I have 10 bins, and each bin contains a specific number of observations, e.g.:
a = [0,0,1,0,0,2,0,0,0,2]
I'd like to subsequently tally how many times any given pair of (non-zero) bins co-occur - based on the number of observations.
Given the above example, bin#3 = 1, bin#6 = 2 and bin#10 = 2.
This means that bin 3 and 6 co-occurred once, bin 3 and 10 co-occurred once, and bin 6 and 10 co-occurred twice (the minimum value of the pair is taken).
My desired output is a full matrix, listing every possible bin combination (columns 1-2) and the tally of what was observed (column 3):
1 2 0
1 3 0
1 4 0
1 5 0
1 6 0
1 7 0
1 8 0
1 9 0
1 10 0
2 3 0
2 4 0
2 5 0
2 6 0
2 7 0
2 8 0
2 9 0
2 10 0
3 4 0
3 5 0
3 6 1
3 7 0
3 8 0
3 9 0
3 10 1
4 5 0
4 6 0
4 7 0
4 8 0
4 9 0
4 10 0
5 6 0
5 7 0
5 8 0
5 9 0
5 10 0
6 7 0
6 8 0
6 9 0
6 10 2
7 8 0
7 9 0
7 10 0
8 9 0
8 10 0
9 10 0
Is there a short and/or fast way of doing this?
You can get all combinations of the bin numbers in many ways. I'll use combvec for ease.
Then it's relatively simple to vectorise this using min...
a = [0,0,1,0,0,2,0,0,0,2];
n = 1:numel(a);
% Use unique and sort to get rid of duplicate pairs when order doesn't matter
M = unique( sort( combvec( n, n ).', 2 ), 'rows' );
% Get rid of rows where columns 1 and 2 are equal
M( M(:,1) == M(:,2), : ) = [];
% Get the overlap count for bins
M( :, 3 ) = min( a(M), [], 2 );
Try this.
bin_output = [....];
bin_matrix = [0,0,1,0,0,2,0,0,0,2];
bin_nonzero = find(bin_matrix);
for j = 1:length(bin_nonzero);
if isequal(j,length(bin_nonzero))
break;
end
for k = (j+1):(length(bin_nonzero))
for m = 1:length(bin_output)
if isequal(bin_output(m,1),j) && isequal(bin_output(m,2),k)
bin_output(m,3) = bin_output(m,3) + min([bin_matrix(1,bin_nonzero(1,j)),bin_matrix(1,bin_nonzero(1,k))]);
end
end
end
end
Related
Suppose I have a matrix with many rows and columns, for example a small subset would be:
1 2 3 4 5 6
1 1 5 6 0 0
1 2 2 3 2 1
1 2 0 3 4 5
1 9 5 7 3 0
I want to find the rows whose columns #4, #5 and #6 contain elements greater than zero, so I in this case would get a vector like this:
1
3
4
I have tried using the find() function this way:
idx = find(y(:, 4:6) > 0)
but I get this:
1
2
3
4
5
6
8
9
10
11
13
14
You can use a combination of find and all like this:
idx = find(all(y(:,4:6) > 0, 2))
This gives:
>> y = [1 2 3 4 5 6; 1 1 5 6 0 0; 1 2 2 3 2 1; 1 2 0 3 4 5; 1 9 5 7 3 0]
y =
1 2 3 4 5 6
1 1 5 6 0 0
1 2 2 3 2 1
1 2 0 3 4 5
1 9 5 7 3 0
>> idx = find(all(y(:,4:6) > 0, 2))
idx =
1
3
4
The idea is that we extract columns 4 to 6, check which values are greater than 0, operate along the 2nd dimension with an AND condition (all), and then extract which indices (rows) are 1/true in the resulting column vector.
[1 2 3 4 5 6 7 8 9 ;
9 8 7 6 5 4 3 2 1 ;
1 2 0 0 1 0 0 0 1 ]
The last row has five columns with zeros. I would like to keep only one column per zero crossing.
like this
[1 2 3 5 8 9 ;
9 8 7 5 2 1 ;
1 2 0 1 0 1 ]
Is this possible with fast Matlab functions or do I have to write some slow complicated for loop ?
You can create a logical array many different ways to find the columns to remove. Something like this would work
% Find the zeros that are not the first zero
cols_to_remove = data(end,:) == 0 & ~diff([false, data(end,:) == 0]) == 1;
% Now remove them
data(:, cols_to_remove) = [];
I have a two long vector. Vector one contains values of 0,1,2,3,4's, 0 represent no action, 1 represent action 1 and 2 represent the second action and so on. Each action is 720 sample point which means that you could find 720 consecutive twos then 720 consecutive 4s for example. Vector two contains raw data corresponding to each action. I need to create a matrix for each action ( 1, 2, 3 and 4) which contains the corresponding data of the second vector. For example matrix 1 should has all the data (vector 2 data) which occurred at the same indices of action 1. Any Help??
Example on small amount of data:
Vector 1: 0 0 1 1 1 0 0 2 2 2 0 0 1 1 1 0 0 2 2 2
Vector 2: 6 7 5 6 4 6 5 9 8 7 9 7 0 5 6 4 1 5 8 0
Result:
Matrix 1:
5 6 4
0 5 6
Matrix 2:
9 8 7
5 8 0
Here is one approach. I used a cell array to store the output matrices, hard-coding names for such variables isn't a good plan.
V1=[0 0 1 1 1 0 0 2 2 2 0 0 1 1 1 0 0 2 2 2]
V2=[6 7 5 6 4 6 5 9 8 7 9 7 0 5 6 4 1 5 8 0]
%// Find length of sequences of 1's/2's
len=find(diff(V1(find(diff(V1)~=0,1)+1:end))~=0,1)
I=unique(V1(V1>0)); %// This just finds how many matrices to make, 1 and 2 in this case
C=bsxfun(#eq,V1,I.'); %// The i-th row of C contains 1's where there are i's in V1
%// Now pick out the elements of V2 based on C, and store them in cell arrays
Matrix=arrayfun(#(m) reshape(V2(C(m,:)),len,[]).',I,'uni',0);
%// Note, the reshape converts from a vector to a matrix
%// Display results
Matrix{1}
Matrix{2}
Since, there is a regular pattern in the lengths of groups within Vector 1, that could be exploited to vectorize many things while proposing a solution. Here's one such implementation -
%// Form new vectors out of input vectors for non-zero elements in vec1
vec1n = vec1(vec1~=0)
vec2n = vec2(vec1~=0)
%// Find positions of group shifts and length of groups
df1 = diff(vec1n)~=0
grp_change = [true df1]
grplen = find(df1,1)
%// Reshape vec2n, so that we end up with N x grplen sized array
vec2nr = reshape(vec2n,grplen,[]).' %//'
%// ID/tag each group change based on their unique vector 2 values
[R,C] = sort(vec1n(grp_change))
%// Re-arrange rows of reshaped vector2, s.t. same ID rows are grouped succesively
vec2nrs = vec2nr(C,:)
%// Find extents of each group & use those extents to have final cell array output
grp_extent = diff(find([1 diff(R) 1]))
out = mat2cell(vec2nrs,grp_extent,grplen)
Sample run for the given inputs -
>> vec1
vec1 =
0 0 1 1 1 0 0 2 2 2 ...
0 0 1 1 1 0 0 2 2 2
>> vec2
vec2 =
6 7 5 6 4 6 5 9 8 7 ...
9 7 0 5 6 4 1 5 8 0
>> celldisp(out)
out{1} =
5 6 4
0 5 6
out{2} =
9 8 7
5 8 0
Here is another solution:
v1 = [0 0 1 1 1 0 0 2 2 2 0 0 1 1 1 0 0 2 2 2];
v2 = [6 7 5 6 4 6 5 9 8 7 9 7 0 5 6 4 1 5 8 0];
m1 = reshape(v2(v1 == 1), 3, [])'
m2 = reshape(v2(v1 == 2), 3, [])'
EDIT: David's solution is more flexible and probably more efficient.
Assuming I'm having a vectors of numbers A, for example: A=[1 3 5 3 9 6](A's length >= 2) and an Integer X=6. Need to find how many pairs (A[i],A[j]) where i<j exist in the vector which answer this condition: A[i]+A[j]=X. The number of pairs is printed.
Not allowed to use for/while. Allowed only ceil,floor,mod,repmat,reshape,size,length,transpose,sort,isempty,all,any,find ,sum,max,min.
With repmat, length and sum -
integer1 = 6; %// One of the paramters
A_ind = 1:length(A) %// Get the indices array
%// Expand A_ind into rows and A_ind' into columns, to form a meshgrid structure
A_ind_mat1 = repmat(A_ind,[length(A) 1])
A_ind_mat2 = repmat(A_ind',[1 length(A)]) %//'
%// Expand A into rows and A' into columns, to form a meshgrid structure
A_mat1 = repmat(A,[length(A) 1])
A_mat2 = repmat(A',[1 length(A)]) %//'
%// Form the binary matrix of -> (A[i],A[j]) where i<j
cond1 = A_ind_mat1 < A_ind_mat2
%// Use the binary matrix as a logical mask to select elements from the two
%// matrices and see which element pairs satisfy -> A[i]+A[j]=X and get a
%// count of those pairs with SUM
pairs_count = sum((A_mat1(cond1) + A_mat2(cond1))==integer1)
Outputs from code run to make it clearer -
A =
1 3 5 3 9 6
A_ind =
1 2 3 4 5 6
A_ind_mat1 =
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
A_ind_mat2 =
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
A_mat1 =
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
A_mat2 =
1 1 1 1 1 1
3 3 3 3 3 3
5 5 5 5 5 5
3 3 3 3 3 3
9 9 9 9 9 9
6 6 6 6 6 6
cond1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 1 1 0 0
1 1 1 1 1 0
pairs_count =
2
A bit more explanation -
Taking few more steps to clarify why pairs_count must be 2 here -
Set all values in A_mat1 and A_mat2 to be zeros that do not satisfy the less than criteria
>> A_mat1(~cond1)=0
A_mat1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 3 0 0 0 0
1 3 5 0 0 0
1 3 5 3 0 0
1 3 5 3 9 0
>> A_mat2(~cond1)=0
A_mat2 =
0 0 0 0 0 0
3 0 0 0 0 0
5 5 0 0 0 0
3 3 3 0 0 0
9 9 9 9 0 0
6 6 6 6 6 0
Now, add A_mat1 and A_mat2 and see how many 6's you got -
>> A_mat1 + A_mat2
ans =
0 0 0 0 0 0
4 0 0 0 0 0
6 8 0 0 0 0
4 6 8 0 0 0
10 12 14 12 0 0
7 9 11 9 15 0
As you can see there are two 6's and thus our result is verified.
I have a matrix
a = [1 2 3 4 5 6 7 8 9 10 11]
I need to split it into rows of 5 and fill the rest of the unset block with zeros like this:
transformed = [ 1 2 3 4 5 ;
6 7 8 9 10;
11 0 0 0 0 ]
You could first expand a to have the required number of elements like this;
a(15) = 0 % Matlab will automatically fill elements 12:14 with 0
then
transformed = reshape(a,[5,3])'
produces
ans =
1 2 3 4 5
6 7 8 9 10
11 0 0 0 0