Having matrix A (n*2) as the source and B as a vector containing a subset of elements A, I'd like to find the row index of items.
A=[1 2;1 3; 4 5];
B=[1 5];
F=arrayfun(#(x)(find(B(x)==A)),1:numel(B),'UniformOutput',false)
gives the following outputs in a cell according to this help page
[2x1 double] [6]
indicating the indices of all occurrence in column-wise. But I'd like to have the indices of rows. i.e. I'd like to know that element 1 happens in row 1 and row 2 and element 5 happens just in row 3. If the indices were row-wise I could use ceil(F{x}/2) to have the desired output. Now with the variable number of rows, what's your suggested solution? As it may happens that there's no complete inclusion tag 'rows' in ismember function does not work. Besides, I'd like to know all indices of specified elements.
Thanks in advance for any help.
Approach 1
To convert F from its current linear-index form into row indices, use mod:
rows = cellfun(#(x) mod(x-1,size(A,1))+1, F, 'UniformOutput', false);
You can combine this with your code into a single line. Note also that you can directly use B as an input to arrayfun, and you avoid one stage of indexing:
rows = arrayfun(#(x) mod(find(x==A)-1,size(A,1))+1, B(:), 'UniformOutput', false);
How this works:
F as given by your code is a linear index in column-major form. This means the index runs down the first column of B, the begins at the top of the second column and runs down again, etc. So the row number can be obtained with just a modulo (mod) operation.
Approach 2
Using bsxfun and accumarray:
t = any(bsxfun(#eq, B(:), reshape(A, 1, size(A,1), size(A,2))), 3); %// occurrence pattern
[ii, jj] = find(t); %// ii indicates an element of B, and jj is row of A where it occurs
rows = accumarray(ii, jj, [], #(x) {x}); %// group results according to ii
How this works:
Assuming A and B as in your example, t is the 2x3 matrix
t =
1 1 0
0 0 1
The m-th row of t contains 1 at column n if the m-th element of B occurs at the n-th row of B. These values are converted into row and column form with find:
ii =
1
1
2
jj =
1
2
3
This means the first element of B ocurrs at rows 1 and 2 of A; and the second occurs at row 3 of B.
Lastly, the values of jj are grouped (with accumarray) according to their corresponding value of ii to generate the desired result.
One approach with bsxfun & accumarray -
%// Create a match of B's in A's with each column of matches representing the
%// rows in A where there is at least one match for each element in B
matches = squeeze(any(bsxfun(#eq,A,permute(B(:),[3 2 1])),2))
%// Get the indices values and the corresponding IDs of B
[indices,B_id] = find(matches)
%// Or directly for performance:
%// [indices,B_id] = find(any(bsxfun(#eq,A,permute(B(:),[3 2 1])),2))
%// Accumulate the indices values using B_id as subscripts
out = accumarray(B_id(:),indices(:),[],#(x) {x})
Sample run -
>> A
A =
1 2
1 3
4 5
>> B
B =
1 5
>> celldisp(out) %// To display the output, out
out{1} =
1
2
out{2} =
3
With arrayfun,ismember and find
[r,c] = arrayfun(#(x) find(ismember(A,x)) , B, 'uni',0);
Where r gives your desired results, you could also use the c variable to get the column of each number in B
Results for the sample input:
>> celldisp(r)
r{1} =
1
2
r{2} =
3
>> celldisp(c)
c{1} =
1
1
c{2} =
2
Related
I have a table like the above attachment. Column A and Column B contains some elements in terms of cell array. I want to create the third column (Level) as the resultant column; based on the following logic.
The row for which, value of cell A = value of cell B will be labeled1. (In the 3rd row, the value of column A=value of column B= 3, hence labeled 1).
Next, the preceding value will be removed from all
the cells of column A; and the step 1 will be repeated until all the
rows are labeled. (In the second step, 3 will be removed from all
the cells, hence both row 1 and row 2 will be labeled as 2;In the
final step, elements {1,2} will be further removed from the last row
resulting the level as 3 )
I am using cell2mat and setdiff functions to compare the values across the cells, but I am not able to frame the above 2 logical steps to run my code successfully. I have just started learning MATLAB, Any help will be highly appreciated.
Here's the simplest answer I could come up with, using a single while loop and assuming the cells of A and B contain row vectors:
Level = zeros(size(A));
index = cellfun(#isequal, A, B);
while any(index)
Level(index) = max(Level)+1;
A = cellfun(#(c) {setdiff(c, unique([A{index}]))}, A);
index = cellfun(#isequal, A, B);
end
The above code first initializes a matrix of zeroes Level the same size as A to store the level values. Then it finds a logical index index of where there are matching cell contents between A and B using cellfun and isequal. It will continue to loop as long as there are any matches indicated by index. The corresponding indices in Level are set to the current maximum value in Level plus one. All the matching cell contents from A are concatenated and the unique values found by unique([A{index}]). A set difference operation is then used (along with cellfun) to remove the matching values from each cell in A, overwriting A with the remaining values. A new index for matches is then computed and the loop restarts.
Given the following sample data from your question:
A = {[1 2 3]; [2 3]; 3; [1 2 3 4]};
B = {[1 2]; 2; 3; 4};
The code returns the expected level vector:
Level =
2
2
1
3
Not my best work, i think it is possible to get rid of the inner loop.
% your testdata
A = {[1 2 3]
[2 3]
3
[1,2,4]};
B = {[1 2]
2
3
4};
Level = NaN(numel(B),1);
temp = A; % copy of A that we are going to remove elements from
k = 0; % loop couter
while any(isnan(Level)) % do until each element of Level is not NaN
k = k+1; % increment counter by 1
% step 1
idx = find(cellfun(#isequal,temp,B)); % determine which cells are equal
Level(idx) = k; % set level of equal cells
% step 2
for k = 1:numel(idx) % for each cell that is equal
%remove values in B from A for each equal cell
temp = cellfun(#setdiff,temp,repmat(B(idx(k)),numel(B),1),'UniformOutput',0);
end
end
I would like to know how to count rows in an matrix in such a way that gives an output for each colum. for example:
X=[1 1 1;
5 5 5]
I would like to find a command that when I input the matrix X the answers is [2 2 2], so that it counts the number of rows per column.
I have already found nunel(X) but the answer is a scalar numel(X)=6, whereas I need per column.
size(X,1) will give you the number of rows in the matrix (a scalar). a matrix has only one number of rows, i.e. each column has the same number of rows.
however if you still want the number of rows per each column you can use:
X = [1 1 1;
5 5 5];
nrows = size(X,1);
ncols = size(X,2);
nrowsPerCol = repmat(nrows, [1 ncols]) % [2 2 2]
Each matrix object in MATLAB has height and width property.
In other words: each column has the same number of rows.
To get this value, use MATLAB's size function:
[numOfRows, numOfCols] = size(X);
I am working with a n x 1 matrix, A, that has repeating values inside it:
A = [0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4]
which correspond to an n x 1 matrix of B values:
B = [2;4;6;8;10; 3;5;7;9;11; 4;6;8;10;12; 5;7;9;11;13]
I am attempting to produce a generalised code to place each repetition into a separate column and store it into Aa and Bb, e.g.:
Aa = [0 0 0 0 Bb = [2 3 4 5
1 1 1 1 4 5 6 7
2 2 2 2 6 7 8 9
3 3 3 3 8 9 10 11
4 4 4 4] 10 11 12 13]
Essentially, each repetition from A and B needs to be copied into the next column and then deleted from the first column
So far I have managed to identify how many repetitions there are and copy the entire column over to the next column and then the next for the amount of repetitions there are but my method doesn't shift the matrix rows to columns as such.
clc;clf;close all
A = [0;1;2;3;4;0;1;2;3;4;0;1;2;3;4;0;1;2;3;4];
B = [2;4;6;8;10;3;5;7;9;11;4;6;8;10;12;5;7;9;11;13];
desiredCol = 1; %next column to go to
destinationCol = 0; %column to start on
n = length(A);
for i = 2:1:n-1
if A == 0;
A = [ A(:, 1:destinationCol)...
A(:, desiredCol+1:destinationCol)...
A(:, desiredCol)...
A(:, destinationCol+1:end) ];
end
end
A = [...] retrieved from Move a set of N-rows to another column in MATLAB
Any hints would be much appreciated. If you need further explanation, let me know!
Thanks!
Given our discussion in the comments, all you need is to use reshape which converts a matrix of known dimensions into an output matrix with specified dimensions provided that the number of elements match. You wish to transform a vector which has a set amount of repeating patterns into a matrix where each column has one of these repeating instances. reshape creates a matrix in column-major order where values are sampled column-wise and the matrix is populated this way. This is perfect for your situation.
Assuming that you already know how many "repeats" you're expecting, we call this An, you simply need to reshape your vector so that it has T = n / An rows where n is the length of the vector. Something like this will work.
n = numel(A); T = n / An;
Aa = reshape(A, T, []);
Bb = reshape(B, T, []);
The third parameter has empty braces and this tells MATLAB to infer how many columns there will be given that there are T rows. Technically, this would simply be An columns but it's nice to show you how flexible MATLAB can be.
If you say you already know the repeated subvector, and the number of times it repeats then it is relatively straight forward:
First make your new A matrix with the repmat function.
Then remap your B vector to the same size as you new A matrix
% Given that you already have the repeated subvector Asub, and the number
% of times it repeats; An:
Asub = [0;1;2;3;4];
An = 4;
lengthAsub = length(Asub);
Anew = repmat(Asub, [1,An]);
% If you can assume that the number of elements in B is equal to the number
% of elements in A:
numberColumns = size(Anew, 2);
newB = zeros(size(Anew));
for i = 1:numberColumns
indexStart = (i-1) * lengthAsub + 1;
indexEnd = indexStart + An;
newB(:,i) = B(indexStart:indexEnd);
end
If you don't know what is in your original A vector, but you do know it is repetitive, if you assume that the pattern has no repeats you can use the find function to find when the first element is repeated:
lengthAsub = find(A(2:end) == A(1), 1);
Asub = A(1:lengthAsub);
An = length(A) / lengthAsub
Hopefully this fits in with your data: the only reason it would not is if your subvector within A is a pattern which does not have unique numbers, such as:
A = [0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0;]
It is worth noting that from the above intuitively you would have lengthAsub = find(A(2:end) == A(1), 1) - 1;, But this is not necessary because you are already effectively taking the one off by only looking in the matrix A(2:end).
If I have a Cell array 2*2 where A{i,j} is a matrix, and I have two vectors v=1:2,c=1:2.
I want A(v,c) to return only A{1,1} and A{2,2} but matlab returns every combination of the two(aka also returns A{1,2} and A{2,1}).
Is there a way without using loops or cellfun ?
What I suspect you are doing is something like this:
B = A(v, c);
When you specify vectors to index into A, it finds the intersection of coordinates and gives you those elements. As such, with your indexing you are basically returning all of the elements in A.
If you want just the top left and lower right elements, use sub2ind instead. You can grab the column-major indices of those locations in your cell array, then slice into your cell array with these indices:
ind = sub2ind(size(A), v, c);
B = A(ind);
Example
Let's create a sample 2 x 2 cell array:
A = cell(2,2);
A{1,1} = ones(2);
A{1,2} = 2*ones(2);
A{2,1} = 3*ones(2);
A{2,2} = 4*ones(2);
Row 1, column 1 is a 2 x 2 matrix of all 1s. Row 1, column 2 is a 2 x 2 matrix of 2s, row 2 column 1 is a 2 x 2 matrix of all 3s and the last entry is a 2 x 2 matrix of all 4s.
With v = 1:2; c=1:2;, running the above code gives us:
>> celldisp(B)
B{1} =
1 1
1 1
B{2} =
4 4
4 4
As you can see, we picked out the top left and bottom right entries exactly.
Minor Note
If it's seriously just a cell array of 2 x 2, and you only want to pick out the top left and lower right elements, you can just do:
B = A([1 4]);
sub2ind would equivalently return 1 and 4 as the column major indices for the top left and lower right elements. This avoids the sub2ind call and still achieves what you want.
Suppose now I have two vectors of same length:
A = [1 2 2 1];
B = [2 1 2 2];
I would like to create a matrix C whose dim=m*n, m=max(A), n=max(B).
C = zeros(m,n);
for i = 1:length(A)
u = A(i);
v = B(i);
C(u,v)=C(u,v)+1;
end
and get
C =[0 2;
1 1]
More precisely, we treat the according indices in A and B as rows and columns in C, and C(u,v) is the number of elements in {k | A(i)=u and B(i)=v, i = 1,2,...,length(A)}
Is there a faster way to do that?
Yes. Use sparse. It assembles (i.e., sums up) the matrix values for repeating row-column pairs for you. You need an additional vector with the values that will be assembled into the matrix entries. If you use ones(size(A)), you will have exactly what you need - counting of repeated row-column pairs
spA=sparse(A, B, ones(size(A)));
full(spA)
ans =
0 2
1 1
The same can be obtained by simply passing scalar 1 to sparse function instead of a vector of values.
For matrices that have a large number of zero entries this is absolutely crucial that you use sparse storage. Another function you could use is accumarray. It can essentially do the same thing, but also works on dense matrix structure:
AA=accumarray([A;B]', 1);
AA =
0 2
1 1
You can pass size argument to accumarray if you want to create a matrix of specific size
AA=accumarray([A;B]', 1, [2 3]);
AA =
0 2 0
1 1 0
Note that you can actually also make it produce sparse matrices, and use a different operator in assembly (i.e., not necessarily a sum)
AA=accumarray([A;B]', 1, [2 3], #sum, 0, true)
will produce a sparse matrix (last parameter set to true) using sum for assembly and 0 as a fill value, i.e. a value which is used in cases a given row-column pair does not exist in A/B.