I want to compare every row of a matrix with its every other row, element by element wise, using MATLAB. If two of the entries match, the result will be stored as 1, and if they don't match, it will be 0. This will give a symmetric matrix consisting of 0s and 1s.
For example, let A = [4 6 7 9 5; 2 6 9 9 1]
Then, the result expected is [1 1 1 1 1; 0 1 0 1 0; 0 1 0 1 0; 1 1 1 1 1]
The code I am using is (for a 1000*1000 random matrix):
A = randi(50,1000,1000);
B = zeros(1000000,1000);
D = zeros(1000000,1);
c=0;
for i=1:1000
for k=1:1000
for j=1:1000
if A(i,j)==A(k,j)
B(k+c,j)=1;
else
B(k+c,j)=0;
end
end
end
c=c+1000;
end
for l=1:1000000
D(l)=0;
for m=1:1000
D(l)=D(l)+(B(l,m)/(1000));
end
end
E=reshape(D,1000,1000);
This goes out of memory. Could anyone please suggest a solution or a more efficient code?
you can try row by row comparison directly as taking a complete row array and comparing with the other row array.
For example,
let
A = [4 6 7 9 5; 2 6 9 9 1];
nA = length(A(:,1));
finalMat = [];
for i = 1:nA
matRow = ones(nA,1)*A(i,:); % create a matrix size of A consists of same row elements
finalMat = [finalMat;matRow == A];
end
see if it is okay for you application.
You can use permute to align dimensions apprpriately and then bsxfun for the comparisons:
reshape(bsxfun(#eq, permute(A, [1 3 2]), permute(A, [3 1 2])), [], size(A,2))
Related
I have a vector M containing single elements and repeats. I want to delete all the single elements. Turning something like [1 1 2 3 4 5 4 4 5] to [1 1 4 5 4 4 5].
I thought I'd try to get the count of each element then use the index to delete what I don't need, something like this:
uniq = unique(M);
list = [uniq histc(M,uniq)];
Though I'm stuck here and not sure how to go forward. Can anyone help?
Here is a solution using unique, histcounts and ismember:
tmp=unique(M) ; %finding unique elements of M
%Now keeping only those elements in tmp which appear only once in M
tmp = tmp(histcounts(M,[tmp tmp(end)])==1); %Thanks to rahnema for his insight on this
[~,ind] = ismember(tmp,M); %finding the indexes of these elements in M
M(ind)=[];
histcounts was introduced in R2014b. For earlier versions, hist can be used by replacing that line with this:
tmp=tmp(hist(M,tmp)==1);
You can get the result with the following code:
A = [a.', ones(length(a),1)];
[C,~,ic] = unique(A(:,1));
result = [C, accumarray(ic,A(:,2))];
a = A(~ismember(A(:,1),result(result(:,2) == 1))).';
The idea is, add ones to the second column of a', then accumarray base on the first column (elements of a). After that, found the elements in first column which have accum sum in the second column. Therefore, these elements repeated once in a. Finally, removing them from the first column of A.
Here is a cheaper alternative:
[s ii] = sort(a);
x = [false s(2:end)==s(1:end-1)]
y = [x(2:end)|x(1:end-1) x(end)]
z(ii) = y;
result = a(z);
Assuming the input is
a =
1 1 8 8 3 1 4 5 4 6 4 5
we sort the list s and get index of the sorted list ii
s=
1 1 1 3 4 4 4 5 5 6 8 8
we can find index of repeated elements and for it we check if an element is equal to the previous element
x =
0 1 1 0 0 1 1 0 1 0 0 1
however in x the first elements of each block is omitted to find it we can apply [or] between each element with the previous element
y =
1 1 1 0 1 1 1 1 1 0 1 1
we now have sorted logical index of repeated elements. It should be reordered to its original order. For it we use index of sorted elements ii :
z =
1 1 1 1 0 1 1 1 1 0 1 1
finally use z to extract only the repeated elements.
result =
1 1 8 8 1 4 5 4 4 5
Here is a result of a test in Octave* for the following input:
a = randi([1 100000],1,10000000);
-------HIST--------
Elapsed time is 5.38654 seconds.
----ACCUMARRAY------
Elapsed time is 2.62602 seconds.
-------SORT--------
Elapsed time is 1.83391 seconds.
-------LOOP--------
Doesn't complete in 15 seconds.
*Since in Octave histcounts hasn't been implemented so instead of histcounts I used hist.
You can test it Online
X = [1 1 2 3 4 5 4 4 5];
Y = X;
A = unique(X);
for i = 1:length(A)
idx = find(X==A(i));
if length(idx) == 1
Y(idx) = NaN;
end
end
Y(isnan(Y)) = [];
Then, Y would be [1 1 4 5 4 4 5]. It detects all single elements, and makes them as NaN, and then remove all NaN elements from the vector.
I have variable matrix :
A = [1 2 8 8 1
4 6 8 1 1
5 3 1 1 8];
and I have variable B :
B=[2 3 1 8 8];
Question is how to find rows and columns (sort by rows) in variable A from variable B.
Example, first index in variable B is 2, and then I want to find value 2 in variable A and get to first rows and columns, and next process until index 5, but if rows and columns has been used so get second position (ex. index 4 & 5 having same value).
rows;
columns;
Result is:
rows = 1 3 1 1 1
columns = 2 2 1 3 4
Use can use find and sub2ind to achieve what you want
but for that you have to take transpose of your A first
A = [1 2 8 8 1
4 6 8 1 1
5 3 1 1 8];
B= [2 3 1 8 8];
TMP = A.';
for i = 1:length(B)
indx = find(TMP== B(i),1,'first') %Finding the element of B present in A
if(~isempty(indx )) % If B(i) is a member of A
[column(i),row(i)] = ind2sub(size(TMP),indx) % store it in row and column matrix
TMP(indx) = nan; % remove that element
end
end
column =
2 2 1 3 4
row =
1 3 1 1 1
As in one of the comments Usama suggested preallocation of memory
you can do that by using
row = zeros(1,sum(ismember(B,A)))
column= zeros(1,sum(ismember(B,A)))
The above code works even if there are some members of B not present in A
Use find. The function could return both a linear index or a row/col index.
Using linear index a solution could be
idx = zeros(size(B));
for i = 1:numel(B)
% Find all indexes
tmpIdx = find(A == B(i));
% Remove those already used
tmpIdx = setdiff(tmpIdx, idx);
% Get the first new unique
idx(i) = tmpIdx(1);
end
% Convert index to row and col
[rows, cols] = ind2sub(size(A),idx)
Giving:
rows = 1 3 1 1 2
cols = 2 2 1 3 3
Note that as the linear indexing goes down column by column, the result here differs from the one in your example (although still a correct index)
rows = 1 3 1 1 1
columns= 2 2 1 3 4
But to get this you could just transpose the A matrix (A.') and flip the rows and cols (the result from ind2sub)
Here is on solution where I use for loop, I tried to optimize the number of iteration and the computational cost. If there is no corresponding value between B and A the row/col index return NaN.
[Bu,~,ord] = unique(B,'stable');
% Index of each different values
[col,row] = arrayfun(#(x) find(A'==x),Bu,'UniformOutput',0)
% For each value in vector B we search the first "non already used" corresponding value in A.
for i = 1:length(B)
if ~isempty(row{ord(i)})
r(i) = row{ord(i)}(1);
row{ord(i)}(1) = [];
c(i) = col{ord(i)}(1);
col{ord(i)}(1) = [];
else
r(i) = NaN;
c(i) = NaN;
end
end
RESULT:
c = [2 2 1 3 4]
r = [1 3 1 1 1]
I've a matrix A and I'd like to find elements of first row which have 1 in the second row. i.e. for following matrix
A=
2 5 6 1
1 0 0 1
I'd like to have output as hits = [2 1] without using loops. and then finds the maximum items in the answer. i.e. (2>1) so my final answer is 2. The response is probably using arrayfun but I've problems and get errors using it. What is the correct syntax?
Thanks
Try this:
out = max(A(1,A(2,:) == 1))
Example:
>> A
A =
2 5 6 1
1 0 0 1
>> out
out =
2
Explanation: (if you need)
%// create a mask of which column you want
mask = A(2,:) == 1 %// by checking all values of 2nd row with 1
%// get only the values of row one, meeting 'the' condition
hits = A(1,mask)
%// Find the maximum from that
maxHits = max(hits)
For Cell Array using cellfun
A = {[2 5 6 1; 1 0 0 1], [2 3 2 5 4; 1 1 3 1 2]} %// eg input
A =
[2x4 double] [2x5 double]
out = cellfun(#(x) max(x(1,x(2,:) == 1)),A)
out =
2 5
I have one matrix like below-
A=[1 1 1 1 1;
0 1 1 1 2;
0 0 1 1 3]
But I want to place all the 0 at the end of the row, so A should be like-
A=[1 1 1 1 1;
1 1 1 2 0;
1 1 3 0 0]
How can I do this? Matlab experts please help me.
There you go. Whole matrix, no loops, works even for non-contiguous zeros:
A = [1 1 1 1 1; 0 1 1 1 2; 0 0 1 1 3];
At = A.'; %// It's easier to work with the transpose
[~, rows] = sort(At~=0,'descend'); %// This is the important part.
%// It sends the zeros to the end of each column
cols = repmat(1:size(At,2),size(At,1),1);
ind = sub2ind(size(At),rows(:),cols(:));
sol = repmat(NaN,size(At,1),size(At,2));
sol(:) = At(ind);
sol = sol.'; %'// undo transpose
As usual, for Matlab versions that do not support the ~ symbol on function return, change ~ by a dummy variable, for example:
[nada, rows] = sort(At~=0,'descend'); %// This is the important part.
A more generic example:
A = [1 3 0 1 1;
0 1 1 1 2;
0 0 1 1 3]
% Sort columns directly
[~,srtcol] = sort(A == 0,2);
% Sorted positions
sz = size(A);
pos = bsxfun(#plus, (srtcol-1)*sz(1), (1:sz(1))'); % or use sub2ind
The result
B = A(pos)
B =
1 3 1 1 0
1 1 1 2 0
1 1 3 0 0
there are many ways to do this. one fast way can be easily like this:
a = [1 2 3 4 0 5 7 0];
idx=(find(a==0));
idx =
5 8
b=a; % save a new copy of the vector
b(idx)=[]; % remove zero elements
b =
1 2 3 4 5 7
c=[b zeros(size(idx))]
c =
1 2 3 4 5 7 0 0
You may modify this code as well.
If your zeros are always together, you could use the circshift command. This shifts values in an array by a specified number of places, and wraps values that run off the edge over to the other side. It looks like you would need to do this separately for each row in A, so in your example above you could try:
A(2,:) = circshift(A(2,:), [1 -1]); % shift the second row one to the left with wrapping
A(3,:) = circshift(A(3,:), [1 -2]); % shift the third row two to the left with wrapping
In general, if your zeros are always at the front of the row in A, you could try something like:
for ii = 1:size(A,1) % iterate over rows in A
numShift = numel(find(A(ii,:) == 0)); % assuming zeros at the front of the row, this is how many times we have to shift the row.
A(ii,:) = circshift(A(ii,:), [1 -numShift]); % shift it
end
Try this (just a fast hack):
for row_k = 1:size(A, 1)
[A_sorted, A_sortmap] = sort(A(row_k, :) == 0, 'ascend');
% update row in A:
A(row_k, :) = A(row_k, A_sortmap);
end
Now optimized for versions of MATLAB not supporting ~ as garbage lhs-identifier.
#LuisMendo's answer is inspiring in its elegance, but I couldn't get it to work (perhaps a matlab version thing). The following (based on his answer) worked for me:
Aaux = fliplr(reshape([1:numel(A)],size(A)));
Aaux(find(A==0))=0;
[Asort iso]=sort(Aaux.',1,'descend');
iso = iso + repmat([0:size(A,1)-1]*size(A,2),size(A,2),1);
A=A.';
A(iso).'
I've also asked this question and got a super elegant answer (non of above answers is same) here:
Optimize deleting matrix leading zeros in MATLAB
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.