I have 2 matrices:
T3(:,:,1) =
0 0 0 0 1 0 0 0 0
0 0 0 0 2 0 0 0 0
0 0 0 0 3 0 0 0 0
0 1 0 1 4 2 0 4 0
0 3 0 2 6 3 0 5 0
2 4 2 5 7 5 4 6 5
4 5 5 7 8 8 5 7 6
5 6 6 8 9 9 8 9 8
T3(:,:,2) =
2 1 1 1 1 1 1 1 1
3 3 2 2 2 2 2 2 2
4 4 4 3 3 3 3 3 3
5 5 5 5 4 4 4 4 4
6 6 6 6 6 5 5 5 5
7 7 7 7 7 7 6 6 6
8 8 8 8 8 8 8 7 7
9 9 9 9 9 9 9 9 8
How do I make values present in T3(:,:,1) turn to zero in T3(:,:,2)?
e.g. in the first column of T3(:,:,1) the values are 2,4,5. I'd like the first column of T3(:,:,2) to have the the values 2,4,5 as zero.
T3(:,:,2) =
0 0 1 0 0 1 1 1 1
3 0 0 0 0 0 2 2 2
0 0 4 3 0 0 3 3 3
0 0 0 0 0 4 0 0 4
6 0 0 6 0 0 0 0 0
7 7 7 0 0 7 6 0 0
8 8 8 0 0 0 0 0 7
9 9 9 9 0 0 9 0 0
I wonder if there is a way to do this using setdiff or unique.
for y=1:H-1
for z=1:H-1
for h=1:H
for d=1:D-1
if T3(y,h,d+1) == T3(z,h,d)
T3(y,h,d+1)=0;
end
end
end
end
end
I can do it as a loop where H=number of columns (9) and D= number of dimensions (2). There must be a better way :)?
Many thanks guys.
Related
I have a vector ("d") which I want to place it's content to a matrix ("dis"), I have the problem to create "dis" matrix. "dis" should be like
dis=[
1 2 3 4 5 6 7 8 9 10 11 12
0 1 2 3 4 5 6 7 8 9 10 11
0 0 1 2 3 4 5 6 7 8 9 10
0 0 0 1 2 3 4 5 6 7 8 9
0 0 0 0 1 2 3 4 5 6 7 8
0 0 0 0 0 1 2 3 4 5 6 7
0 0 0 0 0 0 1 2 3 4 5 6
0 0 0 0 0 0 0 1 2 3 4 5
0 0 0 0 0 0 0 0 1 2 3 4
0 0 0 0 0 0 0 0 0 1 2 3
0 0 0 0 0 0 0 0 0 0 1 2
0 0 0 0 0 0 0 0 0 0 0 1];
n=[0,0;1,0;2,0;3,0;4,0;5,0;6,0;7,0;8,0;9,0;10,0;11,0;12,0];
d=pdist(n,'euclidean');
l=length(n)-1;
dis=[];
for k=1:length(n)-1
dis=[dis;d((k-1)*l-(k*((k-1)/2))+k):d(k*l-((k+1)*k/2)+k)];
end
The problem is that you are not padding dis with the zeros.
You can replace
dis=[];
for k=1:length(n)-1
dis=[dis;d((k-1)*l-(k*((k-1)/2))+k):d(k*l-((k+1)*k/2)+k)];
end
with
dis = zeros(l);
for k=1:l
dis(k,k:end) = d((k-1)*l-(k*((k-1)/2))+k):d(k*l-((k+1)*k/2)+k)
end
Using squareform form #beaker comment above, you can write:
dis = triu(squareform(d));
dis = dis(1:length(dis)-1,2:length(dis));
Does this solves the problem?
Assuming I'm having a vectors of numbers A, for example: A=[1 3 5 3 9 6](A's length >= 2) and an Integer X=6. Need to find how many pairs (A[i],A[j]) where i<j exist in the vector which answer this condition: A[i]+A[j]=X. The number of pairs is printed.
Not allowed to use for/while. Allowed only ceil,floor,mod,repmat,reshape,size,length,transpose,sort,isempty,all,any,find ,sum,max,min.
With repmat, length and sum -
integer1 = 6; %// One of the paramters
A_ind = 1:length(A) %// Get the indices array
%// Expand A_ind into rows and A_ind' into columns, to form a meshgrid structure
A_ind_mat1 = repmat(A_ind,[length(A) 1])
A_ind_mat2 = repmat(A_ind',[1 length(A)]) %//'
%// Expand A into rows and A' into columns, to form a meshgrid structure
A_mat1 = repmat(A,[length(A) 1])
A_mat2 = repmat(A',[1 length(A)]) %//'
%// Form the binary matrix of -> (A[i],A[j]) where i<j
cond1 = A_ind_mat1 < A_ind_mat2
%// Use the binary matrix as a logical mask to select elements from the two
%// matrices and see which element pairs satisfy -> A[i]+A[j]=X and get a
%// count of those pairs with SUM
pairs_count = sum((A_mat1(cond1) + A_mat2(cond1))==integer1)
Outputs from code run to make it clearer -
A =
1 3 5 3 9 6
A_ind =
1 2 3 4 5 6
A_ind_mat1 =
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
A_ind_mat2 =
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
A_mat1 =
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
A_mat2 =
1 1 1 1 1 1
3 3 3 3 3 3
5 5 5 5 5 5
3 3 3 3 3 3
9 9 9 9 9 9
6 6 6 6 6 6
cond1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 1 1 0 0
1 1 1 1 1 0
pairs_count =
2
A bit more explanation -
Taking few more steps to clarify why pairs_count must be 2 here -
Set all values in A_mat1 and A_mat2 to be zeros that do not satisfy the less than criteria
>> A_mat1(~cond1)=0
A_mat1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 3 0 0 0 0
1 3 5 0 0 0
1 3 5 3 0 0
1 3 5 3 9 0
>> A_mat2(~cond1)=0
A_mat2 =
0 0 0 0 0 0
3 0 0 0 0 0
5 5 0 0 0 0
3 3 3 0 0 0
9 9 9 9 0 0
6 6 6 6 6 0
Now, add A_mat1 and A_mat2 and see how many 6's you got -
>> A_mat1 + A_mat2
ans =
0 0 0 0 0 0
4 0 0 0 0 0
6 8 0 0 0 0
4 6 8 0 0 0
10 12 14 12 0 0
7 9 11 9 15 0
As you can see there are two 6's and thus our result is verified.
Is there a way to find the maximum flow between each pair of vertices in matlab?
c = sparse([1 1 2 2 3 4 4 5 5 6 7 8 9 9],[2 3 3 4 5 6 7 6 7 8 9 10 8 10],[15 10 3 8 9 7 5 6 2 12 10 6 10 8],10,10)
a = [2 3 4 5 6 7 8 9 10]
b = arrayfun(#(x)max_flow(c,1,x),a)
OR
b = arrayfun(#(x)graphmaxflow(c,1,x),a)
b =
15 13 8 9 13 7 16 7 13
So, I can take a sparse matrix and get the maximum flow from one vertex to all others. Is there a way to continue this to obtain the max flow for all of the pairs?
I'd eventually like to be able to find the all-pair max flow for a directed, weighted graph. . .
Got it to work:
c = sparse([1 1 2 2 3 4 4 5 5 6 7 8 9 9],[2 3 3 4 5 6 7 6 7 8 9 10 8 10],[15 10 3 8 9 7 5 6 2 12 10 6 10 8],10,10)
for a=1:10
for b=1:10
if a==b
continue
end
t(b,a)=graphmaxflow(c,a,b);
p=t(:);
end
end
I couldn't figure out a way to use arrayfun to do this.
Each maximum flow value:
t =
0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0
13 3 0 0 0 0 0 0 0 0
8 8 0 0 0 0 0 0 0 0
9 3 9 0 0 0 0 0 0 0
13 10 6 7 6 0 0 0 0 0
7 7 2 5 2 0 0 0 0 0
16 11 8 12 8 12 10 0 10 0
7 7 2 5 2 0 10 0 0 0
13 11 8 11 8 6 10 6 14 0
p =
0
15
13
8
9
13
7
...
I have matrix (a) with (1:10),<10 x 1> double. I would like to copy the values and rearrange them column wise into another matrix var. (b). See example below. Also, what method would be most efficient at this task?
matrix a matrix b
1 1
2 2 2
3 3 3 3
4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6 6
7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9
10 10 10 10 10 10 10 10 10 10 10
update:
Hi once again Amro. How about if I wanted to define which values to copy. See below example:
matrix a matrix b
column: 1 2 3 4 5 6 7
1 1
2 2 2
3 3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10 10
Try:
>> a = (1:10)'
a =
1
2
3
4
5
6
7
8
9
10
>> b = tril(repmat(a,1,10))
b =
1 0 0 0 0 0 0 0 0 0
2 2 0 0 0 0 0 0 0 0
3 3 3 0 0 0 0 0 0 0
4 4 4 4 0 0 0 0 0 0
5 5 5 5 5 0 0 0 0 0
6 6 6 6 6 6 0 0 0 0
7 7 7 7 7 7 7 0 0 0
8 8 8 8 8 8 8 8 0 0
9 9 9 9 9 9 9 9 9 0
10 10 10 10 10 10 10 10 10 10
I think in the second matrix you specified you made an error. I'm assuming you wanted to do something like this:
b =
1 0 0 0 0 0
2 2 0 0 0 0
0 3 3 0 0 0
0 0 4 4 0 0
0 0 0 5 5 0
0 0 0 0 6 6
this is simple to do:
%define vector of arbitrary length
a=1:6;
%generate b with shifted diagonal matrices
b=diag(a)+diag(a(2:end),-1);
the second argument of diag just shifts the resulting diagonal.
How can I create the following matrix?
[0 0 0
8 0 0
9 8 0
6 9 8
5 6 9
4 5 6]
How about the toeplitz function?
c=[0 8 9 6 5 4 3]
r=[0 0 0]
t=toeplitz(c,r)
(Disclaimer: untested!)
T should be:
0 0 0
8 0 0
9 8 0
6 9 8
5 6 9
4 5 6
3 4 5