this is my matrix that displays a sample network graph
matrix =
0 1 1 1
1 0 1 0
0 0 0 1
1 1 1 0
where its a 4x4 matrix
1) 2) 3) 4)
1) 0 1 1 1
2) 1 0 1 0
3) 0 0 0 1
4) 1 1 1 0
i want to count this 4x4 matrix like
row 1 counts how many 1s i have and adds column 1 number of 1's to it and returns 1)=5 as total 1's in row 1 and col 1 = 5
i want my output to be like
1=5
2=4
3=4
4=5
This must be it -
out = sum([matrix matrix'],2)
Example run -
matrix =
1 1 1 1
1 0 0 0
0 1 0 1
0 0 1 1
out =
6
3
4
5
The above code would count 1s twice when they appear on the diagonal, which if you don't want, use this -
out1 = sum([matrix matrix'],2) - diag(matrix)
Example run -
matrix =
1 1 1 1
1 0 0 0
0 1 0 1
0 0 1 1
out1 =
5
3
4
4
I agree with the answer of Divakar, but once your graph gets larger and larger, you might not want to transpose the entire matrix. I suggest doing the sum first and then transposing afterwards:
sum(matrix,1)'+sum(matrix,2)-diag(matrix);
matrix =
0 1 1 1
1 0 1 0
0 0 0 1
1 1 1 0
degree=sum(matrix,1)'+sum(matrix,2)-diag(matrix)
degree =
5
4
4
5
Related
I have got the following function for spreading out the number of 1's in a matrix and if there are rows with all 0's or all 1's then that particular row has to be deleted
function ReducedMatrix = ReduceMatrix(result)
D1 = sum(result(:));
NumberOfOnes = floor(D1*0.3);
NewMatrix = zeros(size(result));
NewMatrix(randi(numel(NewMatrix),1,NumberOfOnes)) = 1;
ReducedMatrix = NewMatrix;
while numel(ReducedMatrix)/numel(NewMatrix) > 0.2
IndexOfFullRows = find(all(ReducedMatrix));
if isempty(IndexOfFullRows)
break
end
ReducedMatrix(:,IndexOfFullRows(1)) = [];
end
end
The input of the function and output are as follows
result =
0 1 1 1 1 1 1 1 1 1
1 1 1 1 1 0 1 0 1 1
1 1 0 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 1 0 1 0 1
1 0 1 1 1 1 1 0 1 1
1 1 1 1 0 1 1 1 0 1
1 0 1 1 1 0 1 1 1 1
1 1 1 1 0 1 0 1 1 1
1 1 1 0 1 1 1 1 1 1
ReducedMatrix =
0 1 1 0 0 0 0 0 1 0
0 1 0 0 0 0 0 1 0 0
1 1 1 0 0 0 0 0 0 0
0 0 0 1 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0 0
0 1 0 0 0 0 1 0 1 1
1 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 1
0 0 0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 1
row_sum =
3
2
3
2
1
4
2
2
0
3
col_sum =
3 4 4 1 0 0 3 2 2 3
Now if there exists a row or column with the row_sum/col_sum equal to either 0 or 1 then then the corresponding row has to be deleted.
For Example. Row-R4,R9 and Col-C4,C5,C6 have row_sum and col_sum as either 1,0. So adding them up R4,R9,C4,C5,C6 = 5 rows have to be eliminated from the matrix so my reduced matrix should be of the size 5x5. Please note column should not be eliminated and instead of removing columns having 0 and 1, the corresponding rows can be removed. Similarly this function has to run for larger matrices with the same constraints. I tried doing the above function however i do not possess enough skills to achieve my desired results, Any help is much appreciated
I see a number of potential problems and possible simplifications to your code.
For one thing, the way you construct the original matrix, NewMatrix(randi(numel(NewMatrix),1,NumberOfOnes)) = 1; may not behave the way you would expect. randi does not guarantee that the same index will not appear multiple times in the output, so your new matrix may have fewer ones than the original. To solve this, shuffle the elements using randperm:
ReducedMatrix = [ones(1, NumberOfOnes), zeros(1, numel(result) - NumberOfOnes)];
ReducedMatrix = ReducedMatrix(randperm(numel(ReducedMatrix)));
ReducedMatrix = reshape(ReducedMatrix, size(result));
Secondly, you do not need to construct the new matrix as NewMatrix and then reassign it with ReducedMatrix = NewMatrix;. Just do ReducedMatrix = zeros(size(result)); and skip the reassignment. For the while loop condition, where NewMatrix appears to be "used", remember that numel(NewMatrix) == numel(result).
If you are not removing homogeneous columns, only rows, you do not need a loop to do the removal:
rowSum = sum(ReducedMatrix, 2);
rowMask = (rowSum == size(ReducedMatrix, 2) | rowSum == 0);
ReducedMatrix(rowMask, :) = [];
Your original code seems to swap the row and column indices when removing the rows. It also did not handle the case of all zeros. If you want to remove not more than 30% of rows, you can do something like this before the removal:
rowMask = find(rowMask); % Convert to indices
rowMask = rowMask(1:min(numel(rowMask), round(0.3 * size(ReducedMatrix, 2))));
I'm looking for a way to generate the spans of a given vector in MATLAB.
For example:
if a = [ 0 1 0 1] I need all vectors of the form [0 x 0 y], 1 <= x <= max1, 1 <= y <= max2,.
or if
a = [ 0 1 0 1 1 0] I need all vectors of the form [0 x 0 y z 0], 1 <= x <= max1, 1 <= y <= max2, 1<= z <= max3.
Note that the vector can have a variable number of 1's.
My first impression is that I would need a variable number of for loops, though I don't know if that is doable in MATLAB. Also any other ideas are welcome!
You don't need multiple for loops for this. The code below generates all required vectors as rows of a tall matrix. It actually creates the columns of the matrix one at a time. Each column will have numbers 1:m(i) arranged in the pattern where
each term repeats the number of times equal to the product of all m-numbers after m(i)
the whole pattern repeats the number of times equal to the product of all m-numbers before m(i)
This is what repmat(kron(1:m(i),ones(1,after)),1,before)' does. (Starting with R2015a you can use repelem to simplify this by replacing the kron command, but I don't have that release yet.)
a = [0 1 0 1 1 0];
m = [2 4 3]; // the numbers max1, max2, max3
A = zeros(prod(m), length(a));
i = 1; // runs through elements of m
for j=1:length(a) // runs through elements of a
if (a(j)>0)
before = prod(m(1:i-1));
after = prod(m(i+1:end));
A(:,j) = repmat(kron(1:m(i),ones(1,after)),1,before)';
i = i+1;
end
end
Output:
0 1 0 1 1 0
0 1 0 1 2 0
0 1 0 1 3 0
0 1 0 2 1 0
0 1 0 2 2 0
0 1 0 2 3 0
0 1 0 3 1 0
0 1 0 3 2 0
0 1 0 3 3 0
0 1 0 4 1 0
0 1 0 4 2 0
0 1 0 4 3 0
0 2 0 1 1 0
0 2 0 1 2 0
0 2 0 1 3 0
0 2 0 2 1 0
0 2 0 2 2 0
0 2 0 2 3 0
0 2 0 3 1 0
0 2 0 3 2 0
0 2 0 3 3 0
0 2 0 4 1 0
0 2 0 4 2 0
0 2 0 4 3 0
I am trying to create an 8x8 matrix containing 0s, 1s and 2s. Each row and each column should contain two 0s, three 1s and three 2s.
Previously I have used the below to generate an example containing only 1s and 0s.
output = zeros(8, 8);
for i=1:8
tmp = (1:8) + (i);
tmp = rem(tmp, 4);
output(i,:) = tmp;
output(i,:) = tmp > 0;
end
output =
1 1 0 1 1 1 0 1
1 0 1 1 1 0 1 1
0 1 1 1 0 1 1 1
1 1 1 0 1 1 1 0
1 1 0 1 1 1 0 1
1 0 1 1 1 0 1 1
0 1 1 1 0 1 1 1
1 1 1 0 1 1 1 0
However I would now like something similar to the following:
output =
1 1 0 1 2 2 0 2
1 0 1 2 2 0 2 1
0 1 2 2 0 2 1 1
1 2 2 0 2 1 1 0
2 2 0 2 1 1 0 1
2 0 2 1 1 0 1 2
0 2 1 1 0 1 2 2
2 1 1 0 1 2 2 0
Thanks for your help.
What you have in your example is a Hankel matrix so you could use the hankel function
c = [1 1 0 1 2 2 0 2];
k = [2 1 1 0 1 2 2 0];
A = hankel(c,k)
where c is the first column of the output matrix and k is the last row.
Making your output matrix a Hankel matrix is a good idea (based on your requirements) as it will enforce the row and column frequency counts for each value. You would not necessarily get this just by creating rows that are random permutations of a base row (using randperm for example) as duplicate rows would be possible which would break your column requirements.
As an example, if you want random c with fixed numbers of specific elements, you can randomly permute a base vector containing the required values and frequencies - as per your requirement this would be
c = [0 0 1 1 1 2 2 2];
index = randperm(numel(c));
c = c(index);
c =
0 2 0 2 2 1 1 1
To get the square Hankel structure then choose k to be the next cyclic permutation of c
k = circshift(c',1)'
k =
1 0 2 0 2 2 1 1
and just call hankel with these as mentioned above
A = hankel(c,k)
A =
0 2 0 2 2 1 1 1
2 0 2 2 1 1 1 0
0 2 2 1 1 1 0 2
2 2 1 1 1 0 2 0
2 1 1 1 0 2 0 2
1 1 1 0 2 0 2 2
1 1 0 2 0 2 2 1
1 0 2 0 2 2 1 1
The above output is based on what I got on my machine based on the output from randperm.
Any output matrix generated using the above will meet your requirements specified in the question.
I have a column vector x made up of 4 elements, how can i generate all the possible combinations of the values that x can take such that x*x' is less than or equal to a certain value?
note that the values of x are positive and integers.
To be more clear:
the input is the number of elements of the column vector x and the threshold, the output are the different possible combinations of the values of x respecting the fact that x*x' <=threshold
Example: threshold is 4 and x is a 4*1 column vector.....the output is x=[0 0 0 0].[0 0 0 1],[1 1 1 1]......
See if this works for you -
threshold = 4;
A = 0:threshold
A1 = allcomb(A,A,A,A)
%// Or use: A1 = combvec(A,A,A,A).' from Neural Network Toolbox
combs = A1(sum(A1.^2,2)<=threshold,:)
Please note that the code listed above uses allcomb from MATLAB File-exchange.
Output -
combs =
0 0 0 0
0 0 0 1
0 0 0 2
0 0 1 0
0 0 1 1
0 0 2 0
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
0 2 0 0
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
2 0 0 0
Let's have a M = [10 x 4 x 12] matrix. As example I take the M(:,:,4):
val(:,:,4) =
0 0 1 0
0 1 1 1
0 0 0 1
1 1 1 1
1 1 0 1
0 1 1 1
1 1 1 1
1 1 1 1
0 0 1 1
0 0 1 1
How can I obtain this:
val(:,:,4) =
0 0 3 0
0 2 2 2
0 0 0 4
1 1 1 1
1 1 0 1
0 2 2 2
1 1 1 1
1 1 1 1
0 0 3 3
0 0 3 3
If I have 1 in the first column then all the subsequent 1's should be 1.
If I have 0 in the first column but 1 in the second, all the subsequent 1's should be 2.
If I have 0 in the first and second column but 1 in the third then all the subsequent 1's should be 3.
If I have 0 in the first 3 columns but 1 in the forth then this one should be four.
Note: The logical matrix M is constructed:
Tab = [reshape(Avg_1step.',10,1,[]) reshape(Avg_2step.',10,1,[]) ...
reshape(Avg_4step.',10,1,[]) reshape(Avg_6step.',10,1,[])];
M = Tab>=repmat([20 40 60 80],10,1,size(Tab,3));
This is a very simple approach that works for both 2D and 3D matrices.
%// Find the column index of the first element in each "slice".
[~, idx] = max(val,[],2);
%// Multiply the column index with each row of the initial matrix
bsxfun(#times, val, idx);
This could be one approach -
%// Concatenate input array along dim3 to create a 2D array for easy work ahead
M2d = reshape(permute(M,[1 3 2]),size(M,1)*size(M,3),[]);
%// Find matches for each case, index into each matching row and
%// elementwise multiply all elements with the corresponding multiplying
%// factor of 2 or 3 or 4 and thus obtain the desired output but as 2D array
%// NOTE: Case 1 would not change any value, so it was skipped.
case2m = all(bsxfun(#eq,M2d(:,1:2),[0 1]),2);
M2d(case2m,:) = bsxfun(#times,M2d(case2m,:),2);
case3m = all(bsxfun(#eq,M2d(:,1:3),[0 0 1]),2);
M2d(case3m,:) = bsxfun(#times,M2d(case3m,:),3);
case4m = all(bsxfun(#eq,M2d(:,1:4),[0 0 0 1]),2);
M2d(case4m,:) = bsxfun(#times,M2d(case4m,:),4);
%// Cut the 2D array thus obtained at every size(a,1) to give us back a 3D
%// array version of the expected values
Mout = permute(reshape(M2d,size(M,1),size(M,3),[]),[1 3 2])
Code run with a random 6 x 4 x 2 sized input array -
M(:,:,1) =
1 1 0 1
1 0 1 1
1 0 0 1
0 0 1 1
1 0 0 0
1 0 1 1
M(:,:,2) =
0 1 0 1
1 1 0 0
1 1 0 0
0 0 1 1
0 0 0 1
0 0 1 0
Mout(:,:,1) =
1 1 0 1
1 0 1 1
1 0 0 1
0 0 3 3
1 0 0 0
1 0 1 1
Mout(:,:,2) =
0 2 0 2
1 1 0 0
1 1 0 0
0 0 3 3
0 0 0 4
0 0 3 0