I am creating a sparse matrix
sp = sparse(I,J,Val,X,Y)
My Val matrix is a ones matrix. Much to my surprise the sp matrix does not contain only zeros and ones. I suppose that this happens because in some cases there are duplicates in I,J. I mean the sp(1,1) is set to 1 2 times, and this makes it 2.
Question 1: Is my assumption true? Instead of overwriting the value, does MATLAB really add it?
Question 2: How can we get around this, given that it would be very troublesome to manipulate I and J. Something I can think of, is to use find (thus guaranteeing uniqueness) and then recreate the matrix using ones once more. Any better suggestion?
Question 1: Is my assumption true? Instead of overwriting the value, does Matlab really add it?
Correct. If you have duplicate row and column values each with their own values, MATLAB will aggregate them all into the same row and column location by adding them.
This is clearly seen in the documentation but as a reproducible example, suppose I have the following row and column locations and their associated values at these locations:
i = [6 6 6 5 10 10 9 9].';
j = [1 1 1 2 3 3 10 10].';
v = [100 202 173 305 410 550 323 121].';
Note that these are column vectors as this shape is the expected input. In a neater presentation:
>> [i j v]
ans =
6 1 100
6 1 202
6 1 173
5 2 305
10 3 410
10 3 550
9 10 323
9 10 121
We can see that there are three values that get mapped to location (6, 1), two values that get mapped to location (10, 3) and finally two that get mapped to location (9, 10).
By creating the sparse matrix and displaying it, we thus get:
>> S = sparse(i,j,v)
S =
(6,1) 475
(5,2) 305
(10,3) 960
(9,10) 444
As you can see, the three values mapped to (6, 1) are summed: 100 + 202 + 173 = 475. You can verify this with the other duplicate row and column locations.
Question 2: How can we get around this, given that it would be very troublesome to manipulate I and J. Something I can think of, is to use find (thus guaranteeing uniqueness) and then recreate the matrix using ones once more. Any better suggestion?
There are two possible ways to mitigate this if it is truly your desire to only have a binary matrix.
The first way which may be more preferable to you as you mentioned that manipulating the row and column locations is troublesome is to create the matrix that you have now, but then convert it to logical so that any values that are non-zero are set to 1:
>> S = S ~= 0
S =
10×10 sparse logical array
(6,1) 1
(5,2) 1
(10,3) 1
(9,10) 1
If you require that the precision of the matrix be back in its original double form, cast the result after you convert to logical:
>> S = double(S ~= 0)
S =
(6,1) 1
(5,2) 1
(10,3) 1
(9,10) 1
The second way if you wish is to work on your row and column locations so that you filter out any indices that are non-unique, then create a vector of ones for val that is as long as the unique row and column locations. You can use the unique function to help you do that. Concatenate the row and column locations in a two column matrix and specify that you want to operate on 'rows'. This means that each row is considered an input rather than individual elements in a matrix. Once you find the unique row and column locations, use these as input for creating the sparse matrix:
>> unique_vals = unique([i j], 'rows')
unique_vals =
5 2
6 1
9 10
10 3
>> vals = ones(size(unique_vals, 1));
>> S = sparse(unique_vals(:, 1), unique_vals(:, 2), vals)
S =
(6,1) 1
(5,2) 1
(10,3) 1
(9,10) 1
Related
I would like to create a column vector from the elements of a matrix A of size (3,3) that are not on the diagonal. Thus, I would have 6 elements in that output vector. How can I do this?
Use eye and logical negation, although this is no better than Divakar's original answer, and possibly significantly slower for very large matrices.
>> A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
>> A(~eye(size(A)))
ans =
5
9
4
2
7
14
3
10
15
13
8
12
Use this to get such a column vector, assuming A is the input matrix -
column_vector = A(eye(size(A))==0)
If you don't care about the order of the elements in the output, you can also use a combination of setdiff and diag -
column_vector = setdiff(A,diag(A))
You can also use linear indexing to access the diagonal elements and null them. This will automatically reshape itself to a single vector:
A(1:size(A,1)+1:end) = [];
Bear in mind that this will mutate the original matrix A. If you don't want this to happen, make a copy of your matrix then perform the above operation on that copy. In other words:
Acopy = A;
Acopy(1:size(A,1)+1:end) = [];
Acopy will contain the final result. You need to create a vector starting from 1 and going to the end in increments of the rows of the matrix A added with 1 due to the fact that linear indices are column-major, so the linear indices used to access a matrix progress down each row first for a particular column. size(A,1) will allow us to offset by each column and we add 1 each time to ensure we get the diagonal coefficient for each column in the matrix.
Assuming that the matrix is square,
v = A(mod(0:numel(A)-1, size(A,1)+1) > 0).';
As an example, I have a matrix [1,2,3,4,5]'. This matrix contains one column and 5 rows, and I have to generate a pair of points like (1,2),(1,3)(1,4)(1,5),(2,3)(2,4)(2,5),(3,4)(3,5)(4,5).
I have to store these values in 2 columns in a matrix. I have the following code, but it isn't quite giving me the right answer.
for s = 1:5;
for tb = (s+1):5;
if tb>s
in = sub2ind(size(pairpoints),(tb-1),1);
pairpoints(in) = s;
in = sub2ind(size(pairpoints),(tb-1),2);
pairpoints(in) = tb;
end
end
end
With this code, I got (1,2),(2,3),(3,4),(4,5). What should I do, and what is the general formula for the number of pairs?
One way, though is limited depending upon how many different elements there are to choose from, is to use nchoosek as follows
pairpoints = nchoosek([1:5],2)
pairpoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
See the limitations of this function in the provided link.
An alternative is to just iterate over each element and combine it with the remaining elements in the list (assumes that all are distinct)
pairpoints = [];
data = [1:5]';
len = length(data);
for k=1:len
pairpoints = [pairpoints ; [repmat(data(k),len-k,1) data(k+1:end)]];
end
This method just concatenates each element in data with the remaining elements in the list to get the desired pairs.
Try either of the above and see what happens!
Another suggestion I can add to the mix if you don't want to rely on nchoosek is to generate an upper triangular matrix full of ones, disregarding the diagonal, and use find to generate the rows and columns of where the matrix is equal to 1. You can then concatenate both of these into a single matrix. By generating an upper triangular matrix this way, the locations of the matrix where they're equal to 1 exactly correspond to the row and column pairs that you are seeking. As such:
%// Highest value in your data
N = 5;
[rows,cols] = find(triu(ones(N),1));
pairpoints = [rows,cols]
pairPoints =
1 2
1 3
2 3
1 4
2 4
3 4
1 5
2 5
3 5
4 5
Bear in mind that this will be unsorted (i.e. not in the order that you specified in your question). If order matters to you, then use the sortrows command in MATLAB so that we can get this into the proper order that you're expecting:
pairPoints = sortrows(pairPoints)
pairPoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Take note that I specified an additional parameter to triu which denotes how much of an offset you want away from the diagonal. The default offset is 0, which includes the diagonal when you extract the upper triangular matrix. I specified 1 as the second parameter because I want to move away from the diagonal towards the right by 1 unit so I don't want to include the diagonal as part of the upper triangular decomposition.
for loop approach
If you truly desire the for loop approach, going with your model, you'll need two for loops and you need to keep track of the previous row we are at so that we can just skip over to the next column until the end using this. You can also use #GeoffHayes approach in using just a single for loop to generate your indices, but when you're new to a language, one key advice I will always give is to code for readability and not for efficiency. Once you get it working, if you have some way of measuring performance, you can then try and make the code faster and more efficient. This kind of programming is also endorsed by Jon Skeet, the resident StackOverflow ninja, and I got that from this post here.
As such, you can try this:
pairPoints = []; %// Initialize
N = 5; %// Highest value in your data
for row = 1 : N
for col = row + 1 : N
pairPoints = [pairPoints; [row col]]; %// Add row-column pair to matrix
end
end
We get the equivalent output:
pairPoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Small caveat
This method will only work if your data is enumerated from 1 to N.
Edit - August 20th, 2014
You wish to generalize this to any array of values. You also want to stick with the for loop approach. You can still keep the original for loop code there. You would simply have to add a couple more lines to index your new array. As such, supposing your data array was:
dat = [12, 45, 56, 44, 62];
You would use the pairPoints matrix and use each column to subset the data array to access your values. Also, you need to make sure your data is a column vector, or this won't work. If we didn't, we would be creating a 1D array and concatenating rows and that's not obviously what we're looking for. In other words:
dat = [12, 45, 56, 44, 62];
dat = dat(:); %// Make column vector - Important!
N = numel(dat); %// Total number of elements in your data array
pairPoints = []; %// Initialize
%// Skip if the array is empty
if (N ~= 0)
for row = 1 : N
for col = row + 1 : N
pairPoints = [pairPoints; [row col]]; %// Add row-column pair to matrix
end
end
vals = [dat(pairPoints(:,1)) dat(pairPoints(:,2))];
else
vals = [];
Take note that I have made a provision where if the array is empty, don't even bother doing any calculations. Just output an empty matrix.
We thus get:
vals =
12 45
12 56
12 44
12 62
45 56
45 44
45 62
56 44
56 62
44 62
Assume I have a matrix called A.
The values of the matrix represent coordinates,
so row 2 and column 3 is the coordinate (2,3) in the 2D plan.
How can I map all the values of the matrix to different indices so that (0,0) would get the mapping value of 0 etc.?
(0,0) -> 0
(0,1) -> 1
(0,2) ->2
..
..
and so on.
Thanks.
Assuming that you are okay with the MATLAB indexing that starts with 1, this would work -
A1 = reshape([1:numel(A)],size(A,1),[])'
If you would like to start the mapping from 0, just subtract 1 -
A1 = reshape([1:numel(A)],size(A,1),[])' -1
"The sub2ind command determines the equivalent single index corresponding to a set of subscript values."
For example, if
i = sub2ind(size(A), 2, 3);
then
A(2,3) and A(i) refer to the same element in a matrix A.
In MATLAB you can index matrices linearly. Suppose you have the matrix:
a =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
Now, you can access the element in position (3,2) either using normal subscripts, or using the linear equivalent.
a(3,2)
ans = 7
a(7)
ans = 7
Assuming you have your indices as a list from 1 to numel(a) and don't really need a link between (3,2) and (7), this would be the simplest way to do it.
As you state, you want the element in position (0,0) to have index (0). Since MATLAB indexing starts at 1 you have two alternatives:
If you get a list (for instance from another program) where elements are listed from zero to (numel(a) - 1), such as ind = [0, 3, 6, 8], my suggestion is you simply do ind = ind + 1 (or ind_1 = ind + 1 if you don't want to overwrite the original vector.
Otherwise you can add one every time this way: x = a(ind + 1);.
However, if you really want a link between (3,2) and (7), I believe sub2ind is the way to go.
I have two matrices, 22007x3 and 352x2. The first column in each is an index, most (but not all) of which are shared (i.e. x1 contains indices that aren't in x2).
I would like to combine the two matrices into a 22007x4 matrix, such that column 4 is filled in with the values that correspond to particular indices in both original matrices.
For example:
x1 =
1 1 5
1 2 4
1 3 5
2 1 1
2 2 1
2 3 2
x2 =
1 15.5
2 -5.6
becomes
x3 =
1 1 5 15.5
1 2 4 15.5
1 3 5 15.5
2 1 1 -5.6
2 2 1 -5.6
2 3 2 -5.6
I've tried something along the lines of
x3(1:numel(x1),1:3)=x1;
x3(1:numel(x2(:,2)),4)=x2(:,2);
but firstly I get the error
??? Subscripted assignment dimension mismatch.
and then I can't figure out I would fill the rest of it.
An important point is that there are not necessarily an equal number of rows per index in my data.
How might I make this work?
Taking Amro's answer from here
[~, loc] = ismember(x1(:,1), x2(:,1));
ismember's second argument returns the location in x2 where each element of x1 can be found (or 0 if it can't)
a = x2(loc(loc > 0), 2);
get the relevant values using these row indices but excluding the zeros, hence the loc > 0 mask. You have to exclude these as 1, they are not in x2 and 2 you can't index with 0.
Make a new column of default values to stick on the end of x1. I think NaN() is probably better but zeros() is also fine maybe
newCol = NaN(size(x1,1),1)
Now use logical indexing to get the locations of the non zero elements and put a in those locations
newCol(loc > 0) = a
Finnaly stick it on the end
x3 = [x1, newCol]
I was trying to get the min values of a matrix before the max values of the matrix occurred. I have two matrices: matrix data and matrix a. Matrix a is a subset of matrix data and is composed of the max values of matrix data. I have the following code but obviously doing something wrong.
edit:
Matrix a are the max values of matrix data. I derived it from:
for x=1:size(data,1)
a(x)=max(data(x,:));
end
a=a'
clear x
matrix b code:
for x=1:size(data,1)
b(x)=min(data(x,(x<data==a)));
end
b=b'
clear x
matrix data matrix a matrix b
1 2 3 4 4 1
6 5 4 7 7 4
9 6 12 5 12 6
I need all the min values that occurred before to matrix a occurred in matrix data
Short and simple:
[a,idxmax] = max(data,[],2);
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii))), 1:size(data,1));
which is the same as
b=NaN(1,size(data,1)); % preallocation!
for ii=1:size(data,1)
b(ii) = min(data(ii,1:idxmax(ii)));
end
Ignore maximum itself
If you want minimum of everything really before (and not including the maximum), it's possible that the maximum is the first number, and you try taking minimum of an empty matrix. Solution then is to use cell output, which can be empty:
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii)-1)), 1:size(data,1),'uni',false);
Replace empty cells with NaN
If you want to replace empty cells to Nan and then back to a matrix use this:
b(cellfun(#isempty,b))={NaN};
b=cell2mat(b);
or simply use the earlier version and replace b(ii) with NaN when it is equal to a(ii) same outcome:
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii))), 1:size(data,1));
b(b'==a) = NaN
Example:
data=magic(4)
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
outputs:
a' = 16 11 12 15
b =
16 5 6 4
and
b =[1x0 double] [5] [6] [4]
for the 2nd solution using cell output and ignoring the maximum itself also.
And btw:
for x=1:size(data,1)
a(x)=max(data(x,:));
end
a=a'
clear x
can be replaced with
a=max(data,[],2);
It's not pretty but this is the only way I found so far of doing this kind of thing without a loop.
If loops are ok I would recommend Gunther Struyf answer as the most compact use of matlab's in-built array looping function, arrayfun.
Some of the transposition etc may be superfluous if you're wanting column mins instead of row...
[mx, imx] = max(data');
inds = repmat(1:size(data,2), [size(data,1),1]);
imx2 = repmat(imx', [1, size(data,2)]);
data2 = data;
data2(inds >= imx2) = inf;
min(data2');
NOTE: if data is not needed we can remove the additional data2 variable, and reduce the line count.
So to demonstrate what this does, (and see if I understood the question correctly):
for input
>> data = [1,3,-1; 5,2,1]
I get minima:
>> min(data2')
ans = [1, inf]
I.e. it only found the min values before the max values for each row, and anything else was set to inf.
In words:
For each row get index of maximum
Generate matrix of column indices
Use repmat to generate a matrix, same size as data where each row is index of maximum
Set data to infinity where column index > max_index matrix
find min as usual.