I have an array a of size ZxW.
Z = 20;
W = 30;
A = 40; %will be used below
size(a)
20 30
Then I apply to a two different transformations, and then after those I delete a and I cannot go back to it.
First transformation:
b = repelem(a(:,1),A,A);
Second transformation:
c = repmat(a,[1,1,A,A]);
d = c(:,1,:,:);
After those transformations and deleting a (which cannot be used for the following), I want to compare d and b using
assert( isequal(b,f) )
Where f is a transformation of d that makes the assertion true.
My first idea was a simple reshape:
f = reshape(squeeze(d),[Z*A,A]);
Which does not work as repelem and repmat move entries differently. How can I do this?
Thanks for the attention.
Sincerely
Luca
EDIT: changed
c = repmat(a,[A,A]);
with
c = repmat(a,[1,1,A,A]);
The answer (by Jan Simon) is:
f = reshape(permute(d, [3,1,4,2]), [Z*A,A]);
isequal(b, f) % 1: equal
Thanks for the help.
Luca
NOTE: This solution is obsoleted with the edited question, please refer to the other solutions posted.
Look closely at what the pattern of b and d are:
b = [1 1 1 2 2 2 3 3 3].'
d = [1 2 3 1 2 3 1 2 3].'
Hence, the transformation f can be:
f = reshape(d, [A, A]).'
f = f(:)
This will convert d to be exactly b or vice versa.
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)
In MATLAB I have a very large matrix (matrix A). Now I would like to find the row-index of the row which contain certain values in the second column. These values - which I'm looking for in Matrix A - are stored in anonther Matrix (Matrix B) with consists out of a row (800 numbers).
Besides I would like to redo this calculation for the same matrix A, but for ten different matrices, with different sizes (which contain the values I'm looking for in different columns of matrix A).
Because of the sizes of the matrix I think i need a loop to extract the row in matrix A which contain te value of Matrix B. How can I do this?
regards,
V
Thanks for the quick response! Indeed the problem is maybe a bit complex to answer without an example, and indeed duplicate entries cause some problems. Therefore hereby an example
For example I have a -simplified- matrix A:
1 2 3 4
9 9 9 9
4 3 2 1
And a -simplified- matrix (row) B: [9 3]
And a -simplified- matrix (row) C: [9 2]
Then I would like to get matrix D and matrix E.These matrices should contain in the first column the numbers from the original matrix D(or E) and in the second column the corresponding row-location of this value in matrix A.
So, matrix D =
9 2
3 3
matrix E =
9 2
2 3
As represented in this example matrix B and matrix C can contain data which is present in several column of matrix A (like the nine). However, martix B should "search" in column 2 of matrix A. Likewise, should matrix C "search" in column 3 of Matrix A, resulting in matrix D and E as given in the example.
As mentionned by Shai in his comment, your question is quite vague and a lot of special case could arise (duplicate entries, relative size of A and B, etc.). But in all generality I tried a small piece of code that seems to do what you want. There are certainly quicker ways of doing it, and certainly more information on your problem could help optimize this.
colA=2;
% Example
nmax=10;
nA=5;
A=randi(nmax,[nA nA]);
nB=3;
B=randi(nmax,[1 nB]);
% Find rows
rows=cell(size(B));
for i=1:numel(B)
rows(i)={find(A(:,colA)==B(i))};
end
The input / output was:
A =
3 7 8 5 4
9 7 3 7 5
8 2 9 9 8
9 5 9 7 9
3 3 4 6 8
B =
1 7 5
rows =
[0x1 double] [1;2] [4]
Assuming you have two vectors, largeDataIndex (the second column of your matrix) and interestingIndex (your b) and you want the following:
For each value of interestingIndex , find the position in largeDataIndex
Then an easy method would be this:
result = zeros(size(interestingIndex))
for i = 1:length(result)
result(i) = find(interestingIndex(i) == largeDataIndex)
end
Note that this assumes there is always just one entry that matches, otherwise you should define result as a cell array rather than a vector.
In Matlab, sort returns both the sorted vector and an index vector showing which vector element has been moved where:
[v, ix] = sort(u);
Here, v is a vector containing all the elements of u, but sorted. ix is a vector showing the original position of each element of v in u. Using Matlab's syntax, u(ix) == v.
My question: How do I obtain u from v and ix?
Of course, I could simply use:
w = zero(size(v));
for i = 1:length(v)
w(ix(i)) = v(i)
end
if nnz(w == u) == length(u)
print('Success!');
else
print('Failed!');
end
But I am having this tip-of-the-tongue feeling that there is a more elegant, single-statement, vectorized way of doing this.
If you are wondering why one would need to do this instead of just using u: I was trying to implement the Benjamini-Hochberg procedure which adjusts each element of the vector based on its position after sorting, but recovering the original order after adjusting was important for me.
The solution is:
w(ix) = v;
This is a valid Matlab operation provided that w is either at least as big as v, or not yet declared.
Example:
>> u = [4 8 10 6 2];
>> [v, ix] = sort(u)
v = 2 4 6 8 10
ix = 5 1 4 2 3
>> u(ix)
ans = 2 4 6 8 10
>> w(ix) = v
w = 4 8 10 6 2
(Apologies for the trivial question-answer, but I realized the solution as I was typing the question, and thought it might be useful to someone.)
In matlab I have a 128 by n matrix, which we can call
[A B C]
where each letter is an 128 by 1 matrix.
So what I want to do is concat the above matrix with another matrix,
[A~ D E].
Where A~ is similar in its values to A.
What I want to get as the result of the concat would be:
[A B C D E],
where A~ is omitted.
What is the best way to do this? Note that I do not know beforehand that A~ is similar.
To clarify, my problem is how would I determine if two columns are similar? By similar I mean where between two columns, many of the row values are close in value.
Maybe an illustration would help as well
Vector A: [1 2 3 4 5 6 7 8 9]'
| | | | | | | | |
Vector B: [20 2.4 4 5 0 7 7 7.6 10]'
where there are some instances where the values are completely different, but for the most part the values are close. I don't have a defined threshold for this, but ideally it would be something that I could experiment with.
If you want to omit only identical columns, this is one way to do it:
%# Define the example matrices.
Matrix1 = [ 1 2 3; 4 5 6; 7 8 9 ]';
Matrix2 = [ 4 5 6; 7 8 10 ]';
%# Concatenate the matrices and keep only unique columns.
OutputMatrix = unique([ Matrix1, Matrix2 ]', 'rows')';
To solve this, a matching algorithm called vl_ubcmatch can be used.
[matches, scores] = vl_ubcmatch(da, db) ; For each descriptor in da,
vl_ubcmatch finds the closest descriptor in db (as measured by the L2
norm of the difference between them). The index of the original match
and the closest descriptor is stored in each column of matches and the
distance between the pair is stored in scores.
source:
http://www.vlfeat.org/overview/sift.html
Thus, the solution is to find the matched columns with the highest scores and eliminate them before concatenating.
I think it's pdist2 you need.
Consider the following example:
>> X = rand(25, 5);
>> Y = rand(100, 5);
>> Y(22, : ) = 0.99*X(22,:);
>> D = pdist2(X,Y, 'euclidean');
>> [~,ind] = min(D(:));
>> [i,j]=ind2sub(size(D),ind)
i =
22
j =
22
which is indeed the entry we manipulated to be similar. Read help pdist2 or doc pdist2 for more background.