I have two matrices:
X =
1 2 3
4 5 6
7 8 9
`Y` =
1 10 11
4 12 13
7 14 15
I know that if I want to find the index of a specific element in X or Y, I can use the function find. For example:
index_3 = find(X==3)
What I want is to find or search in a very automatic way if a column in X is also present in Y. In other terms, I want a function which can tell me if a column in X is equal to a column in Y. In fact to to try this, one can use the function ismember which indeed has an optional flag to compare rows:
rowsX = ismember(X, Y, 'rows');
So a simple way to get columns is just by taking the transpose of both matrices:
rowsX = ismember(X.', Y.', 'rows')
rowsX =
1
0
0
But how can I do that in other manner?
Any help will be very appreciated!
You can do that with bsxfun and permute:
rowsX = any(all(bsxfun(#eq, X, permute(Y, [1 3 2])), 1), 3);
With
X = [ 1 2 3
4 5 6
7 8 9 ];
Y = [ 1 10 11
4 12 13
7 14 15 ];
this gives
rowsX =
1 0 0
How it works
permute "turns Y 90 degrees" along a vertical axis, so columns of Y are kept aligned with columns of X, but rows of Y are moved to the third dimension. Testing for equality with bsxfun and applying all(...,1) gives a matrix that tells which columns of X equal which columns of Y. Then any(...,3) produces the desired result: true if a column of X equals any column of Y.
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...
Given a logical column vector (size n x 1) v and an array a (size m x n) how do I generate a new array consisting of all the columns in a where the numerical index of said column (1...n) is 1 at the corresponding location in v.
So for example if v was
1
0
0
1
and a was
1 4 7 10
2 5 8 11
3 6 9 12
the new array would be
1 10
2 11
3 12
because the first and fourth elements of v are 1 (true), so the new array should contain the first and fourth columns of a.
I have tried a bunch of things involving normal logical indexing and transpose but I can't get it to work. All help is appreciated
You want to use the logical indexing to select the columns and select all rows. In the example below, I have explicitly cast v as a logical just in case it's not a logical matrix already.
new = a(:, logical(v))
1 10
2 11
3 12
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 am trying to allocate (x, y) points to the cells of a non-uniform rectangular grid. Simply speaking, I have a grid defined as a sorted non-equidistant array
xGrid = [x1, x2, x3, x4];
and an array of numbers x lying between x1 and x4. For each x, I want to find its position in xGrid, i.e. such i that
xGrid(i) <= xi <= xGrid(i+1)
Is there a better (faster/simpler) way to do it than arrayfun(#(x) find(xGrid <= x, 1, 'last'), x)?
You are looking for the second output of histc:
[~,where] = histc(x, xGrid)
This returns the array where such that xGrid(where(i)) <= x(i) < xGrid(where(i)+1) holds.
Example:
xGrid = [2,4,6,8,10];
x = [3,5,6,9,11];
[~,where] = histc(x, xGrid)
Yields the following output:
where =
1 2 3 4 0
If you want xGrid(where(i)) < x(i) <= xGrid(where(i)+1), you need to do some trickery of negating the values:
[~,where] = histc(-x,-flip(xGrid));
where(where~=0) = numel(xGrid)-where(where~=0)
This yields:
where =
1 2 2 4 0
Because x(3)==6 is now counted for the second interval (4,6] instead of [6,8) as before.
Using bsxfun for the comparisons and exploiting find-like capabilities of max's second output:
xGrid = [2 4 6 8]; %// example data
x = [3 6 5.5 10 -10]; %// example data
comp = bsxfun(#gt, xGrid(:), x(:).'); %'// see if "x" > "xGrid"
[~, result] = max(comp, [], 1); %// index of first "xGrid" that exceeds each "x"
result = result-1; %// subtract 1 to find the last "xGrid" that is <= "x"
This approach gives 0 for values of x that lie outside xGrid. With the above example values,
result =
1 3 2 0 0
See if this works for you -
matches = bsxfun(#le,xGrid(1:end-1),x(:)) & bsxfun(#ge,xGrid(2:end),x(:))
[valid,pos] = max(cumsum(matches,2),[],2)
pos = pos.*(valid~=0)
Sample run -
xGrid =
5 2 1 6 8 9 2 1 6
x =
3 7 14
pos =
8
4
0
Explanation on the sample run -
First element of x, 3 occurs last between ...1 6 with the criteria of xGrid(i) <= xi <= xGrid(i+1) at the backend of xGrid and that 1 is at the eight position, so the first element of the output pos is 8. This continues for the second element 7, which is found between 6 and 8 and that 6 is at the fourth place in xGrid, so the second element of the output is 4. For the third element 14 which doesn't find any neighbours to satisfy the criteria xGrid(i) <= xi <= xGrid(i+1) and is therefore outputted as 0.
If x is a column this might help
xg1=meshgrid(xGrid,1:length(x));
xg2=ndgrid(x,1:length(xGrid));
[~,I]=min(floor(abs(xg1-xg2)),[],2);
or a single line implementation
[~,I]=min(floor(abs(meshgrid(xGrid,1:length(x))-ndgrid(x,1:length(xGrid)))),[],2);
Example: xGrid=[1 2 3 4 5], x=[2.5; 1.3; 1.7; 4.8; 3.3]
Result:
I =
2
1
1
4
3