I don't know how to do it. I have some matrices with repeated values and I'd like to sort them and get a matrix of sorted indices. For example:
a = [1 4 3 10 8 2];
b = sort(a);
% This doesn't work but I wish it did - that's what I'm looking for.
% idx = find(a==b); idx = [1 6 3 2 5 4];
[v idx] = ismember(b,a);
However, when there are repeated values and NaN this produces errors. Try this:
a = [1 NaN 4 2 10 8 2];
b=sort(a);
[v, i] = ismember(b,a);
Gives [1 7 7 3 6 5 0] which is valid but I need it to be [ 1 4 7 3 6 5 0].
I can deal with this later but it's going to be much more elegant if it were to return the above result.
sort has a two output value version that gives you exactly the indices you're looking for:
a = [1 4 3 10 8 2];
[b, idx] = sort(a);
Outputs
idx =
1 6 3 2 5 4
This works correctly with NaNs as well:
a = [1 NaN 4 2 10 8 2];
[b, idx] =sort(a);
>> idx
idx =
1 4 7 3 6 5 2
>> b
b =
1 2 2 4 8 10 NaN
Related
I have a matrix, L, with two columns. I want to find its sub-matrices have equal values on their 2nd column. I want to do that using MATLAB without any for loop.
example:
L=[1 2;3 2;4 6;5 3;7 3;1 3;2 7;9 7]
then the sub-matrices are:
[1 2;3 2] , [4 6] , [5 3;7 3;1 3] and [2 7;9 7]
You can use a combination of arrayfun + unique to get that -
[~,~,labels] = unique(L(:,2),'stable')
idx = arrayfun(#(x) L(labels==x,:),1:max(labels),'Uniform',0)
Display output -
>> celldisp(idx)
idx{1} =
1 2
3 2
idx{2} =
4 6
idx{3} =
5 3
7 3
1 3
idx{4} =
2 7
9 7
You can use accumarray directly or with a sorted array, depending on you want the order of the rows to be stable, or the order of the submatricxes to be stable.
Say you want the rows to be stable:
>> [L2s,inds] = sort(L(:,2));
>> M = accumarray(L2s,inds,[],#(v){L(v,:)});
>> M(cellfun(#isempty,M)) = []; % remove empty cells
>> celldisp(M)
M{1} =
1 2
3 2
M{2} =
5 3
7 3
1 3
M{3} =
4 6
M{4} =
2 7
9 7
This question already has answers here:
Got confused with a vector indexed by a matrix, in Matlab
(2 answers)
Closed 8 years ago.
Suppose:
a =
1 2 3
4 5 6
2 3 4
and
b =
1 3 2
6 4 8
In MATLABa(b) gives:
>> a(b)
ans =
1 2 4
3 2 6
What is the reason for this output?
when you have a matrix a:
a =
1 2 3
4 5 6
7 8 9
and b:
b =
1 3 4
3 2 6
then a(b) is a way of adressing items in a and gives you:
>> a(b)
ans =
1 7 2
7 4 8
to understand this you have to think of a als a single column vector
>> a(:)
ans =
1
4
7
2
5
8
3
6
9
now the first row of b (1 3 4) addresses elements in this vector so the first, the 3rd and the forth element of that single column vector which are 1 7 and 2 are adressed. Next the secound row of b is used as adresses for a secound line in the output so the 3rd, the 2nd and the 6th elements are taken from a, those are 7 4 and 8.
It's just a kind of matrix indexing.
Matrix indexes numeration in 'a' matrix is:
1 4 7
2 5 8
3 6 9
This is a possible duplicate to this post where I gave an answer: Got confused with a vector indexed by a matrix, in Matlab
However, I would like to duplicate my answer here as I think it is informative.
That's a very standard MATLAB operation that you're doing. When you have a vector or a matrix, you can provide another vector or matrix in order to access specific values. Accessing values in MATLAB is not just limited to single indices (i.e. A(1), A(2) and so on).
For example, let's say we had a vector a = [1 2 3 4]. Let's also say we had b as a matrix such that it was b = [1 2 3; 1 2 3; 1 2 3]. By doing a(b) to access the vector, what you are essentially doing is a lookup. The output is basically the same size as b, and you are creating a matrix where there are 3 rows, and each element accesses the first, second and third element. Not only can you do this for a vector, but you can do this for a matrix as well.
Bear in mind that when you're doing this for a matrix, you access the elements in column major format. For example, supposing we had this matrix:
A = [1 2
3 4
5 6
7 8]
A(1) would be 1, A(2) would be 3, A(3) would be 5 and so on. You would start with the first column, and increasing indices will traverse down the first column. Once you hit the 5th index, it skips over to the next column. So A(5) would be 2, A(6) would be 4 and so on.
Here are some examples to further your understanding. Let's define a matrix A such that:
A = [5 1 3
7 8 0
4 6 2]
Here is some MATLAB code to strengthen your understanding for this kind of indexing:
A = [5 1 3; 7 8 0; 4 6 2]; % 3 x 3 matrix
B = [1 2 3 4];
C = A(B); % C should give [5 7 4 1]
D = [5 6 7; 1 2 3; 4 5 6];
E = A(D); % E should give [8 6 3; 5 7 4; 1 8 6]
F = [9 8; 7 6; 1 2];
G = A(F); % G should give [2 0; 3 6; 5 7]
As such, the output when you access elements this way is whatever the size of the vector or matrix that you specify as the argument.
In order to be complete, let's do this for a vector:
V = [-1 9 7 3 0 5]; % A 6 x 1 vector
B = [1 2 3 4];
C = V(B); % C should give [-1 9 7 3]
D = [1 3 5 2];
E = V(D); % E should give [-1 7 0 9]
F = [1 2; 4 5; 6 3];
G = V(F); % G should give [-1 9; 3 0; 5 7]
NB: You have to make sure that you are not providing indexes that would make the accessing out of bounds. For example if you tried to specify the index of 5 in your example, it would give you an error. Also, if you tried anything bigger than 9 in my example, it would also give you an error. There are 9 elements in that 3 x 3 matrix, so specifying a column major index of anything bigger than 9 will give you an out of bounds error.
The following codes runs in Matlab:
a = [1 2 3 4]
b = [ 1 2 3; 1 2 3; 1 2 3]
a(b)
The result of a(b) is a matrix:
[ 1 2 3; 1 2 3; 1 2 3]
Can anyone explain what happened here? Why a vector can be indexed by a matrix, how to interpret the result?
That's a very standard MATLAB operation that you're doing. When you have a vector or a matrix, you can provide another vector or matrix in order to access specific values. Accessing values in MATLAB is not just limited to single indices (i.e. A(1), A(2) and so on).
For example, what you have there is a vector of a = [1 2 3 4]. When you try to use b to access the vector, what you are essentially doing is a lookup. The output is basically the same size as b, and what you are doing is creating a matrix where there are 3 rows, and each element accesses the first, second and third element. Not only can you do this for a vector, but you can do this for a matrix as well.
Bear in mind that when you're doing this for a matrix, you access the elements in column major format. For example, supposing we had this matrix:
A = [1 2
3 4
5 6
7 8]
A(1) would be 1, A(2) would be 3, A(3) would be 5 and so on. You would start with the first column, and increasing indices will traverse down the first column. Once you hit the 5th index, it skips over to the next column. So A(5) would be 2, A(6) would be 4 and so on.
Here are some examples to further your understanding. Let's define a matrix A such that:
A = [5 1 3
7 8 0
4 6 2]
Here is some MATLAB code to strengthen your understanding for this kind of indexing:
A = [5 1 3; 7 8 0; 4 6 2]; % 3 x 3 matrix
B = [1 2 3 4];
C = A(B); % C should give [5 7 4 1]
D = [5 6 7; 1 2 3; 4 5 6];
E = A(D); % E should give [8 6 3; 5 7 4; 1 8 6]
F = [9 8; 7 6; 1 2];
G = A(F); % G should give [2 0; 3 6; 5 7]
As such, the output when you access elements this way is whatever the size of the vector or matrix that you specify as the argument.
In order to be complete, let's do this for a vector:
V = [-1 9 7 3 0 5]; % A 6 x 1 vector
B = [1 2 3 4];
C = V(B); % C should give [-1 9 7 3]
D = [1 3 5 2];
E = V(D); % E should give [-1 7 0 9]
F = [1 2; 4 5; 6 3];
G = V(F); % G should give [-1 9; 3 0; 5 7]
NB: You have to make sure that you are not providing indexes that would make the accessing out of bounds. For example if you tried to specify the index of 5 in your example, it would give you an error. Also, if you tried anything bigger than 9 in my example, it would also give you an error. There are 9 elements in that 3 x 3 matrix, so specifying a column major index of anything bigger than 9 will give you an out of bounds error.
Notice that the return value of a(b) is the same size as b.
a(b) simply takes each element of b, call it b(i,j), as an index and returns the outputs a(b(i,j)) as a matrix the same size as b. You should play around with other examples to get a more intuitive feel for this:
b = [4 4 4; 4 4 4];
a(b) % Will return [4 4 4; 4 4 4]
c = [5; 5];
a(c) % Will error as 5 is out of a's index range
I have two vectors:
a = [1 3 5 7 9];
b = [2 4 6 8 10];
That I need to combine together element wise. Meaning that I need the first element of vector a, then the first element of vector b, second of a, second of b, and so forth until I get the following:
combined = [1 2 3 4 5 6 7 8 9 10]
How do I do this within MatLab?
Edit
I ran a test of the top three answers (Josh, Marc, & Kronos) and compared the time it took to run them. I ran each 100 times after doing a 10 iteration warmup. The vectors created were exactly the same size in length (16e+6) and were random values ranging from 1 to 100:
Test Results
Test: Total Time (100 runs): Avg Time Per Exec:
Josh B 21.3687 0.2137
Marc C 21.4273 0.2143
Kronos 31.1897 0.3119
It appears that both Josh's and Marc's solutions are similar in execution time.
a = [1 3 5 7 9];
b = [2 4 6 8 10];
temp = [a; b];
combined = temp(:)';
This can be done by the following:
a = [1 3 5 7 9];
b = [2 4 6 8 10];
combinedSize = size(a, 2) * 2;
combined(1:2:combinedSize) = a;
combined(2:2:combinedSize) = b;
This is obviously assuming that your vectors are exactly the same size. If by chance you want to merge two vectors that are not the same size then you can do the following:
combinedSize = max(size(a, 2), size(b, 2)) * 2;
combined = NaN(1,combinedSize);
combined(1:2:size(a,2)*2) = a;
combined(2:2:size(b,2)*2) = b;
This will place a NaN for the remaining elements of the smaller vector. For example, given the following sample vectors:
a = [1 3 5 7 9 11];
b = [2 4 6 8];
will result in the combined vector:
combined =
1 2 3 4 5 6 7 8 9 NaN 11 NaN
Place the vectors below eachother in a matrix and use reshape. For example:
>> A=[1 2 3]
A =
1 2 3
>> B=[4 5 6]
B =
4 5 6
>> C=reshape([A;B],1,size(A,2)+size(B,2))
C =
1 4 2 5 3 6
It's straightforward to generalize to more than 2 vectors.
You can also give a try to looping, for example:
a=[1 2 3 4 5];
b=[11 12 13 14 15];
for i = 1:N
{
if (i%2==0)
{ c[i] = b[i]; }
else
{ c[i] = a[i]; }
This shall work!
All the answers above only work if the two vectors have the same number of elements. The following will work even if they have different number of elements:
>>
A = [1 3 5];
B = [2 4 6 7 8];
C = [1 3 5 7 8];
D = [2 4 6];
AB = nan(1,2*max(numel(A),numel(B)));
CD = nan(1,2*max(numel(C),numel(D)));
AB(2*(1:length(A))) = A;
AB(1+2*(1:length(B))) = B;
CD(2*(1:length(C))) = C;
CD(1+2*(1:length(D))) = D;
>>
AB = AB(~isnan(AB))
CD = CD(~isnan(CD))
The result would be:
AB =
1 2 3 4 5 6 7 8
CD =
1 2 3 4 5 6 7 8
I need some help in converting a 2X2 matrix to a 4X4 matrix in the following manner:
A = [2 6;
8 4]
should become:
B = [2 2 6 6;
2 2 6 6;
8 8 4 4;
8 8 4 4]
How would I do this?
In newer versions of MATLAB (R2015a and later) the easiest way to do this is using the repelem function:
B = repelem(A, 2, 2);
For older versions, a short alternative to the other (largely) indexing-based solutions is to use the functions kron and ones:
>> A = [2 6; 8 4];
>> B = kron(A, ones(2))
B =
2 2 6 6
2 2 6 6
8 8 4 4
8 8 4 4
Can be done even easier than Jason's solution:
B = A([1 1 2 2], :); % replicate the rows
B = B(:, [1 1 2 2]); % replicate the columns
Here's one more solution:
A = [2 6; 8 4];
B = A( ceil( 0.5:0.5:end ), ceil( 0.5:0.5:end ) );
which uses indexing to do everything and doesn't rely on the size or shape of A.
This works:
A = [2 6; 8 4];
[X,Y] = meshgrid(1:2);
[XI,YI] = meshgrid(0.5:0.5:2);
B = interp2(X,Y,A,XI,YI,'nearest');
This is just two-dimensional nearest-neighbor interpolation of A(x,y) from x,y ∈ {1,2} to x,y ∈ {0.5, 1, 1.5, 2}.
Edit: Springboarding off of Jason S and Martijn's solutions, I think this is probably the shortest and clearest solution:
A = [2 6; 8 4];
B = A([1 1 2 2], [1 1 2 2]);
A = [2 6; 8 4];
% arbitrary 2x2 input matrix
B = repmat(A,2,2);
% replicates rows & columns but not in the way you want
B = B([1 3 2 4], :);
% swaps rows 2 and 3
B = B(:, [1 3 2 4]);
% swaps columns 2 and 3, and you're done!
Here's a method based on simple indexing that works for an arbitrary matrix. We want each element to be expanded to an MxN submatrix:
A(repmat(1:end,[M 1]),repmat(1:end,[N 1]))
Example:
>> A=reshape(1:6,[2,3])
A =
1 3 5
2 4 6
>> A(repmat(1:end,[3 1]),repmat(1:end,[4 1]))
ans =
1 1 1 1 3 3 3 3 5 5 5 5
1 1 1 1 3 3 3 3 5 5 5 5
1 1 1 1 3 3 3 3 5 5 5 5
2 2 2 2 4 4 4 4 6 6 6 6
2 2 2 2 4 4 4 4 6 6 6 6
2 2 2 2 4 4 4 4 6 6 6 6
To see how the method works, let's take a closer look at the indexing. We start with a simple row vector of consecutive numbers
>> m=3; 1:m
ans =
1 2 3
Next, we extend it to a matrix, by repeating it M times in the first dimension
>> M=4; I=repmat(1:m,[M 1])
I =
1 2 3
1 2 3
1 2 3
1 2 3
If we use a matrix to index an array, then the matrix elements are used consecutively in the standard Matlab order:
>> I(:)
ans =
1
1
1
1
2
2
2
2
3
3
3
3
Finally, when indexing an array, the 'end' keyword evaluates to the size of the array in the corresponding dimension. As a result, in the example the following are equivalent:
>> A(repmat(1:end,[3 1]),repmat(1:end,[4 1]))
>> A(repmat(1:2,[3 1]),repmat(1:3,[4 1]))
>> A(repmat([1 2],[3 1]),repmat([1 2 3],[4 1]))
>> A([1 2;1 2;1 2],[1 2 3;1 2 3;1 2 3;1 2 3])
>> A([1 1 1 2 2 2],[1 1 1 1 2 2 2 2 3 3 3 3])
There is a Reshape() function that allows you to do this...
For example:
reshape(array, [64, 16])
And you can find a great video tutorial here
Cheers