I have a Matrix of size M by N in which each row has some zero entries. I want to create M row vectors such that each of the vector contains the non zero elements of each row. For example if I have the following Matrix
A=[0 0 0 5;0 0 4 6;0 1 2 3;9 10 2 3]
I want four different row vectors of the following form
[5]
[4 6]
[1 2 3]
[9 10 2 3]
This can be done with accumarray using an anonymous function as fourth input argument. To make sure that the results are in the same order as in A, the grouping values used as first input should be sorted. This requires using (a linearized version of) A transposed as second input.
ind = repmat((1:size(A,2)).',1,size(A,2)).';
B = A.';
result = accumarray(ind(:), B(:), [], #(x){nonzeros(x).'});
With A = [0 0 0 5; 0 0 4 6; 0 1 2 3; 9 10 2 3]; this gives
result{1} =
5
result{2} =
4 6
result{3} =
1 2 3
result{4} =
9 10 2 3
Since Matlab doesn't support non-rectangular double arrays, you'll want to settle on a cell array. One quick way to get the desired output is to combine arrayfun with logical indexing:
nonZeroVectors = arrayfun(#(k) A(k,A(k,:)~=0),1:size(A,1),'UniformOutput',false);
I used the ('UniformOutput',false) name-value pair for the reasons indicated in the documentation (I'll note that the pair ('uni',0) also works, but I prefer verbosity). This input produces a cell array with the entries
>> nonZerosVectors{:}
ans =
5
ans =
4 6
ans =
1 2 3
ans =
9 10 2 3
Related
I want to find all possible variations (combinations) of a vector, choosing various numbers of elements from that vector.
For example, suppose I have the vector:
x = [1 2 3 4 5];
I can determine the number of combinations for each number of chosen elements:
x = [1 2 3 4 5]';
n = numel(x);
for k = 1:n
combs(k) = nchoosek(n,k);
end
sum(combs)
This results in:
combs = 5 10 10 5 1
sum(combs) = 31
I want a way to store all 31 of these combinations in an array, for example a cell array, with n cells, within each is an array in which each row is a vector combination of the elements.
e.g. at k = 4:
combs{4} =
1 2 3 4
1 2 3 5
1 2 4 5
1 3 4 5
2 3 4 5
Is there an existing function that does this, or what would be the most simple approach to this?
Call nchoosek with a vector as first input, using arrayfun (or equivalently for) to loop over the number of picked elements:
n = 5;
combs = arrayfun(#(k) nchoosek(1:n,k), 1:n, 'UniformOutput', false);
Here is an approach using dec2bin , find and accumarray:
x = [1 2 3 4 5];
[a b] = find(dec2bin(1:2^numel(x)-1)=='1');
combs = accumarray(a,x(b),[],#(c){c});
Let z = [1 3 5 6] and by getting all the difference between each elements:
we get:
bsxfun(#minus, z', z)
ans =
0 -2 -4 -5
2 0 -2 -3
4 2 0 -1
5 3 1 0
I now want to order these values in ascending order and remove the duplicates. So:
sort(reshape(bsxfun(#minus, z', z),1,16))
ans =
Columns 1 through 13
-5 -4 -3 -2 -2 -1 0 0 0 0 1 2 2
Columns 14 through 16
3 4 5
C = unique(sort(reshape(bsxfun(#minus, z', z),1,16)))
C =
-5 -4 -3 -2 -1 0 1 2 3 4 5
But by looking at -5 in [-5 -4 -3 -2 -1 0 1 2 3 4 5],
how can I tell where -5 comes from. By reading myself the matrix,
0 -2 -4 -5
2 0 -2 -3
4 2 0 -1
5 3 1 0
I know it comes from z(1) - z(4), i.e. row 1 column 4.
Also 2 comes from both z(3) - z(2) and z(2) - z(1), which comes from two cases. Without reading the originally matrix itself, how can we know that the 2 in [-5 -4 -3 -2 -1 0 1 2 3 4 5] is originally in row 3 column 2 and row 2 column 1 of the original matrix?
So by looking at each element in [-5 -4 -3 -2 -1 0 1 2 3 4 5], how do we know, for example, where -5 comes from in the original matrix index efficiently. I want to know as I need to do operation on ,e.g.,-5 and two indices that produce this: for example, for each difference, say -5, i do (-5)*1*6, as z(1)- z(6) = -5. But for 2, I need to do 2*(3*2+2*1) as z(3) - z(2) = 2, z(2) - z(1) = 2 which is not distinct.
Thinking hard, I think i should not reshape bsxfun(#minus, z', z) to array. I will also create two index array such that I can do operations like (-5)*1*6 stated above effectively. However, this is easier said than done and I also have to take care of nondistinct sources. Or should I do the desired operations first?
Use the third output from unique. And don't sort, unique will do that for you.
[sortedOutput,~,linearIndices] = unique(reshape(bsxfun(#minus, z', z),[1 16]))
You can reconstruct the result from bsxfun like so:
distances = reshape(sortedOutput(linearIndices),[4 4]);
If you want to know where a certain value appears, you write
targetValue = -5;
targetValueIdx = find(sortedOutput==targetValue);
linearIndexIntoDistances = find(targetValueIdx==linearIndices);
[row,col] = ind2sub([4 4],linearIndexIntoDistances);
Because linearIndices is 1 wherever the first value in sortedOutput appears in the original vector.
If you save the result of bsxfun in an intermediate variable:
distances=bsxfun(#minus, z', z)
Then you can look for the values of C in distances using find iteratively.
[rows,cols]=find(C(i)==distances)
This will give all rows and cols if the values are repeated. You just need to then use them for your equation.
You can use accumarray to collect all row and column indices that correspond to the same value in the matrix of differences:
z = [1 3 5 6]; % data vector
zd = bsxfun(#minus, z.', z); % matrix of differences
[C, ~, ind] = unique(zd); % unique values and indices
[rr, cc] = ndgrid(1:numel(z)); % template for row and col indices
f = #(x){x}; % anonymous function to collect row and col indices
row = accumarray(ind, rr(:), [], f); % group row indices according to ind
col = accumarray(ind, cc(:), [], f); % same for col indices
For example, C(6) is value 0, which appears four times in zd, at positions given by row{6} and col{6}:
>> row{6}.'
ans =
3 2 1 4
>> col{6}.'
ans =
3 2 1 4
As you see, the results are not guaranteed to be sorted. If you need to sort them in linear order:
rowcol = cellfun(#(r,c)sortrows([r c]), row, col, 'UniformOutput', false);
so now
>> rowcol{6}
ans =
1 1
2 2
3 3
4 4
I'm not sure I've followed exactly but some points to consider:
unique will sort the data for you by default so you don't need to call sort first
unique actually has three outputs and you can recover your original vector (i.e. with duplicates) using the third output so
[C,~,ic] = unique(reshape(bsxfun(#minus, z', z),1,16))
now you can get back to bsxfun(#minus, z', z),1,16) by calling
reshape(C(ic), numel(z), numel(z))
You might be more interested in the second output of unique which tells you what index each unique value was at in your 1-by-16 vector. It really depends on what you're trying to do though. But with this you could get a list of row column pairs to match your unique values:
[rows, cols] = ndgrid(1:4);
coords = [rows(:), cols(:)];
[C, ia] = unique(reshape(bsxfun(#minus, z', z),1,16));
coords_pairs = coords(ia,:)
which results in
coords_pairs =
1 4
1 3
2 4
2 3
3 4
4 4
4 3
3 2
4 2
3 1
4 1
I have got a problem with splitting a vector by zeros.
I have a vector for example
v=[1 3 2 6 4 0 0 2 4 6 0 0 0 3 1]
I need to get vectors like
v1=[1 3 2 6 4]
v2=[2 4 6]
v3=[3 1]
Is there any way to do this by using MATLAB functions?
Of course I don't know of how many subvectors are included in main vector v and how many zeros delimits vectors.
I'm not a programmer and also I'm not a pro in MATLAB.
I know a procedural way to do this but want do it by MATLAB somehow.
I found a function A = strsplit(str,delimiter) but I don't have string I have a vector.
So I searched for conversion function. I found S = char(V) but when I executed it it crashed.
It's better to have the output as a cell array, not as separate variables. That way the output will be easier to handle.
Try this:
v = [1 3 2 6 4 0 0 2 4 6 0 0 0 3 1]; %// data
w = [false v~=0 false]; %// "close" v with zeros, and transform to logical
starts = find(w(2:end) & ~w(1:end-1)); %// find starts of runs of non-zeros
ends = find(~w(2:end) & w(1:end-1))-1; %// find ends of runs of non-zeros
result = arrayfun(#(s,e) v(s:e), starts, ends, 'uniformout', false); %// build result
Result (for your example):
>> result{:}
ans =
1 3 2 6 4
ans =
2 4 6
ans =
3 1
The strsplit() solution for a vector of whole numbers smaller than 9 (so a very specific solution, for a general solution see Luis Mendo's). Split and convert back to number:
res = strsplit(char(v), char(0));
res = cellfun(#(x) x - 0,res,'un',0);
celldisp(res)
res{1} =
1 3 2 6 4
res{2} =
2 4 6
res{3} =
3 1
I can't figure out how to create a vector based on condition on more than one other vectors. I have three vectors and I need values of one vector if values on other vectors comply to condition.
As an example below I would like to choose values from vector a if values on vector b==2 and values on vector c==0 obviously I expect [2 4]
a = [1 2 3 4 5 6 7 8 9 10];
b = [1 2 1 2 1 2 1 2 1 2];
c = [0 0 0 0 0 1 1 1 1 1]
I thought something like:
d = a(b==2) & a(c==0)
but I have d = 1 1 1 1 1 not sure why.
It seems to be basic problem but I can find solution for it.
In your case you can consider using a(b==2 & c==0)
Use ismember to find the matching indices along the rows after concatenating b and c and then index to a.
Code
a(ismember([b;c]',[2 0],'rows'))
Output
ans =
2
4
You may use bsxfun too for the same result -
a(all(bsxfun(#eq,[b;c],[2 0]'),1))
Or you may just tweak your method to get the correct result -
a(b==2 & c==0)
I’ve a matrix A = (4*4) and a cell array B {4,1}. I’d like to find all the values of B in A, searching row by row and after I’d like to delete the correspondent column associated to this particular value. I’ve a problem using bsxfun o cellfun and find function with a matrix and a cell array. I‘ve tried to convert the cell array into a matrix but I don’t have more the exact correspondence.
For example:
A =
1 5 10 23
2 4 2 18
3 3 5 14
1 9 10 4
B =
1
2 4
3 3 14
1
To obtain:
C =
10
2
5
10
Thanks in advance,
L.
Here's how:
C = cellfun(#(x, y){sparse(1,find(ismember(x,y),numel(y)),true,1,size(A,2))}, ...
mat2cell(A, ones(size(A, 1), 1), size(A, 2)), B(:));
C = A(:, all(~vertcat(C{:})));
The cellfun is fed with two cell arrays: the first one contains the rows of A and second one is B. The anonymous function is the tricky part; it operates on a pair of two corresponding rows as follows:
It employs ismember to check which columns in A contain any of the elements in B.
It uses find to pick only the first N ones, with respect to the number of elements in the B.
It uses sparse as a fancy way of zeroing out the rest of the elements.
For your example it would look like this:
A = [1 5 10 23; 2 4 2 18; 3 3 5 14; 1 9 10 4];
B = {1; [2 4]; [3 3 14]; 1};
C = cellfun(#(x, y){sparse(1,find(ismember(x,y),numel(y)),true,1,size(A,2))}, ...
mat2cell(A, ones(size(A, 1), 1), size(A, 2)), B(:));
which yields:
C =
{
[1 0 0 0]
[1 1 0 0]
[1 1 0 1]
[1 0 0 0]
}
After that, it's a simple matter of logical indexing to pick the resulting columns:
C = A(:, all(~vertcat(C{:})));
which in this case should be:
C =
10
2
5
10