Say I have a cell array in the following format:
A = {4;[22 16 4]; 23; 51; [16 22]; 32; [4 50]};
I want to output the unique columns gained by any permutations of the vectors in the rows.
For example, for the example in the above, the only column vectors that would satisfy this would be
[4; 22; 23; 51; 16; 32; 50] and [4; 16; 23; 51; 22; 32; 50].
I can't choose 4 from the second or last rows since 4 is the only option in the first row. Moreover, I can't choose 22 in both the second and fifth rows since this would make the column non-unique. Although empty choices in some rows are not allowed, if there are no unique columns, then I would need to output an empty column.
Does anyone have a smart way of doing this (fairly quickly) in Matlab? Any help would be much appreciated.
Many thanks!
Here's a brute-force approach:
Generate all possible columns obtained by choosing one element from each vector (Cartesian product). This can be done with this approach.
Keep only those columns for which all elements are different. This is easily done with sort, diff, all, and logical indexing.
Code:
A = {4; [22 16 4]; 23; 51; [16 22]; 32; [4 50]}; % Data
n = numel(A); % Step 1
c = cell(1,n);
[c{end:-1:1}] = ndgrid(A{:});
c = cat(n+1, c{end:-1:1});
c = reshape(c,[],n).'; % Step 1
result = c(:,all(diff(sort(c,1),[],1),1)); % Step 2
Related
Imagine I have two matrices with different sizes (let's say 6x2 and 5x2) as follows:
A = [47 10;
29 10;
23 10;
34 10;
12 10;
64 10];
B = [23 20;
12 20;
54 20
47 20;
31 20];
I need to compare A(:,1) with B(:,1) and delete the rows in matrix A whose first-column-element is different from matrix B's first-column-element (so my focus is only on first columns of the matrices). So I should eventually get something like this:
A = [47 10;
12 10;
23 10];
as "47", "12", and "23" are the only first-column-elements in A that also exist in B! I have written this but I get the error "Matrix dimensions must agree."!
TF = A(:,1) ~= B(:,1); %define indexes in A that A(:,1) is not equal to B(:,1)
A(TF,:) = [];
Any ideas how I could fix this?
You can use ismember:
result = A(ismember(A(:,1), B(:,1)), :);
Replacing this line
TF = A(:,1) ~= B(:,1);
with this line
[~,TF] = setdiff(A(:,1),B(:,1));
yields the desired result.
So, I have this cell array contains n x 2 matrix in each cell. Here is the sample data :
[16 17;17 17]
<6x2 double>
<52x2 double>
[17 17;17 18]
[17 18;17 17]
What I am going to do is eliminate the duplicated matrix (matrices with same values or reversed values). in this case is [17 18; 17 17] (5th row), because we already have [17 17; 17 18] (4th row)
I tried using unique function but it says that the function just worked for strings. I also tried to find it with cellfun like this
cellfun(#(x) x==whatToSearch, lineTraced, 'UniformOutput', false)
but it says 'Matrix dimension must agree'
Thanks in advance.
Here is a solution. Given a m x 1 column cell array C of matrices, this code deletes the equivalent duplicates. (Here it will delete the 4th and 5th matrices, which are equivalent to the 1st).
C{1} = magic(3);
C{2} = magic(4);
C{3} = magic(5);
C{4} = C{1};
C{5} = flipud(C{1});
myEq = #(A,B) isequal(A,B) | isequal(A,flipud(B)); %// equivalence operator tests for same or up-down flipped matrix
C = C(:); %// ensure the cell array is a column
Crep = repmat(C,1,size(C,1)); %// repeat cell array along rows to get a square
comp = cellfun(myEq,Crep,Crep'); %'//get a comparison matrix by comparing with transpose
comp = tril(comp) - eye(size(comp)); %// ignore upper triangle and diagonal
idx = find( sum(comp,2)==0 ); %// get index of matrices we want to keep
result = C(idx); %// get result
The output is deletes the 4th and 5th matrices, leaving the first three magic matrices:
>> result
result =
[3x3 double] [4x4 double] [5x5 double]
>> result{1}, result{2}, result{3}
ans =
8 1 6
3 5 7
4 9 2
ans =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
ans =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Here's code that does what you want.
mydup = rand(5,2);
mycell = {mydup;mydup;rand(7,2);rand(3,2);rand(5,2)}
myNewCell = trimCell(mycell)
Where trimCell is the function below:
function myNewCell = trimCell(myCell)
exclude = false(size(myCell));
for iter = 1:size(myCell,1)
toCompare = myCell{iter};
comparisons = cellfun(#(x) all(size(x)==size(toCompare)) && myEquals(x, toCompare),myCell);
% comparisons has at least 1 true in it, because toCompare==toCompare
exclude(iter) = sum(comparisons)>1;
end
myNewCell = myCell(~exclude);
end
function boolValue = myEquals(x,y)
assert(all(size(x)==size(y)));
boolValue = true;
for iter = 1:size(x,1)
thisRow = all(sort(x(iter,:))==sort(y(iter,:)));
boolValue = boolValue && thisRow;
if ~boolValue
return
end
end
end
Basically, what this function does, is, for each matrix in the cell it checks first if the sizes are equal. If the sizes aren't equal, then we know the matrices aren't equal, even if we reorder the rows.
If they are the same size, then we need to check if they're equal after row reordering. That's accomplished by the function myEquals which goes through row by row and sorts the rows before comparing them. It exits with false on the first row it finds that is not equal after reordering.
Hope this helps,
Andrew
How to select a submatrix (not in any pattern) in Matlab? For example, for a matrix of size 10 by 10, how to select the submatrix consisting of intersection of the 1st 2nd and 9th rows and the 4th and 6th columns?
Thanks for any helpful answers!
TLDR: Short Answer
As for your question, suppose you have an arbitrary 10-by-10 matrix A. The simplest way to extract the desired sub-matrix would be with an index vector:
B = A([1 2 9], [4 6]);
Indexing in MATLAB
There's an interesting article in the official documentation that comprehensively explains indexing in MATLAB.
Basically, there are several ways to extract a subset of values, I'll summarize them for you:
1. Indexing Vectors
Indexing vectors indicate the indices of the element to be extracted. They can either contain a single index or several, like so:
A = [10 20 30 40 50 60 70 80 90]
%# Extracts the third and the ninth element
B = A([3 9]) %# B = [30 90]
Indexing vectors can be specified for each dimension separately, for instance:
A = [10 20 30; 40 50 60; 70 80 90];
%# Extract the first and third rows, and the first and second columns
B = A([1 3], [1 2]) %# B = [10 30; 40 60]
There are also two special subscripts: end and the colon (:):
end simply indicates the last index in that dimension.
The colon is just a short-hand notation for "1:end".
For example, instead of writing A([1 2 3], [2 3]), you can write A(:, 2:end). This is especially useful for large matrices.
2. Linear Indexing
Linear indexing treats any matrix as if it were a column vector by concatenating the columns into one column vector and assigning indices to the elements respectively. For instance, we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to compute b = A(2). The equivalent column vector is:
A = [10;
40;
70;
20;
50;
80;
30;
60;
90]
and thus b equals 40.
The special colon and end subscripts are also allowed, of course. For that reason, A(:) converts any matrix A into a column vector.
Linear indexing with matrix subscripts:
It is also possible to use another matrix for linear indexing. The subscript matrix is simply converted into a column vector, and used for linear indexing. The resulting matrix is, however always of the same dimensions as the subscript matrix.
For instance, if I = [1 3; 1 2], then A(I) is the same as writing reshape(A(I(:)), size(I)).
Converting from matrix subscripts to linear indices and vice versa:
For that you have sub2ind and ind2sub, respectively. For example, if you want to convert the subscripts [1, 3] in matrix A (corresponding to element 30) into a linear index, you can write sub2ind(size(A), 1, 3) (the result in this case should be 7, of course).
3. Logical Indexing
In logical indexing the subscripts are binary, where a logical 1 indicates that the corresponding element is selected, and 0 means it is not. The subscript vector must be either of the same dimensions as the original matrix or a vector with the same number of elements. For instance, if we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to extract A([1 3], [1 2]) using logical indexing, we can do either this:
Ir = logical([1 1 0]);
Ic = logical([1 0 1]);
B = A(Ir, Ic)
or this:
I = logical([1 0 1; 1 0 1; 0 0 0]);
B = A(I)
or this:
I = logical([1 1 0 0 0 0 1 1 0]);
B = A(I)
Note that in the latter two cases is a one-dimensional vector, and should be reshaped back into a matrix if necessary (for example, using reshape).
Let me explain with an example:
Let's define a 6x6 matrix
A = magic(6)
A =
35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16
4 36 29 13 18 11
From this matrix you want the elements in rows 1, 2 and 5, and in the columns 4 and 6
B = A([1 2 5],[4 6])
B =
26 24
21 25
12 16
Hope this helps.
function f = sub(A,i,j)
[m,n] = size(A);
row = 1:m;
col = 1:n;
x = row;
x(i) = [];
y=col;
y(j) = [];
f= A(x,y);
Returns the matrix A, with the ith row and jth column removed.
I have a m-by-n matrix and I want to shift each row elements k no. of times (" one resultant matrix for each one shift so a total of k matrices corresponding to each row shifts ")(k can be different for different rows and 0<=k<=n) and want to index all the resultant matrices corresponding to each individual shift.
Eg: I have the matrix: [1 2 3 4; 5 6 7 8; 2 3 4 5]. Now, say, I want to shift row1 by 2 times (i.e. k=2 for row1) and row2 by 3times (i.e. k=3 for row2) and want to index all the shifted versions of matrices (It is similar to combinatorics of rows but with limited and diffeent no. of shifts to each row).
Can someone help to write up the code? (please help to write the general code but not for the example I mentioned here)
I found the following question useful to some extent, but it won't solve my problem as my problem looks like a special case of this problem:
Matlab: How to get all the possible different matrices by shifting it's rows (Update: each row has a different step)
See if this works for you -
%// Input m-by-n matrix
A = rand(2,5) %// Edit this to your data
%// Initialize shifts, k for each row. The number of elements would be m.
sr = [2 3]; %// Edit this to your data
[m,n] = size(A); %// Get size
%// Get all the shits in one go
sr_ind = arrayfun(#(x) 0:x,sr,'un',0); %//'
shifts = allcomb(sr_ind{:},'matlab')'; %//'
for k1 = 1:size(shifts,2)
%// Get shift to be used for each row for each iteration
shift1 = shifts(:,k1);
%// Get circularly shifted column indices
t2 = mod(bsxfun(#minus,1:n,shift1),n);
t2(t2==0) = n;
%// Get the linear indices and use them to index into input to get the output
ind = bsxfun(#plus,[1:m]',(t2-1)*m); %//'
all_matrices = A(ind) %// outputs
end
Please note that this code uses MATLAB file-exchange code allcomb.
If your problem in reality is not more complex than what you showed us, it can be done by a double loop. However, i don't like my solution, because you would need another nested loop for each row you want to shift. Also it generates all shift-combinations from your given k-numbers, so it has alot of overhead. But this can be a start:
% input
m = [1 2 3 4; 5 6 7 8; 2 3 4 5];
shift_times = {0:2, 0:3}; % 2 times for row 1, 3 times for row 2
% desird results
desired_matrices{1} = [4 1 2 3; 5 6 7 8; 2 3 4 5];
desired_matrices{2} = [3 4 1 2; 5 6 7 8; 2 3 4 5];
desired_matrices{3} = [1 2 3 4; 8 5 6 7; 2 3 4 5];
desired_matrices{4} = [4 1 2 3; 8 5 6 7; 2 3 4 5];
desired_matrices{5} = [3 4 1 2; 8 5 6 7; 2 3 4 5];
% info needed:
[rows, cols] = size(m);
count = 0;
% make all shift combinations
for shift1 = shift_times{1}
% shift row 1
m_shifted = m;
idx_shifted = [circshift([1:cols]',shift1)]';
m_shifted(1, :) = m_shifted(1, idx_shifted);
for shift2 = shift_times{2}
% shift row 2
idx_shifted = [circshift([1:cols]',shift2)]';
m_shifted(2, :) = m_shifted(r_s, idx_shifted);
% store them
store{shift1+1, shift2+1} = m_shifted;
end
end
% store{i+1, j+1} stores row 1 shifted by i and row 2 shifted by j
% example
all(all(store{2,1} == desired_matrices{1})) % row1: 1, row2: 0
all(all(store{2,2} == desired_matrices{4})) % row1: 1, row2: 1
all(all(store{3,2} == desired_matrices{5})) % row1: 2, row2: 1
How to select a submatrix (not in any pattern) in Matlab? For example, for a matrix of size 10 by 10, how to select the submatrix consisting of intersection of the 1st 2nd and 9th rows and the 4th and 6th columns?
Thanks for any helpful answers!
TLDR: Short Answer
As for your question, suppose you have an arbitrary 10-by-10 matrix A. The simplest way to extract the desired sub-matrix would be with an index vector:
B = A([1 2 9], [4 6]);
Indexing in MATLAB
There's an interesting article in the official documentation that comprehensively explains indexing in MATLAB.
Basically, there are several ways to extract a subset of values, I'll summarize them for you:
1. Indexing Vectors
Indexing vectors indicate the indices of the element to be extracted. They can either contain a single index or several, like so:
A = [10 20 30 40 50 60 70 80 90]
%# Extracts the third and the ninth element
B = A([3 9]) %# B = [30 90]
Indexing vectors can be specified for each dimension separately, for instance:
A = [10 20 30; 40 50 60; 70 80 90];
%# Extract the first and third rows, and the first and second columns
B = A([1 3], [1 2]) %# B = [10 30; 40 60]
There are also two special subscripts: end and the colon (:):
end simply indicates the last index in that dimension.
The colon is just a short-hand notation for "1:end".
For example, instead of writing A([1 2 3], [2 3]), you can write A(:, 2:end). This is especially useful for large matrices.
2. Linear Indexing
Linear indexing treats any matrix as if it were a column vector by concatenating the columns into one column vector and assigning indices to the elements respectively. For instance, we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to compute b = A(2). The equivalent column vector is:
A = [10;
40;
70;
20;
50;
80;
30;
60;
90]
and thus b equals 40.
The special colon and end subscripts are also allowed, of course. For that reason, A(:) converts any matrix A into a column vector.
Linear indexing with matrix subscripts:
It is also possible to use another matrix for linear indexing. The subscript matrix is simply converted into a column vector, and used for linear indexing. The resulting matrix is, however always of the same dimensions as the subscript matrix.
For instance, if I = [1 3; 1 2], then A(I) is the same as writing reshape(A(I(:)), size(I)).
Converting from matrix subscripts to linear indices and vice versa:
For that you have sub2ind and ind2sub, respectively. For example, if you want to convert the subscripts [1, 3] in matrix A (corresponding to element 30) into a linear index, you can write sub2ind(size(A), 1, 3) (the result in this case should be 7, of course).
3. Logical Indexing
In logical indexing the subscripts are binary, where a logical 1 indicates that the corresponding element is selected, and 0 means it is not. The subscript vector must be either of the same dimensions as the original matrix or a vector with the same number of elements. For instance, if we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to extract A([1 3], [1 2]) using logical indexing, we can do either this:
Ir = logical([1 1 0]);
Ic = logical([1 0 1]);
B = A(Ir, Ic)
or this:
I = logical([1 0 1; 1 0 1; 0 0 0]);
B = A(I)
or this:
I = logical([1 1 0 0 0 0 1 1 0]);
B = A(I)
Note that in the latter two cases is a one-dimensional vector, and should be reshaped back into a matrix if necessary (for example, using reshape).
Let me explain with an example:
Let's define a 6x6 matrix
A = magic(6)
A =
35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16
4 36 29 13 18 11
From this matrix you want the elements in rows 1, 2 and 5, and in the columns 4 and 6
B = A([1 2 5],[4 6])
B =
26 24
21 25
12 16
Hope this helps.
function f = sub(A,i,j)
[m,n] = size(A);
row = 1:m;
col = 1:n;
x = row;
x(i) = [];
y=col;
y(j) = [];
f= A(x,y);
Returns the matrix A, with the ith row and jth column removed.