I want to change the individual elements in a matrix by different values simultaneously.
How do I do that?
For example: I want to change the first element in matrix A by certain amount and the second element by a different amount simultaneously.
{ A = [1; 2]
% instead of doing A(1) = .....
A(2) = .....
}
You can access the elements of a vector or matrix and replace them.
For a vector this is intuitive.
octave:16> A = 1:9
A =
1 2 3 4 5 6 7 8 9
octave:17> A([1 3 5 7 9]) = 0
A =
0 2 0 4 0 6 0 8 0
This can be done for a matrix as well. The elements of a matrix are arranged in a column-first manner. You can use a single index to access the elements of a matrix.
octave:18> A = [1 2 3; 4 5 6; 7 8 9]
A =
1 2 3
4 5 6
7 8 9
The 2nd element of A is the same as A(2, 1). The 4th element of A is the same as A(1, 2).
octave:21> A(2)
ans = 4
octave:22> A(4)
ans = 2
So, you can set all the odd elements of A to 0 in one go like this:
octave:19> A([1 3 5 7 9]) = 0
A =
0 2 0
4 0 6
0 8 0
Just add a vector with the differences. A += [0.1; 0.2]
octave:1> A = [1; 2];
octave:2> A += [0.1; 0.2]
A =
1.1000
2.2000
Related
I have a Matrix A. I want to iterate over the inner part of the matrix (B), while also working with the rows and columns that are not part of B.
A = [1 4 5 6 7 1; B = [2 2 2 2;
8 2 2 2 2 1; 2 3 3 2;
9 2 3 3 2 1; 2 8 2 2];
0 2 8 2 2 1;
1 1 1 1 1 1];
I know it is possible to select the part of A like this:
[rows,columns] = size(A);
B = A([2:1:rows-1],[2:1:columns-1]);
for i = 1:(rows*columns)
%do loop stuff
endfor
This however wont work because I also need the outer rows and columns for my calculations. How can I achieve a loop without altering A?
So, why do not use two indexes for the inner matrix?
%....
for i=2:rows-1
for j=2:cols-1
% here, A(i,j) are the B elements, but you
% can still access to A(i-1, j+1) if you want.
end
end
%....
It is the first time I deal with column-compress storage (CCS) format to store matrices. After googling a bit, if I am right, in a matrix having n nonzero elements the CCS is as follows:
-we define a vector A_v of dimensions n x 1 storing the n non-zero elements
of the matrix
- we define a second vector A_ir of dimensions n x 1 storing the rows of the
non-zero elements of the matrix
-we finally define a third vector A_jc whose elements are the indices of the
elements of A_v which corresponds to the beginning of new column, plus a
final value which is by convention equal t0 n+1, and identifies the end of
the matrix (pointing theoretically to a virtual extra-column).
So for instance,
if
M = [1 0 4 0 0;
0 3 5 2 0;
2 0 0 4 6;
0 0 7 0 8]
we get
A_v = [1 2 3 4 5 7 2 4 6 8];
A_ir = [1 3 2 1 2 4 2 3 3 4];
A_jc = [1 3 4 7 9 11];
my questions are
I) is what I wrote correct, or I misunderstood anything?
II) what if I want to represent a matri with some columns which are zeroes, e.g.,
M2 = [0 1 0 0 4 0 0;
0 0 3 0 5 2 0;
0 2 0 0 0 4 6;
0 0 0 0 7 0 8]
wouldn't the representation of M2 in CCS be identical to the one of M?
Thanks for the help!
I) is what I wrote correct, or I misunderstood anything?
You are perfectly correct. However, you have to take care that if you use a C or C++ library offsets and indices should start at 0. Here, I guess you read some Fortran doc for which indices are starting at 1. To be clear, here is below the C version, which simply translates the indices of your Fortran-style correct answer:
A_v = unmodified
A_ir = [0 2 1 0 1 3 1 2 2 4] (in short [1 3 2 1 2 4 2 3 3 4] - 1)
A_jc = [0 2 3 6 8 10] (in short [1 3 4 7 9 11] - 1)
II) what if I want to represent a matri with some columns which are
zeroes, e.g., M2 = [0 1 0 0 4 0 0;
0 0 3 0 5 2 0;
0 2 0 0 0 4 6;
0 0 0 0 7 0 8]
wouldn't the representation of M2 in CCS be identical to the one of M?
I you have an empty column, simply add a new entry in the offset table A_jc. As this column contains no element this new entry value is simply the value of the previous entry. For instance for M2 (with index starting at 0) you have:
A_v = unmodified
A_ir = unmodified
A_jc = [0 0 2 3 6 8 10] (to be compared to [0 2 3 6 8 10])
Hence the two representations are differents.
If you just start learning about sparse matrices there is an excelllent free book here: http://www-users.cs.umn.edu/~saad/IterMethBook_2ndEd.pdf
i would like to sum all the values from my 2nd column which have the same value in the first column.
So my matrix looks maybe like this:
column: [1 1 1 2 2 3 3 3 3 4 5 5]
column: [3 5 8 2 6 4 0 6 1 0 2 6]
now i would like to have for the value 1 in the 1st column a sum of 3, 5 and 8 in the 2nd column, the same goes for 2, 3 and so from the 1st column.
Like this for example:
[1 2 3 4 5],
[16 8 11 0 8]
i'm thankful for any suggestions!
Sum all values when values are equal :
Just to init :
a = [1 1 1 2 2 3 3 3 3 4 5 5 ; 3 5 8 2 6 4 0 6 1 0 2 6];
a = a.';
Let's go :
n=0
for i=1:size(a,1)
if a(i,1) == a(i,2)
n = n + a(i,1)
end
end
n
For the second question :
mat=0
for j = 1:max(a(:,1))
n=0
for i=1:size(a,1)
if j == a(i,1)
n = n + a(i,2)
end
end
mat(j,1) = j
mat(j,2) = n
end
mat
Result :
mat =
1 16
2 8
3 11
4 0
5 8
Suppose I have an array X of size n by p by q. I would like to reshape it as a matrix with p rows, and in each row put the concatenation of the n rows of size q, resulting in a matrix of size p by nq.
I managed to do it with a loop but it takes a while say if n=1000, p=300, q=300.
F0=[];
for k=1:size(F,1)
F0=[F0,squeeze(X(k,:,:))];
end
Is there a faster way?
I think this is what you want:
Y = reshape(permute(X, [2 1 3]), size(X,2), []);
Example with n=2, p=3, q=4:
>> X
X(:,:,1) =
0 6 9
8 3 0
X(:,:,2) =
4 7 1
3 7 4
X(:,:,3) =
4 7 2
6 7 6
X(:,:,4) =
6 1 9
1 4 3
>> Y = reshape(permute(X, [2 1 3]), size(X,2), [])
Y =
0 8 4 3 4 6 6 1
6 3 7 7 7 7 1 4
9 0 1 4 2 6 9 3
Try this -
reshape(permute(X,[2 3 1]),p,[])
Thus, for code verification, one can look into a sample case run -
n = 2;
p = 3;
q = 4;
X = rand(n,p,q)
F0=[];
for k=1:n
F0=[F0,squeeze(X(k,:,:))];
end
F0
F0_noloop = reshape(permute(X,[2 3 1]),p,[])
Output is -
F0 =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
F0_noloop =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
Rather than using vectorization to solve the problem, you could look at the code to try and figure out what may improve performance. In this case, since you know the size of your output matrix F0 should be px(n*q), you could pre-allocate memory to F0 and avoid the constant resizing of the matrix at each iteration of the for loop
n=1000;
p=300;
q=300;
F0=zeros(p,n*q);
for k=1:size(F,1)
F0(:,(k-1)*q+1:k*q) = squeeze(F(k,:,:));
end
While probably not as efficient as the other two solutions, it is an alternative. Try the above and see what happens!
I have a 6 * 6 matrix
A=
3 8 8 8 8 8
4 6 1 0 7 -1
9 7 0 2 6 -1
7 0 0 5 4 4
4 -1 0 2 8 1
1 -1 0 8 3 9
I am interested in finding row and column number of neighbors starting from A(4,4)=5. But They will be linked to A(4,4) as neighbor only if A(4,4) has element 4 on right, 6 on left, 2 on top, 8 on bottom 1 on top left diagonally, 3 on top right diagonally, 7 on bottom left diagonally and 9 on bottom right diagonally. TO be more clear A(4,4) will have neighbors if the neighbors are surrounding A(4,4) as follows:
1 2 3;
6 5 4;
7 8 9;
And this will continue as each neighbor is found.
Also 0 and -1 will be ignored. In the end I want to have these cells' row and column number as shown in figure below. Is there any way to visualize this network as well. This is sample only. I really have a huge matrix.
A = [3 8 8 8 8 8;
4 6 1 0 7 -1;
9 7 0 2 6 -1;
7 0 0 5 4 4;
4 -1 0 2 8 1;
1 -1 0 8 3 9];
test = [1 2 3;
6 5 4;
7 8 9];
%//Pad A with zeros on each side so that comparing with test never overruns the boundries
%//BTW if you have the image processing toolbox you can use the padarray() function to handle this
P = zeros(size(A) + 2);
P(2:end-1, 2:end-1) = A;
current = zeros(size(A) + 2);
past = zeros(size(A) + 2);
%//Initial state (starting point)
current(5,5) = 1; %//This is A(4,4) but shifted up 1 because of the padding
condition = 1;
while sum(condition(:)) > 0;
%//get the coordinates of any new values added to current
[x, y] = find(current - past);
%//update past to last iterations current
past = current;
%//loop through all the coordinates returned by find above
for ii=1:size(x);
%//Make coord vectors that represent the current coordinate plus it 8 immediate neighbours.
%//Note that this is why we padded the side in the beginning, so if we hit a coordinate on an edge, we can still get 8 neighbours for it!
xcoords = x(ii)-1:x(ii)+1;
ycoords = y(ii)-1:y(ii)+1;
%//Update current based on comparing the coord and its neighbours against the test matrix, be sure to keep the past found points hence the OR
current(xcoords, ycoords) = (P(xcoords, ycoords) == test) | current(xcoords, ycoords);
end
%//The stopping condition is when current == past
condition = current - past;
end
%//Strip off the padded sides
FinalAnswer = current(2:end-1, 2:end-1)
[R, C] = find(FinalAnswer);
coords = [R C] %//This line is unnecessary, it just prints out the results at the end for you.
OK cool you got very close, so here is the final solution with the loops. It runs in about 0.002 seconds so it's pretty quick I think. The output is
FinalAnswer =
0 0 0 0 0 0
0 1 1 0 0 0
0 1 0 1 0 0
1 0 0 1 1 1
0 0 0 0 1 0
0 0 0 0 0 1
coords =
4 1
2 2
3 2
2 3
3 4
4 4
4 5
5 5
4 6
6 6