Position of integers in vector - matlab

Question

I think the question is pretty basic, but still keeps me busy since some time.
Lets assume we have a vector containing 4 integers randomly repetetive, like:
v = [ 1 3 3 3 4 2 1 2 3 4 3 2 1 4 3 3 4 2 2]
I am searching for the vector of all positions of each integer, e.g. for 1 it should be a vector like:
position_one = [1 7 13]
Since I want to search every row of a 100x10000 matrix I was not able to deal with linear indeces.
Thanks in advance!

Answer 1

Rows and columns
Since your output for every integer changes, a cell array will fit the whole task. For the whole matrix, you can do something like:
A = randi(4,10,30); % some data
Row = repmat((1:size(A,1)).',1,size(A,2)); % index of the row
Col = repmat((1:size(A,2)),size(A,1),1); % index of the column
pos = #(n) [Row(A==n) Col(A==n)]; % Anonymous function to find the indices of 'n'
than for every n you can write:
>> pos(3)
ans =
1 1
2 1
5 1
6 1
9 1
8 2
3 3
. .
. .
. .
where the first column is the row, and the second is the column for every instance of n in A.
And for all ns you can use an arrayfun:
positions = arrayfun(pos,1:max(A(:)),'UniformOutput',false) % a loop that goes over all n's
or a simple for loop (faster):
positions = cell(1,max(A(:)));
for n = 1:max(A(:))
positions(n) = {pos(n)};
end
The output in both cases would be a cell array:
positions =
[70x2 double] [78x2 double] [76x2 double] [76x2 double]
and for every n you can write positions{n}, to get for example:
>> positions{1}
ans =
10 1
2 3
5 3
3 4
5 4
1 5
4 5
. .
. .
. .
Only rows
If all you want in the column index per a given row and n, you can write this:
A = randi(4,10,30);
row_pos = #(k,n) A(k,:)==n;
positions = false(size(A,1),max(A(:)),size(A,2));
for n = 1:max(A(:))
positions(:,n,:) = row_pos(1:size(A,1),n);
end
now, positions is a logical 3-D array, that every row corresponds to a row in A, every column corresponds to a value of n, and the third dimension is the presence vector for the combination of row and n. this way, we can define R to be the column index:
R = 1:size(A,2);
and then find the relevant positions for a given row and n. For instance, the column indices of n=3 in row 9 is:
>> R(positions(9,3,:))
ans =
2 6 18 19 23 24 26 27
this would be just like calling find(A(9,:)==3), but if you need to perform this many times, the finding all indices and store them in positions (which is logical so it is not so big) would be faster.

Answer 2

Find linear indexes in a matrix: I = find(A == 1).
Find two dimensional indexes in matrix A: [row, col] = find(A == 1).
%Create sample matrix, with elements equal one:
A = zeros(5, 4);
A([2 10 14]) = 1
A =
0 0 0 0
1 0 0 0
0 0 0 0
0 0 1 0
0 1 0 0
Find ones as linear indexes in A:
find(A == 1)
ans =
2
10
14
%This is the same as reshaping A to a vector and find ones in the vector:
B = A(:);
find(B == 1);
B' =
0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
Find two dimensional indexes:
[row, col] = find(A == 1)
row =
2
5
4
col =
1
2
3

Answer 3

You can do that with accumarray using an anonymous function as follows:
positions = accumarray(v(:), 1:numel(v), [], #(x) {sort(x.')});
For
v = [ 1 3 3 3 4 2 1 2 3 4 3 2 1 4 3 3 4 2 2];
this gives
positions{1} =
1 7 13
positions{2} =
6 8 12 18 19
positions{3} =
2 3 4 9 11 15 16
positions{4} =
5 10 14 17