I have a matrix as shown in below:
A=[2 4;1 3;8 6;5 1;4 9]
now i need to extract the matrix A into 2 parts:
newpoint=[2 4];
rest=[1 3;8 6;5 1;4 9];
then apply loop again to extract the second column as new point :
newpoint=[1 3];
rest=[2 4;8 6;5 1;4 9];
Applying loop again to take third column number as new point :
newpoint=[8 6];
rest=[2 4;1 3;5 1;4 9];
Take the number in row sequence until the last row . Can someone be kind enough to help.Thanks~
Apart from HebeleHododo's answer, if you have big matrices maybe you can try this:
A = [2 4; 1 3; 8 6; 5 1; 4 9];
B = zeros(size(A,1)-1,size(A,2));
for idx = 1:size(A, 1)
newpoint = A(idx, :);
B(1:idx-1,:) = A(1:idx-1,:);
B(idx:end,:) = A(idx+1:end,:);
% do stuff
end
It doesn't get rid of the for loop, but the temporary B matrix is pre-allocated and the copy between A and B is clear, which makes it quicker.
For A = rand(100000,2); HebeleHododo's method takes ~123 seconds in my computer
and the one above takes ~85 seconds.
Edit: Just for reference, the timing is done using Intel Core i5-3450 CPU # 3.10GHz and Matlab R2011b
You said you want to extract columns, but gave examples with rows. I am going ahead and assuming you meant rows.
You can do it with a for loop.
A = [2 4; 1 3; 8 6; 5 1; 4 9];
for idx = 1:size(A, 1)
newpoint = A(idx, :);
rest = A; % Copy A to rest
rest(idx, :) = []; % Remove newpoint line
% do stuff
end
Results of first two iterations:
newpoint =
2 4
rest =
1 3
8 6
5 1
4 9
newpoint =
1 3
rest =
2 4
8 6
5 1
4 9
This is not a good method if your A matrix is big.
Edit: In fact, do not use this method. George Aprilis timed it and found 123 seconds for a 100000x2 matrix. I guess my computer is much slower. It took 216 seconds. I repeat, do not use this.
Related
For example:
a = [1 2 3; 4 5 6; 7 8 9];
b = [2 4]; %//Indices I got
How can I set to zero every element of a not indexed in b in order to obtain:
0 2 0
4 0 0
0 0 0
I tried for loop:
for i = 1:numel(a)
if i ~= b
a(i) = 0;
end
end
but the matrix I cope with is really large and it takes ridiculously long time to finish running.
Is there any smart way to do it? Thank you.
Try this:
a = [1 2 3; 4 5 6; 7 8 9];
b = [2 4];
a(setdiff(1:length(a(:)),b)) = 0;
UPDATE
As proposed by #Daniel, for large matrices is better to use
a(setdiff(1:numel(a),b)) = 0;
An alternative to Anton's direct solution is one based on copying:
a = [1 2 3; 4 5 6; 7 8 9];
b = [2 4];
atmp = a(b);
a = zeros(size(a));
a(b) = atmp; %// copy needed elements
I guess efficiency of the two approaches boils down to allocation vs setdiff. Also, if your resulting matrix has many zeroes, you should perhaps consider using a sparse matrix.
How to effectively vectorize the following MATLAB code, which performs permutation of each row of matrix R by indices in corresponding row of matrix P?
for i = 1:size(P,1)
pP(i,:) = R(i,P(i,:));
end
example:
P = [3 2 1;
3 1 2;
2 3 1;
2 1 3;
1 2 3;
1 3 2]
R = [6 5 4;
6 4 5;
5 6 4;
5 4 6;
4 5 6;
4 6 5]
produce following matrix pR:
4 5 6
5 6 4
6 4 5
4 5 6
4 5 6
4 5 6
One approach with bsxfun -
nrows = size(R,1)
pP = R(bsxfun(#plus,[1:nrows]',(P-1)*nrows))
Or with ndgrid -
[m,n] = size(R)
pP = R(sub2ind([m n],ndgrid(1:m,1:n),P))
Or replace ndgrid(1:m,1:n) with repmat: repmat([1:m]',[1 n]) or with meshgrid:meshgrid(1:m,1:n).'.
This might not be the best way to do it, but you could do something like:
IND1 = P(:,1)
Q(:,1) = diag(R(:,IND));
and repeat for P(:,2), P(:,3) in a similar fashion.
You can use arrayfun to avoid the loop but probably won't gain in performance if that it is the reason for vectorizing it:
cell2mat(arrayfun(#(k) R(k, P(k,:)), (1:size(P,1)).', 'uni', 0))
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
I got a 4-by-n matrix, like
A =
1 5 9
3 0 6
2 3 10
7 8 4
What I want to do with A is getting each half column of A as
Line1Point1 = [1 3]
Line1Point2 = [2 7]
Line2Point1 = [5 0]
Line2Point2 = [3 8]
Line3Point1 = [9 6]
Line3Point2 = [10 4]
How could I do that? I’m pretty new to matlab coding.. Any help is really appreciated..
Cheers
Use reshape function, for example:
>> A = [1 5 9;
3 0 6;
2 3 10;
7 8 4];
>> reshape(A,2,6)
ans =
1 2 5 3 9 10
3 7 0 8 6 4
Storing such information as many variables is generally a bad idea
Some options for storing and accessing are
Cell array
Line=mat2cell(A,[2,2],ones(1,size(A,2))).'
access with
Line{2,1}
ans =
5
0
Indexing
as other answers
Anonymous Function
Line=#(l,p)A(2*p-1:2*p,l)
access with
Line(2,1)
ans =
5
0
Structure
Not really a useful solution, more for interests sake
for ii=1:size(A,2);for jj=1:2;Line(ii).Point(jj).Value=A(2*jj-1:2*jj,ii);end;end
access with
Line(2).Point(1).Value
ans =
5
0
A(1:2,1) will give you first half of the first column.
A(3:4,1) will give you second half of the first column.
A(1:2,2) will give you first half of the second column.
A(3:4,2) will give you second half of the second column.
A(1:2,3) will give you first half of the third column.
A(3:4,3) will give you second half of the third column.
You can create the variables with the eval function, which executes the input string. Using eval is commonly regarded as bad practice since it is horrible to debug.
Nevertheless, here's the code:
A = [1 5 9; 3 0 6; 2 3 10; 7 8 4];
for ii = 1:length(A(1,:))
eval(['Line' num2str(ii) 'Point1 = A(1:2, ii)' ]);
eval(['Line' num2str(ii) 'Point2 = A(3:4, ii)' ]);
end
% Now all variables are created - for example: Line2Point1
A more elegant solution could be to store the vectors in a cell array. You can acces the first vectors for example by typing: c{1,1}
c = cell(length(A(1,:)),2)
for ii = 1:length(A(1,:))
c{ii,1} = A(1:2, ii);
c{ii,2} = A(3:4, ii);
end
I would suggest using 3D arrays to store and then access those values.
Code
N = size(A,1)/2;
LinePoint = permute(reshape(A,N,size(A,1)/N,[]),[1 3 2])
Here,
2nd dimension indices (columns) would represent Line IDs
3rd dimension indices would represent Point IDs.
Thus, the representative 3D array would be - LinePoint(:,LineID,PointID).
Example run
For your given A, we would have LinePoint as -
LinePoint(:,:,1) =
1 5 9
3 0 6
LinePoint(:,:,2) =
2 3 10
7 8 4
Thus,
Line1Point1 would be denoted by LinePoint(:,1,1)
Line1Point2 would be denoted by LinePoint(:,1,2)
Line2Point1 would be denoted by LinePoint(:,2,1)
Line2Point2 would be denoted by LinePoint(:,2,2)
Line3Point1 would be denoted by LinePoint(:,3,1)
Line3Point2 would be denoted by LinePoint(:,3,2)
just lets make it simple, assume that I have a 10x3 matrix in matlab. The numbers in the first two columns in each row represent the x and y (position) and the number in 3rd columns show the corresponding value. For instance, [1 4 12] shows that the value of function in x=1 and y=4 is equal to 12. I also have same x, and y in different rows, and I want to average the values with same x,y. and replace all of them with averaged one.
For example :
A = [1 4 12
1 4 14
1 4 10
1 5 5
1 5 7];
I want to have
B = [1 4 12
1 5 6]
I really appreciate your help
Thanks
Ali
Like this?
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7];
[x,y] = consolidator(A(:,1:2),A(:,3),#mean);
B = [x,y]
B =
1 4 12
1 5 6
Consolidator is on the File Exchange.
Using built-in functions:
sparsemean = accumarray(A(:,1:2), A(:,3).', [], #mean, 0, true);
[i,j,v] = find(sparsemean);
B = [i.' j.' v.'];
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7]; %your example data
B = unique(A(:, 1:2), 'rows'); %find the unique xy pairs
C = nan(length(B), 1);
% calculate means
for ii = 1:length(B)
C(ii) = mean(A(A(:, 1) == B(ii, 1) & A(:, 2) == B(ii, 2), 3));
end
C =
12
6
The step inside the for loop uses logical indexing to find the mean of rows that match the current xy pair in the loop.
Use unique to get the unique rows and use the returned indexing array to find the ones that should be averaged and ask accumarray to do the averaging part:
[C,~,J]=unique(A(:,1:2), 'rows');
B=[C, accumarray(J,A(:,3),[],#mean)];
For your example
>> [C,~,J]=unique(A(:,1:2), 'rows')
C =
1 4
1 5
J =
1
1
1
2
2
C contains the unique rows and J shows which rows in the original matrix correspond to the rows in C then
>> accumarray(J,A(:,3),[],#mean)
ans =
12
6
returns the desired averages and
>> B=[C, accumarray(J,A(:,3),[],#mean)]
B =
1 4 12
1 5 6
is the answer.