Let's say I've got a vector a = [1 2 4]. I want it converted into a vector which looks like this b = [1 2 0 4], i.e. each number is placed into a correct position and since 3 is not included into the vector a, it is replaced by 0 in vector b. This can be done the following way:
a = [1 2 4]
b = zeros(1, size(a, 2));
b(1, a) = a;
I can't figure out a way to do the same for a matrix. For example,
c = [1 4 2 0; 3 1 0 0; 4 0 0 0; 1 3 4 0];
I need to convert into a matrix that looks like this:
d = [1 2 0 4; 1 0 3 0; 0 0 0 4; 1 0 3 4];
Any tips? How can this be done? How can I do this without using loops?
Here's a vectorized solution:
a = [1 4 2 0; 3 1 0 0; 4 0 0 0; 1 3 4 0];
b = zeros(size(a,1),max(a(:)));
[rowIdx,~] = find(a);
vals = a(a>0);
b( sub2ind(size(b),rowIdx,vals) ) = vals;
Does this work? (Edited: fixed mistake.)
[m,n] = size(c)
d = zeros(m,n)
for i=1:m
d(i,c(i,c(i,:)>0)) = c(i,c(i,:)>0)
end
Related
I trying to make a program that uses the for loop to calculate the set of index D. But I have a problem because the length of index are not the same.
Example:
z = [0 0 0 0 0 0 1]
v(1,:) = [1 0 0 0 1 0 1]
v(2,:) = [0 1 0 0 1 1 1]
v(3,:) = [0 0 1 0 1 1 0]
v(4,:) = [0 0 0 1 0 1 1]
v(1,:) = find(v(1,:)~=z);
v(2,:) = find(v(2,:)~=z);
v(3,:) = find(v(3,:)~=z);
v(4,:) = find(v(4,:)~=z);
we obtain :
D(1,:) = [1 5];
D(2,:) = [2 5 6];
D(3,:) = [3 5 6 7];
D(4,:) = [4 6];
Code :
for aa = 1:4
D(aa,:) = [find(v(aa,:)~=z)];
end
not work because length(D(1,:))~=length(D(2,:))~=length(D(3,:))
How I can use a loop to determine set of index D?
Thank you for any help!
One solution can be using cell like the following:
for aa = 1:4
D{aa} = [find(v(aa,:)~=z)];
end
You can use the matrix D, but initialize it beforehand like:
D = ones(size(v)) + length(z)
Then fill it like:
for ii = 1:size(z,1)
D(ii,v(ii,:)~=z) = find(v(ii,:)~=z);
end
Notice, I added the length of v to the matrix of ones, such that you are sure that the predefined numbers in the matrix are larger than any index, hence the min() will not freak out.
if i have a matrix, say:
A = [ 0 2 4 0
2 0 5 0
4 5 0 3
0 0 3 0 ]
and i want to find the maximum value in the matrix i can type:
max(max(A))
or
max(A(:))
if i only want to find the maximum of rows 1 and 2 and columns 3 and 4 i can do this:
a = [1 2]
b = [3 4]
max(max(A(a,b))
but what if i want to find the indices of the rows and columns that correspond to that value?
according to the matlab documentation, if i am using the whole matrix i can use the ind2sub function:
[val,idx] = max(A(:))
[row,col] = ind2sub(size(A),idx)
but how can i get that working for my example where i am using vectors a and b to determine the rows and columns it finds the values over?
here is the only way i have been able to work it out so far:
max_val = 0;
max_idx = [1 1];
for ii = a
[val,idx] = max(A(ii,b))
if val > max_val
max_val = val
max_idx = [ii idx]
but that seems rather clunky to me.. any ideas?
Assuming that the submatrix A(a,b) is contiguous (like in your example):
A = [ 0 2 4 0
2 0 5 0
4 5 0 3
0 0 3 0 ]
a = [1 2]; b = [3 4];
B = A(a,b)
[val,idx] = max(B(:));
[row,col] = ind2sub(size(B),idx);
maxrow = row + a(1) - 1;
maxcol = col + b(1) - 1;
You are finding the relative index in the submatrix B. Which is equivalent to the additional rows and columns from the upper left corner of the submatrix.
Now assuming that a and b result in a set of rows and columns that are NOT a contiguous submatrix, e.g. a = [1 3], b = [3 4], the result is very similar. "row" and "col" are the index in the a and b vectors:
A = [ 0 2 4 0
2 0 5 0
4 5 0 3
0 0 3 0 ]
a = [1 3]; b = [3 4];
B = A(a,b)
[val,idx] = max(B(:));
[row,col] = ind2sub(size(B),idx);
maxrow = a(row);
maxcol = b(col);
Now you're working in the index of indexes.
Given this vector
a = [1 2 3 4]
I want to create a matrix like this
b = [1 0 0 0;
2 1 0 0;
3 2 1 0;
4 3 2 1;
0 4 3 2;
0 0 4 3;
0 0 0 4]
in a vectorized way not using loops.
Hint: use conv2 (hover mouse to see code):
a = [1 2 3 4];
b = conv2(a(:), eye(numel(a)));
Or, in a similar mood, you can use convmtx (from the Signal Processing Toolbox):
a = [1 2 3 4];
b = convmtx(a(:), numel(a));
One way to do it:
a = [1 2 3 4]
n = numel(a);
%// create circulant matrix from input vector
b = gallery('circul',[a zeros(1,n-1)]).' %'
%// crop the result
c = b(:,1:n)
Another way:
b = union( tril(toeplitz(a)), triu(toeplitz(fliplr(a))),'rows','stable')
or its slightly variation
b = union( toeplitz(a,a.*0),toeplitz(fliplr(a),a.*0).','rows','stable')
and probably even faster:
b = [ toeplitz(a,a.*0) ; toeplitz(fliplr(a),a.*0).' ]
b(numel(a),:) = []
With bsxfun -
na = numel(a)
b = zeros(2*na-1,na)
b(bsxfun(#plus,[1:na]',[0:na-1]*2*na)) = repmat(a(:),1,na)
If you are looking for a faster pre-allocation, you can do -
b(2*na-1,na) = 0;.
Another bsxfun -
a=[1 2 3 4];
m=numel(a);
b=[a,zeros(1,m-1)].';
Q=bsxfun(#circshift, b, [0:m-1])
0I have on matrix-
A=[1 2 2 3 5 5;
1 5 5 8 8 7;
2 9 9 3 3 5];
From matrix i need to count now many nonzero elements ,how any 1,how many 2 and how many 3 in each row of given matrix"A".For these i have written one code like:
[Ar Ac]=size(A);
for j=1:Ar
for k=1:Ac
count(:,j)=nnz(A(j,:));
d(:,j)=sum(A(j,:)== 1);
e(:,j)=sum(A(j,:)==2);
f(:,j)=sum(A(j,:)==3);
end
end
but i need to write these using on loop i.e. here i manually use sum(A(j,:)== 1),sum(A(j,:)== 2) and sum(A(j,:)== 3) but is there any option where i can only write sum(A(j,:)== 1:3) and store all the values in the different row i.e, the result will be like-
b=[1 2 1;
1 0 0;
0 1 2];
Matlab experts need your valuable suggestions
Sounds like you're looking for a histogram count:
U = unique(A);
counts = histc(A', U)';
b = counts(:, ismember(U, [1 2 3]));
Example
%// Input matrix and vector of values to count
A = [1 2 2 3 5 5; 1 5 5 8 8 7; 2 9 9 3 3 5];
vals = [1 2 3];
%// Count values
U = unique(A);
counts = histc(A', U)';
b = counts(:, ismember(U, vals));
The result is:
b =
1 2 1
1 0 0
0 1 2
Generalizing the sought values, as required by asker:
values = [ 1 2 3 ]; % or whichever values are sought
B = squeeze(sum(bsxfun(#(x,y) sum(x==y,2), A, shiftdim(values,-1)),2));
Here is a simple and general way. Just change n to however high you want to count. n=max(A(:)) is probably a good general value.
result = [];
n = 3;
for col= 1:n
result = [result, sum(A==col, 2)];
end
result
e.g. for n = 10
result =
1 2 1 0 2 0 0 0 0 0
1 0 0 0 2 0 1 2 0 0
0 1 2 0 1 0 0 0 2 0
Why not use this?
B=[];
for x=1:size(A,1)
B=[B;sum(A(x,:)==1),sum(A(x,:)==2),sum(A(x,:)==3)];
end
I'd do this way:
B = [arrayfun(#(i) find(A(i,:) == 1) , 1:3 , 'UniformOutput', false)',arrayfun(#(i) find(A(i,:) == 2) , 1:3 , 'UniformOutput', false)',arrayfun(#(i) find(A(i,:) == 3) , 1:3 , 'UniformOutput', false)'];
res = cellfun(#numel, B);
Here is a compact one:
sum(bsxfun(#eq, permute(A, [1 3 2]), 1:3),3)
You can replace 1:3 with any array.
you can make an anonymous function for it
rowcnt = #(M, R) sum(bsxfun(#eq, permute(M, [1 3 2]), R),3);
then running it on your data returns
>> rowcnt(A,1:3)
ans =
1 2 1
1 0 0
0 1 2
and for more generalized case
>> rowcnt(A,[1 2 5 8])
ans =
1 2 2 0
1 0 2 2
0 1 1 0
Suppose I have a matrix A, and I'd like to get the matrix [A 0; 0 1]. Is there a built function to do this?
So if my matrix is [2 3; 1 4], I'd get back [2 3 0; 1 4 0; 0 0 1]
The easiest way is:
newA = A;
newA(end+1,end+1) = 1;
This works because you can index outside an array for assignments, because end indicates the last element (here in row and column), and because Matlab pads with zeros when you grow an array. If you just want to grow A, you can even skip the creation of newA, of course.
I always use matrix concatenation for problems like this
So for your example:
A = [2 3; 1 4]
A = [A A(:,1)*0; A(1,:)*0 1]
produces
A =
2 3 0
1 4 0
0 0 1
The nice thing about this trick is that its very flexible and you can do all sorts of tranformations
very easily. For example
A = [2 3; 1 4]
A = [1 A(1,:)*0; A(:,1)*0 A]
produces
A =
1 0 0
0 2 3
0 1 4