I have a numerical array CentroidBins which is 3694x4. Columns 3 and 4 are arbitrary X and Y bins with a range of 1-20. My goal in the last bit of code was to go through columns 3 and 4 to count the number of times a particular pair appeared (ie. 1,1 or 1,2....etc) and place that into a 20x20 array with rows being Y bins and columns being X bins. I managed to construct something which looks like what a want, but the output is 18x17, I am assuming it is deleting rows and columns populated by "0". How can I make sure this produces 20x20?
bin20 = centroids_array / 20 %create 20 bins
imRound = round(bin20)
CentroidBins = [centroids_array , imRound]
save("CentroidBins.mat", "CentroidBins");
disp(CentroidBins)
nrow = size(CentroidBins, 1);
B = CentroidBins(:,[3 4]);
NumF = full(sparse(B(1:end-nrow),B(nrow+1:end),1))
to count the occurrence of pairs, you use hist and unique
a=[1 2; 1 2; 2 3; 8 1; 2 3];
[foo,ix,jx]=unique(a,'rows');
count=hist(jx,unique(jx)) % report the repeated counts of each unique pair
foo % lists the unique pairs
Related
Definition of Problem
I have two arrays called weights and indices respectively:
weights = [1, 3, 2];
indices = [1, 1, 2, 3, 2, 4];
m = 4; % Number of Rows in Matrix
n = 3; % Number of Columns in Matrix
M = zeros(m, n);
The array called indices is storing the indices where I need to store a 1 in every column.
For instance for the first column at row 1 which is indicated in indices(1) I need to store a 1 and this is indicated by weights(1) which is equal to 1.
M(indices(1), 1) = 1;
For column 2, at rows 1 to 3 (indices(2:4)) I need to store a 1. The range of indices for column 2 are again indicated by weights(2).
M(indices(2:4),2) = 1;
Similarly, for column 3, at rows 2 and 4 (indices(5:6)) I need to store a 1. The range of indices for column 3 are again indicated by weights(3).
M(indices(5:6),3) = 1;
Expected Binary Matrix
The expected and resulting binary matrix is:
1 1 0
0 1 1
0 1 0
0 0 1
Solution
Is there any way I can do this in generalized manner by using both the weights and indices arrays rather than doing it in a hard coded manner to create create the binary matrix M?
You just have a weird way of describing your indices, so you just need to convert them to something standard.
columsn_idx=repelem(1:n,weights); % repeat column number as much as needed
row_idx=indices ; % just for clarity
M(sub2ind([m,n],row_idx,columsn_idx))=1;% Use linear indices
I have a matrix of growing length for example a 4-by-x matrix A where x is increasing in a loop. I want to find the smallest column c where all columns before that, each, carry one single number. The matrix A can look like:
A = [1 2 3 4;
1 2 3 5;
1 2 3 1;
1 2 3 0];
where c=3, and x=4.
At each iteration of the loop where A grows in length, the value of index c grows as well. Therefore, at each iteration, I want to update the value of c. How efficiently can I code this in Matlab?
Let's say you had the matrix A and you wanted to check a particular column iito see if all its elements are the same. The code would be:
all(A(:, ii)==A(1, ii)) % checks if all elements in column are same as first element
Also, keep in mind that once the condition is broken, x cannot be updated anymore. Therefore, your code should be:
x = 0;
while true
%% expand A by one column
if ~all(A(:, x)==A(1, x)) % true if all elements in column are not the same as first element
break;
end
x = x+1;
end
You could use this:
c = find(arrayfun(#(ind)all(A(1, ind)==A(:, ind)), 1:x), 1, 'first');
This finds the first column where not all values are the same. If you run this in a loop, you can detect when entries in a column start to differ:
for x = 1:maxX
% grow A
c = find(arrayfun(#(ind)~all(A(1, ind)==A(:, ind)), 1:x), 1, 'first');
% If c is empty, all columns have values equal to first row.
% Otherwise, we have to subtract 1 to get the number of columns with equal values
if isempty(c)
c = x;
else
c = c - 1;
end
end
Let me give a try as well:
% Find the columns which's elements are same and sum the logical array up
c = sum(A(1,:) == power(prod(A,1), 1/size(A,1)))
d=size(A,2)
To find the last column such that each column up to that one consists of equal values:
c = find(any(diff(A,1,1),1),1)-1;
or
c = find(any(bsxfun(#ne, A, A(1,:)),1),1)-1;
For example:
>> A = [1 2 3 4 5 6;
1 2 3 5 5 7;
1 2 3 1 5 0;
1 2 3 0 5 8];
>> c = find(any(diff(A,1,1),1),1)-1
c =
3
You can try this (easy and fast):
Equal_test = A(1,:)==A(2,:)& A(2,:)==A(3,:)&A(3,:)==A(4,:);
c=find(Equal_test==false,1,'first')-1;
You can also check the result of find if you want.
Given a 2-column matrix, for instance. The input is:
[ 1,2;
3,4;
5,5]
The expected output is:
[1,2;
3,4;]
Does anyone know how do accomplish this? Many thanks for your time and attention.
You could use logical indexing:
A = [1 2;3 4;5 5];
match = A(:,1) == A(:,2); // 1 where row has the same elements in both columns
A(match,:) = []; // make the match columns empty
You would need to make this more generic for another case, but for two columns and your example this will work.
Your question suggests your matrix may have an arbitrary number of columns. In that case you may want to delete a row if it has (a) any two elements equal, or (b) all elements equal.
One possible approach is:
Apply sort along each row;
Use diff to compute differences between consecutive elements;
Generate a logical index with all to (a) keep rows for which all such differences are non-zero, or with any to (b) keep rows for which any such difference is non-zero:
So:
X = [1 2 3;
3 4 3;
5 5 5];
Y = X(all(diff(sort(X,2),[],2),2),:);
Z = X(any(diff(sort(X,2),[],2),2),:);
gives
Y =
1 2 3
Z =
1 2 3
3 4 3
I have a matlab problem to solve. In have two vectores that limit my space, x_low and x_high. The matrix pos needs to have values within this spaces and each column of the matrix has different bounds given by the two vectores. Now my problem is that randi gives valus between two integers but i need to change the bounds for each columns. There is another way to use randi or a different matlab function to do this?
I know there are better codes to do this but i'm starting to use matlab and i know to do it this way, any aid is welcome
x_low = [Io_low, Iirr_low, Rp_low, Rs_low, n_low]; % vector of constant values
x_high = [Io_high, Iirr_high, Rp_high, Rs_high, n_high]; % vector of constant values
pos = rand(particles, var);
var = length(x_high);
for i = 1: particles % rows
for k = 1: var %columns
if pos(i, k) < x_low(k) || pos(i, k) > x_high(k) % if the position is out of bounder
pos(i, k) = randi(x_low(k), x_high(k), 1); % fill it with a particle whithin the bounderies
end
end
end
If I understand correctly, you need to generate a matrix with integer values such that each column has different lower and upper limits; and those lower and upper limits are inclusive.
This can be done very simply with
rand (to generate random numbers between 0 and 1 ),
bsxfun (to take care of the lower and upper limits on a column basis), and
round (so that the results are integer values).
Let the input data be defined as
x_low = [1 6 11]; %// lower limits
x_high = [3 10 100]; %// upper limits
n_rows = 7; %// number of columns
Then:
r = rand(n_rows, numel(x_low)); %// random numbers between 0 and 1
r = floor(bsxfun(#times, r, x_high-x_low+1)); %// adjust span and round to integers
r = bsxfun(#plus, r, x_low); %// adjust lower limit
gives something like
r =
2 7 83
3 6 93
2 6 22
3 10 85
3 7 96
1 10 90
2 8 57
If you need to fill in values only at specific entries of matrix pos, you can use something like
ind = bsxfun(#lt, pos, x_low) | bsxfun(#gt, pos, x_high); %// index of values to replace
pos(ind) = r(ind);
This a little overkill, because the whole matrixd r is generated only to use some of its entries. To generate only the needed values the best way is probably to use loops.
You can use cellfun for this. Something like:
x_low = [Io_low, Iirr_low, Rp_low, Rs_low, n_low];
x_high = [Io_high, Iirr_high, Rp_high, Rs_high, n_high];
pos = cell2mat(cellfun(#randi, mat2cell([x_low' x_high'], ones(numel(x_low),1), 1), repmat({[particles 1]}, [numel(x_low) 1)])))';
Best,
I have quite big array. To make things simple lets simplify it to:
A = [1 1 1 1 2 2 3 3 3 3 4 4 5 5 5 5 5 5 5 5];
So, there is a group of 1's (4 elements), 2's (2 elements), 3's (4 elements), 4's (2 elements) and 5's (8 elements). Now, I want to keep only columns, which belong to group of 3 or more elements. So it will be like:
B = [1 1 1 1 3 3 3 3 5 5 5 5 5 5 5 5];
I was doing it using for loop, scanning separately 1's, 2's, 3's and so on, but its extremely slow with big arrays...
Thanks for any suggestions how to do it in more efficient way :)
Art.
A general approach
If your vector is not necessarily sorted, then you need to run to count the number of occurrences of each element in the vector. You have histc just for that:
elem = unique(A);
counts = histc(A, elem);
B = A;
B(ismember(A, elem(counts < 3))) = []
The last line picks the elements that have less than 3 occurrences and deletes them.
An approach for a grouped vector
If your vector is "semi-sorted", that is if similar elements in the vector are grouped together (as in your example), you can speed things up a little by doing the following:
start_idx = find(diff([0, A]))
counts = diff([start_idx, numel(A) + 1]);
B = A;
B(ismember(A, A(start_idx(counts < 3)))) = []
Again, note that the vector need not to be entirely sorted, just that similar elements are adjacent to each other.
Here is my two-liner
counts = accumarray(A', 1);
B = A(ismember(A, find(counts>=3)));
accumarray is used to count the individual members of A. find extracts the ones that meet your '3 or more elements' criterion. Finally, ismember tells you where they are in A. Note that A needs not be sorted. Of course, accumarray only works for integer values in A.
What you are describing is called run-length encoding.
There is software for this in Matlab on the FileExchange. Or you can do it directly as follows:
len = diff([ 0 find(A(1:end-1) ~= A(2:end)) length(A) ]);
val = A(logical([ A(1:end-1) ~= A(2:end) 1 ]));
Once you have your run-length encoding you can remove elements based on the length. i.e.
idx = (len>=3)
len = len(idx);
val = val(idx);
And then decode to get the array you want:
i = cumsum(len);
j = zeros(1, i(end));
j(i(1:end-1)+1) = 1;
j(1) = 1;
B = val(cumsum(j));
Here's another way to do it using matlab built-ins.
% Set up
A=[1 1 1 1 2 2 3 3 3 3 4 4 5 5 5 5 5];
threshold=2;
% Get the unique elements of the array
uniqueElements=unique(A);
% Count haw many times each unique element occurs
counts=histc(A,uniqueElements);
% Write which elements should be kept
toKeep=uniqueElements(counts>threshold);
% Make a logical index
indexer=false(size(A));
for i=1:length(toKeep)
% For every unique element we want to keep select the indices in A that
% are equal
indexer=indexer|(toKeep(i)==A);
end
% Apply index
B=A(indexer);