This question already has answers here:
Normalizing rows of matrix, so that their norm is equal to 1 (MATLAB)
(2 answers)
Closed 6 years ago.
I have a vector:
vector = [1 2 3;4 5 6; 7 9 0]
vector =
1 2 3
4 5 6
7 9 0
I want to take this and create a unit vector. We can get the magnitude by doing:
mag = sqrt(sum(vector'.^2))'
mag =
3.7417
8.7750
11.4018
When we try to divide each element by the magnitude I get an error:
vector./mag
Error using ./
Matrix dimensions must agree.
Essentially I must divide every vector element in each row by every row in the mag vector. How can I do this?
The other answers give the correct result but you can vectorize the calculation for faster calculation.
ret = bsxfun(#rdivide, vector, mag)
I recommend using the bsxfun, it is a very useful function for matrix calculation.
The problem is that, as the error message says, the dimensions of vector and mag don't match.
You want to divide every element of the first row of vector by mag(1).
What you need is repmat(), which "repeats copies of array".
Writing
repmat(mag,1,3)
returns a 3x3 matrix such that every column is an exact copy of mag:
3.7417 3.7417 3.7417
8.7750 8.7750 8.7750
11.4018 11.4018 11.4018
So you can use the one-liner:
vector./repmat(mag,1,3)
ans =
0.26726 0.53452 0.80178
0.45584 0.56980 0.68376
0.61394 0.78935 0.00000
That way, the first row of vector, i.e., [1 2 3], is divided element-by-element by [3.7417 3.7417 3.7417].
In other words, every element of vector is divided by the correct magnitude.
You can use matrix operators in MATLAB:
result = diag(1./mag)*vector;
If dimension of mag can be too big you can use sparse version of it:
result = spdiags(1./mag,0,speye(numel(mag)))*vector;
A simple solution is the use of a for-loop:
vector = [1 2 3; 4 5 6; 7 9 0];
mag = sqrt(sum(vector'.^2))';
A = [];
for i = 1:numel(mag)
A(i,:) = vector(i,:)./mag(i);
end
vector = [1 2 3; 4 5 6 ;7 9 0] ;
[n,m]=size(vector);
for i=1:n
normv=norm(vector(i,:),2);
nvector(i,:)=vector(i,:)/normv;
end
the nvector will contain the normalized vector of each line
Lets say I have this matrice A: [3 x 4]
1 4 7 10
2 5 8 11
3 6 9 12
I want to permute the element of in each column, but they can't change to a different column, so 1 2 3 need to always be part of the first column. So for exemple I want:
3 4 8 10
1 5 7 11
2 6 9 12
3 4 8 11
1 6 7 10
2 5 9 12
1 6 9 11
. . . .
So in one matrix I would like to have all the possible permutation, in this case, there are 3 different choices 3x3x3x3=81possibilities.So my result matrixe should be 81x4, because I only need each time one [1x4]line vector answer, and that 81 time.
An other way to as the question would be (for the same end for me), would be, if I have 4 column vector:
a=[1;2;3]
b=[4;5;6]
c=[7;8;9]
d=[10;11;12;13]
Compare to my previous exemple, each column vector can have a different number of row. Then is like I have 4 boxes, A, B C, D and I can only put one element of a in A, b in B and so on; so I would like to get all the permutation possible with the answer [A B C D] beeing a [1x4] row, and in this case, I would have 3x3x3x4=108 different row. So where I have been missunderstood (my fault), is that I don't want all the different [3x4] matrix answers but just [1x4]lines.
so in this case the answer would be:
1 4 7 10
and 1 4 7 11
and 1 4 7 12
and 1 4 7 13
and 2 4 8 10
and ...
until there are the 108 combinations
The fonction perms in Matlab can't do that since I don't want to permute all the matrix (and btw, this is already a too big matrix to do so).
So do you have any idea how I could do this or is there is a fonction which can do that? I, off course, also could have matrix which have different size. Thank you
Basically you want to get all combinations of 4x the permutations of 1:3.
You could generate these with combvec from the Neural Networks Toolbox (like #brainkz did), or with permn from the File Exchange.
After that it's a matter of managing indices, applying sub2ind (with the correct column index) and rearranging until everything is in the order you want.
a = [1 4 7 10
2 5 8 11
3 6 9 12];
siz = size(a);
perm1 = perms(1:siz(1));
Nperm1 = size(perm1,1); % = factorial(siz(1))
perm2 = permn(1:Nperm1, siz(2) );
Nperm2 = size(perm2,1);
permidx = reshape(perm1(perm2,:)', [Nperm2 siz(1), siz(2)]); % reshape unnecessary, easier for debugging
col_base_idx = 1:siz(2);
col_idx = col_base_idx(ones(Nperm2*siz(1) ,1),:);
lin_idx = reshape(sub2ind(size(a), permidx(:), col_idx(:)), [Nperm2*siz(1) siz(2)]);
result = a(lin_idx);
This avoids any loops or cell concatenation and uses straigh indexing instead.
Permutations per column, unique rows
Same method:
siz = size(a);
permidx = permn(1:siz(1), siz(2) );
Npermidx = size(permidx, 1);
col_base_idx = 1:siz(2);
col_idx = col_base_idx(ones(Npermidx, 1),:);
lin_idx = reshape(sub2ind(size(a), permidx(:), col_idx(:)), [Npermidx siz(2)]);
result = a(lin_idx);
Your question appeared to be a very interesting brain-teaser. I suggest the following:
in = [1,2,3;4,5,6;7,8,9;10,11,12]';
b = perms(1:3);
a = 1:size(b,1);
c = combvec(a,a,a,a);
for k = 1:length(c(1,:))
out{k} = [in(b(c(1,k),:),1),in(b(c(2,k),:),2),in(b(c(3,k),:),3),in(b(c(4,k),:),4)];
end
%and if you want your result as an ordinary array:
out = vertcat(out{:});
b is a 6x3 array that contains all possible permutations of [1,2,3]. c is 4x1296 array that contains all possible combinations of elements in a = 1:6. In the for loop we use number from 1 to 6 to get the permutation in b, and that permutation is used as indices to the column.
Hope that helps
this is another octave friendly solution:
function result = Tuples(A)
[P,n]= size(A);
M = reshape(repmat(1:P, 1, P ^(n-1)), repmat(P, 1, n));
result = zeros(P^ n, n);
for i = 1:n
result(:, i) = A(reshape(permute(M, circshift((1:n)', i)), P ^ n, 1), i);
end
end
%%%example
A = [...
1 4 7 10;...
2 5 8 11;...
3 6 9 12];
result = Tuples(A)
Update:
Question updated that: given n vectors of different length generates a list of all possible tuples whose ith element is from vector i:
function result = Tuples( A)
if exist('repelem') ==0
repelem = #(v,n) repelems(v,[1:numel(v);n]);
end
n = numel(A);
siz = [ cell2mat(cellfun(#numel, A , 'UniformOutput', false))];
tot_prd = prod(siz);
cum_prd=cumprod(siz);
tot_cum = tot_prd ./ cum_prd;
cum_siz = cum_prd ./ siz;
result = zeros(tot_prd, n);
for i = 1: n
result(:, i) = repmat(repelem(A{i},repmat(tot_cum(i),1,siz(i))) ,1,cum_siz(i));
end
end
%%%%example
a = {...
[1;2;3],...
[4;5;6],...
[7;8;9],...
[10;11;12;13]...
};
result =Tuples(a)
This is a little complicated but it works without the need for any additional toolboxes:
You basically want a b element 'truth table' which you can generate like this (adapted from here) if you were applying it to each element:
[b, n] = size(A)
truthtable = dec2base(0:power(b,n)-1, b) - '0'
Now you need to convert the truth table to linear indexes by adding the column number times the total number of rows:
idx = bsxfun(#plus, b*(0:n-1)+1, truthtable)
now you instead of applying this truth table to each element you actually want to apply it to each permutation. There are 6 permutations so b becomes 6. The trick is to then create a 6-by-1 cell array where each element has a distinct permutation of [1,2,3] and then apply the truth table idea to that:
[m,n] = size(A);
b = factorial(m);
permutations = reshape(perms(1:m)',[],1);
permCell = mat2cell(permutations,ones(b,1)*m,1);
truthtable = dec2base(0:power(b,n)-1, b) - '0';
expandedTT = cell2mat(permCell(truthtable + 1));
idx = bsxfun(#plus, m*(0:n-1), expandedTT);
A(idx)
Another answer. Rather specific just to demonstrate the concept, but can easily be adapted.
A = [1,4,7,10;2,5,8,11;3,6,9,12];
P = perms(1:3)'
[X,Y,Z,W] = ndgrid(1:6,1:6,1:6,1:6);
You now have 1296 permutations. If you wanted to access, say, the 400th one:
Permutation_within_column = [P(:,X(400)), P(:,Y(400)), P(:,Z(400)), P(:,W(400))];
ColumnOffset = repmat([0:3]*3,[3,1])
My_permutation = Permutation_within_column + ColumnOffset; % results in valid linear indices
A(My_permutation)
This approach allows you to obtain the 400th permutation on demand; if you prefer to have all possible permutations concatenated in the 3rd dimension, (i.e. a 3x4x1296 matrix), you can either do this with a for loop, or simply adapt the above and vectorise; for example, if you wanted to create a 3x4x2 matrix holding the first two permutations along the 3rd dimension:
Permutations_within_columns = reshape(P(:,X(1:2)),3,1,[]);
Permutations_within_columns = cat(2, Permutations_within_columns, reshape(P(:,Y(1:2)),3,1,[]));
Permutations_within_columns = cat(2, Permutations_within_columns, reshape(P(:,Z(1:2)),3,1,[]));
Permutations_within_columns = cat(2, Permutations_within_columns, reshape(P(:,W(1:2)),3,1,[]));
ColumnOffsets = repmat([0:3]*3,[3,1,2]);
My_permutations = Permutations_within_columns + ColumnOffsets;
A(My_permutations)
This approach enables you to collect a specific subrange, which may be useful if available memory is a concern (i.e. for larger matrices) and you'd prefer to perform your operations by blocks. If memory isn't a concern you can get all 1296 permutations at once in one giant matrix if you wish; just adapt as appropriate (e.g. replicate ColumnOffsets the right number of times in the 3rd dimension)
This question already has answers here:
Vector norm of an array of vectors in MATLAB
(4 answers)
Closed 5 years ago.
I have an input matrix that has 3 rows and 1000 columns. Each column represents and x, y, z variable. I want to find the magnitude of each column and store that in an output matrix that has 1 row and 1000 columns.
This is my current attempt but it doesn't seem to be working:
output(1,:) = norm(input(3,:));
my input matrix looks like:
x1, x2,...,x1000
y1, y2,...,y1000
z1, z2,...,z1000
I want my output matrix to look like:
[magnitude(x1,y1,z1), magnitude(x2,y2,z2),...,magnitude(x1000,y1000,z1000)]
Any help would be greatly appreciated.
norm(input(3,:)) will give you the norm of the 1000 elements of the third row.
Easy solution is to just run a for loop.
output = zeros(1,1000); %Preallocate space
for i = 1:length(output)
output(i) = norm(input(:, i));
end
MATLAB's norm function only works for single vectors. Let A be the name of the matrix which columns you want to find the norm to. Then this command does the job:
norm_A = sqrt(sum(A.*A));
Here is an example:
>> A = [1:5; 1:5; 1:5]
A =
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
>> norm_A = sqrt(sum(A.*A))
norm_A =
1.7321 3.4641 5.1962 6.9282 8.6603
Short Version
How can I do concatMap in MATLAB? I'm trying to build a single vector from a series of smaller, differently sized vectors. I know I can do:
result = [];
for i=1:N
result = [result nextPart(i)];
end
but that has a serious speed impact and there must be a smarter way to do concatMap.
Long Version
I'm trying to write a MATLAB function that returns the counterdiagonals of a block. For example, if you have the block:
1 2 4
3 5 7
6 8 9
then counterDiagonals(block) should return [1 2 3 4 5 6 7 8 9].
I have a function that will find a single counter diagonal of a block. i.e. counterDiagonal(x, 3) will return [4 5 6].
Therefore, counterDiagonals should be as simple as concatMap counterDiagonal(x, i) (1:N) where N is (2*length(block)-1). How can I do this in MATLAB in an efficient way?
One problem with the accepted answer: if the matrix A had zeros, they will be incorrectly removed from the result.. Instead you should work on the indices of the elements:
A = [0 2 4; 3 5 7; 6 8 9]; %# Sample matrix (contains zeros)
ind = reshape(1:numel(A), size(A)); %# indices of elements
ind = fliplr( spdiags( fliplr(ind) ) ); %# get the anti-diagonals (or use ROT90)
ind(ind==0) = []; %# keep non-zero indices
result = A(ind); %# get elements in desired order
This is very similar to this answer I gave in a previous question (the difference was that the anti-digaonals were in reverse order).
I believe what you want to do can be accomplished using the functions ROT90 and SPDIAGS:
A = [1 2 4; 3 5 7; 6 8 9]; %# Sample matrix
result = rot90(A); %# Rotate the matrix counter-clockwise
result = spdiags(result); %# Find all the diagonals
result = result(result ~= 0).'; %'# Remove zero padding and format the results
%# into a row vector
And you should end up with result = [1 2 3 4 5 6 7 8 9].
EDIT: As Amro mentions in a comment, the above code assumes that there are no zeroes in the original matrix A. If there are zeroes in the original matrix, one solution is to replace them with a non-zero flag value that you know doesn't appear in the original matrix (like, for example, NaN), run the above code, then replace the flag values in the result:
A = [0 2 4; 3 0 7; 6 8 0]; %# Sample matrix
result = rot90(A); %# Rotate the matrix counter-clockwise
result(result == 0) = nan; %# Replace zeroes with NaN
result = spdiags(result); %# Find all the diagonals
result = result(result ~= 0).'; %'# Remove zero padding and format the results
%# into a row vector
result(isnan(result)) = 0; %# Put the original zeroes back
Short version:
If you preassign your result array, everything will be a lot faster.
result = zeros(1,knownLengthOfResultsArray); %# such as "numel(block)"
ct = 1;
for i=1:N
tmp = nextPart(i);
nTmp = length(tmp);
result(ct:ct+nTmp-1) = tmp;
ct = ct + nTmp;
end
Long version:
However, it may be even more efficient to rewrite your algorithm. See e.g. the answers to this question (use fliplr on your array first), or #gnovice's answer.