
% Input
x = [1 2 3 4 5]'
y = [6 7 8 9 10]'
% Position
pos = 8;
% Add some code for checking numel(x) >= pos here...
% Output
z = [x; NaN(pos-numel(x)-1, 1); y]
x =
1
2
3
4
5
y =
6
7
8
9
10
z =
1
2
3
4
5
NaN
NaN
6
7
8
9
10
Meh, after some editing, I realized, that a comment would've been sufficient...
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
Suppose I have the following array:
x = [a b
c d
e f
g h
i j];
I want to "swipe a window of two rows" progressively (one row at a time) along the array to generate the following array:
y = [a b c d e f g h
c d e f g h i j];
What is the most efficient way to do this? I don't want to use cellfun or arrayfun or for loops.
im2col is going to be your best bet here if you have the Image Processing Toolbox.
x = [1 2
3 4
5 6
7 8];
im2col(x.', [1 2])
% 1 2 3 4 5 6
% 3 4 5 6 7 8
If you don't have the Image Processing Toolbox, you can also easily do this with built-ins.
reshape(permute(cat(3, x(1:end-1,:), x(2:end,:)), [3 2 1]), 2, [])
% 1 2 3 4 5 6
% 3 4 5 6 7 8
This combines the all rows with the next row by concatenating a row-shifted version along the third dimension. Then we use permute to shift the dimensions around and then we reshape it to be the desired size.
If you don't have the Image Processing Toolbox, you can do this using simple indexing:
x =
1 2
3 4
5 6
7 8
9 10
y = x.'; %% Transpose it, for simplicity
z = [y(1:end-2); y(3:end)] %% Take elements 1:end-2 and 3:end and concatenate them
z =
1 2 3 4 5 6 7 8
3 4 5 6 7 8 9 10
You can do the transposing and reshaping in a simple step (see Suever's edit), but the above might be easier to read, understand and debug for beginners.
Here's an approach to solve it for a generic case of selecting L rows per window -
[m,n] = size(x) % Store size
% Extend rows by indexing into them with a progressive array of indices
x_ext = x(bsxfun(#plus,(1:L)',0:m-L),:);
% Split the first dim at L into two dims, out of which "push" back the
% second dim thus created as the last dim. This would bring in the columns
% as the second dimension. Then, using linear indexing reshape into the
% desired shape of L rows for output.
out = reshape(permute(reshape(x_ext,L,[],n),[1,3,2]),L,[])
Sample run -
x = % Input array
9 1
3 1
7 5
7 8
4 9
6 2
L = % Window length
3
out =
9 1 3 1 7 5 7 8
3 1 7 5 7 8 4 9
7 5 7 8 4 9 6 2
Suppose A is a 3-D matrix as below (2 rows-2 columns-2 pages).
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
I want to have a vector, say "a", whose inputs are the average of diagonal elements of matrices on each page. So in this simple case, a=[(1+4)/2;(5+8)/2].
But I have difficulties in matlab to do so. I tried the codes below but failed.
mean(A(1,1,:),A(2,2,:))
You can use "partially linear indexing" in the two dimensions that define the diagonal, as follows:
Since partially linear indexing can only be applied on trailing dimensions, you first need to apply permute to rearrange dimensions, so that the first and second dimensions become second and third.
Now you leave the first dimension untouched, linearly-index the diagonals in the second and third dimensions (which effectly reduces those two dimensions to one), and apply mean along the (combined) second dimension.
Code:
B = permute(A, [3 1 2]); %// step 1: permute
result = mean(B(:,1:size(A,1)+1:size(A,1)*size(A,2)), 2); %// step 2: index and mean
In your example,
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
this gives
result =
2.5000
6.5000
You can use bsxfun for a generic solution -
[m,n,r] = size(A)
mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
Sample run -
>> A
A(:,:,1) =
8 4 1
7 6 3
1 5 8
A(:,:,2) =
1 7 6
8 5 2
1 2 7
A(:,:,3) =
6 2 8
1 1 6
1 4 5
A(:,:,4) =
8 1 6
1 5 1
9 2 7
>> [m,n,r] = size(A);
>> sum(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
22 13 12 20
>> mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
7.3333 4.3333 4 6.6667
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