Overview
An n×m matrix A and an n×1 vector Date are the inputs of the function S = sumdate(A,Date).
The function returns an n×m vector S such that all rows in S correspond to the sum of the rows of A from the same date.
For example, if
A = [1 2 7 3 7 3 4 1 9
6 4 3 0 -1 2 8 7 5]';
Date = [161012 161223 161223 170222 160801 170222 161012 161012 161012]';
Then I would expect the returned matrix S is
S = [15 9 9 6 7 6 15 15 15;
26 7 7 2 -1 2 26 26 26]';
Because the elements Date(2) and Date(3) are the same, we have
S(2,1) and S(3,1) are both equal to the sum of A(2,1) and A(3,1)
S(2,2) and S(3,2) are both equal to the sum of A(2,2) and A(3,2).
Because the elements Date(1), Date(7), Date(8) and Date(9) are the same, we have
S(1,1), S(7,1), S(8,1), S(9,1) equal the sum of A(1,1), A(7,1), A(8,1), A(9,1)
S(1,2), S(7,2), S(8,2), S(9,2) equal the sum of A(1,2), A(7,2), A(8,2), A(9,2)
The same for S([4,6],1) and S([4,6],2)
As the element Date(5) does not repeat, so S(5,1) = A(5,1) = 7 and S(5,2) = A(5,2) = -1.
The code I have written so far
Here is my try on the code for this task.
function S = sumdate(A,Date)
S = A; %Pre-assign S as a matrix in the same size of A.
Dlist = unique(Date); %Sort out a non-repeating list from Date
for J = 1 : length(Dlist)
loc = (Date == Dlist(J)); %Compute a logical indexing vector for locating the J-th element in Dlist
S(loc,:) = repmat(sum(S(loc,:)),sum(loc),1); %Replace the located rows of S by the sum of them
end
end
I tested it on my computer using A and Date with these attributes:
size(A) = [33055 400];
size(Date) = [33055 1];
length(unique(Date)) = 2645;
It took my PC about 1.25 seconds to perform the task.
This task is performed hundreds of thousands of times in my project, therefore my code is too time-consuming. I think the performance will be boosted up if I can eliminate the for-loop above.
I have found some built-in functions which do special types of sums like accumarray or cumsum, but I still do not have any ideas on how to eliminate the for-loop.
I would appreciate your help.
You can do this with accumarray, but you'll need to generate a set of row and column subscripts into A to do it. Here's how:
[~, ~, index] = unique(Date); % Get indices of unique dates
subs = [repmat(index, size(A, 2), 1) ... % repmat to create row subscript
repelem((1:size(A, 2)).', size(A, 1))]; % repelem to create column subscript
S = accumarray(subs, A(:)); % Reshape A into column vector for accumarray
S = S(index, :); % Use index to expand S to original size of A
S =
15 26
9 7
9 7
6 2
7 -1
6 2
15 26
15 26
15 26
Note #1: This will use more memory than your for loop solution (subs will have twice the number of element as A), but may give you a significant speed-up.
Note #2: If you are using a version of MATLAB older than R2015a, you won't have repelem. Instead you can replace that line using kron (or one of the other solutions here):
kron((1:size(A, 2)).', ones(size(A, 1), 1))
i want to get this: if
a=1 2 3
4 5 6
7 8 9 i want a at the end to be
a= 1 6 9
12 25 18
49 24 81
which mean the diagonal i did it in square also the opposite diagonal
and the other elemnts in the matrix i multpplied them with 3.. so i
have this so far :
a=rand(n)
c=fliplr(diag(diag(a)).^2)+(tril(a,-1)+triu(a,1)*3)
i=(diag(diag(fliplr(a)))).^2
c(1:n+1:n^2)=0
result=fliplr(c+i)
this works if i do it in a comand window but i want to write it as a
function! thanks in advance.
it's easier to use logical indexing. if you want to use it as a function just open a new file with function A = myfunc(A)
% random matrix
n = 5;
A = randi(10,n);
% here you can do:
% function A = myfunc(A)
% diagonal indexes
diagIdxs1 = eye(size(A),'logical');
diagIdxs2 = fliplr(diagIdxs1);
diagIdxs = diagIdxs1 | diagIdxs2;
% do operations on diagonals and on non-diagonals
A(diagIdxs) = A(diagIdxs).^2;
A(~diagIdxs) = A(~diagIdxs).*3;
A shorter and computationally faster alternative to using logical indexing would be to use element-wise multiplication and element-wise powers.
% function b = myfun(a)
n = 10000
a = randi(10,n);
I = eye(size(a,1),'logical');
b = a.^2.*I + a.*3.*~I;
% end
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)
I am attempting to set up some code to extract certain elements of a matrix, and keeping only these values in another matrix, in the order they were extracted.
Example: If I have a random 1X20 matrix, but want only every Nth = 5th element beginning with 4 and 5, I would want it to construct a new matrix (1x8) consisting only of 4, 5, 9, 10, 14, 15, 19, 20.
What I have so far is:
r = rand(1,20);
n = 5;
a = r(4 : n : end);
b = r(5 : n : end);
So instead of two separate matrices, I instead want one matrix in its original chronological order (again, a 1x8 matrix consisting of the elements in the order of 4,5,9,10,14,15,19,20). Essentially, I'd like to be able to do this for any number of values while still maintaining the original order the elements were in.
Create all the indices to index into r separately for indices starting with 4 and 5 and then sort them to keep the order of elements as it was originally in r.
So, this should work -
ab = r(sort([4:n:numel(r) 5:n:numel(r)]))
A more generic solution for a variable number of starting values:
% example
A = 1:20;
% starting values and number of values to skip
a = [4,5];
n = 5;
% index vector
idx = bsxfun(#plus,a',(0:numel(A)/n-1)*n)
% indexing
result = A(idx(:))
returns:
idx(:)' = 4 5 9 10 14 15 19 20
Another example:
A = 1:40;
a = [3,4,7];
n = 10;
idx = bsxfun(#plus,a',(0:numel(A)/n-1)*n)
returns:
idx(:)' = 3 4 7 13 14 17 23 24 27 33 34 37
You can do it using ndgrid (this idea is taken from the code of kron, which does more or less what you want but with products instead of sums):
a = [4 5]; %// initial values
M = 20; %// maximum value
s = 5; %// step
[ii jj] = ndgrid(a,0:s:M-max(a));
ind = (ii(:)+jj(:)).';
How can I interpolate a vector in MATLAB?
For example, I have the following matrix:
M=
1 10
2 20
3 30
4 40
The first column of M denotes the independent parameter of x coordinate while the second column of M denotes the output or y coordinate.
I also have the following input vector:
a =
2.3
2.1
3.5
For each value of a, I wish to determine what the output interpolated result would be. In this case, given a, I wish to return
23
21
35
Here's the answer to the question after the edit, i.e. "how to interpolate"
You want to use interp1
M = [1 10;2 20;3 30;4 40];
a = [2.3;2.1;3.5;1.2];
interpolatedVector = interp1(M(:,1),M(:,2),a)
interpolatedVector =
23
21
35
12
Here's the answer to the question "find the two closest entries in a vector", i.e. the original question before the edit.
x=[1,2,3,4,5]'; %'#
a =3.3;
%# sort the absolute difference
[~,idx] = sort(abs(x-a));
%# find the two closest entries
twoClosestIdx = idx(1:2);
%# turn it into a logical array
%# if linear indices aren't good enough
twoClosestIdxLogical = false(size(x));
twoClosestIdxLogical(twoClosestIdx) = true;
twoClosestIdxLogical =
0
0
1
1
0