I would like to remove some columns and rows from a big matrix. Those are the columns and the rows which have all zeros values. Is there any function in MATLAB that can do it for you quite fast? My matrices are sparse. I am doing this way:
% To remove all zero columns from A
ind = find(sum(A,1)==0) ;
A(:,ind) = [] ;
% To remove all zeros rows from A
ind = find(sum(A,2)==0) ;
A(ind,:) = [] ;
It would be nice to have a line of code for this as I may do this kind of task repeatedly. Thanks
A single line of code would be:
A=A(any(A,2),any(A,1))
There is no need to use find like you did, you can directly index using logical vectors. The any function finds the rows or columns with any non-zero elements.
1 Dimension:
I'll first show a simpler example based on another duplicate question, asking to to remove only the rows containing zeros elements.
Given the matrix A=[1,2;0,0];
To remove the rows of 0, you can:
sum the absolute value of each rows (to avoid having a zero sum from a mix of negative and positive numbers), which gives you a column vector of the row sums.
keep the index of each line where the sum is non-zero.
in code:
A=[1,2;0,0];
% sum each row of the matrix, then find rows with non-zero sum
idx_nonzerolines = sum(abs(A),2)>0 ;
% Create matrix B containing only the non-zero lines of A
B = A(idx_nonzerolines,:) ;
will output:
>> idx_nonzerolines = sum(abs(A),2)>0
idx_nonzerolines =
1
0
>> B = A(idx_nonzerolines,:)
B =
1 2
2 Dimensions:
The same method can be used for 2 dimensions:
A=[ 1,2,0,4;
0,0,0,0;
1,3,0,5];
idx2keep_columns = sum(abs(A),1)>0 ;
idx2keep_rows = sum(abs(A),2)>0 ;
B = A(idx2keep_rows,idx2keep_columns) ;
outputs:
>> B = A(idx2keep_rows,idx2keep_columns)
B =
1 2 4
1 3 5
Thanks to #Adriaan in comments for spotting the edge case ;)
Related
I have a vector T of length n and m other vectors of the same length with 0 or 1 used as condition to select elements of T. The condition vectors are combined into a matrix I of size n x m.
Is there a one liner to extract a matrix M of values from Tsuch that the i-th column of M are those elements in T that are selected by the condition elements of the i-th column in I?
Example:
T = (1:10)'
I = mod(T,2) == 0
T(I)'
yields
2 4 6 8 10
However
I = mod(T,2:4) == 0
T(I)'
yields an error in the last statement. I see that the columns might select a different number of elements which results in vectors of different lengths (as in the example). However, even this example doesn't work:
I = zeros(10,2)
I(:,1) = mod(T,2)==0
I(:,2) = mod(T,2)==1
Is there any way to achieve the solution in a one liner?
The easiest way I can think of to do something like this is to take advantage of the element-wise multiplication operator .* with your matrix I. Take this as an example:
% these lines are just setup of your problem
m = 10;
n = 10;
T = [1:m]';
I = randi([0 1], m, n);
% 1 liner to create M
M = repmat(T, 1, n) .* I;
What this does is expand T to be the same size as I using repmat and then multiplies all the elements together using .*.
Here is a one linear solution
mat2cell(T(nonzeros(bsxfun(#times,I,(1:numel(T)).'))),sum(I))
First logical index should be converted to numeric index for it we multiply T by each column of I
idx = bsxfun(#times,I,(1:numel(T)).');
But that index contain zeros we should extract those values that correspond to 1s in matrix I:
idx = nonzeros(idx);
Then we extract repeated elements of T :
T2 = T(idx);
so we need to split T2 to 3 parts size of each part is equal to sum of elements of corresponding column of I and mat2cell is very helpful
result = mat2cell(T2,sum(I));
result
ans =
{
[1,1] =
2
4
6
8
10
[2,1] =
3
6
9
[3,1] =
4
8
}
One line solution using cellfun and mat2cell
nColumns = size(I,2); nRows = size(T,1); % Take the liberty of a line to write cleaner code
cellfun(#(i)T(i),mat2cell(I,nRows,ones(nColumns,1)),'uni',0)
What is going on:
#(i)T(i) % defines a function handle that takes a logical index and returns elements from T for those indexes
mat2cell(I,nRows,ones(nColumns,1)) % Split I such that every column is a cell
'uni',0 % Tell cellfun that the function returns non uniform output
For example, I have a matrix A with 400 lines and 4000 columns which is mostly composed by zeros but have three ones. I do need to know exactly what position these ones occupy, let's say A(30,4000), A(400,3050) and A(50,200).
Simply do:
[row,col] = find(A);
This will give you the row and column locations of all non-zero entries. row would contain all of the row locations and col would contain all of the column locations as N x 1 vectors, where N is the number of non-zero elements. In your example, the above output would be equivalent to:
row = [30; 400; 50];
col = [4000; 3050; 200];
You just need to use find(.) function.
For example if you have:
q=[1 2 3;1 2 4];
[r c]=find(q==2)
in this case r includes rows and c includes columns of desired value.
I have a vector whose elements identify the indices (per column) that I need to set in a different matrix. Specifically, I have:
A = 7
1
2
and I need to create a matrix B with some number of rows of zeros, except for the elements identified by A. In other words, I want B:
B = zeros(10, 3); % number of rows is known; num columns = size(A)
B(A(1), 1) = 1
B(A(2), 2) = 1
B(A(3), 3) = 1
I would like to do this without having to write a loop.
Any pointers would be appreciated.
Thanks.
Use linear indexing:
B = zeros(10, 3);
B(A(:).'+ (0:numel(A)-1)*size(B,1)) = 1;
The second line can be written equivalently with sub2ind (may be a little slower):
B(sub2ind(size(B), A(:).', 1:numel(A))) = 1;
I have a quite big (107 x n) matrix X. Within these n columns, each three columns belong to each other. So, the first three columns of matrix X build a block, then columns 4,5,6 and so on.
Within each block, the first 100 row elements of the first column are important X(1:100,1:3:end). Whenever in this first column the number of zeros or NaNs is greater or equal 20, it should delete the whole block.
Is there a way to do this without a loop?
Thanks for any advice!
Assuming the number of columns of the input to be a multiple of 3, there could be two approaches here.
Approach #1
%// parameters
rl = 100; %// row limit
cl = 20; %// count limit
X1 = X(1:rl,1:3:end) %// Important elements from input
match_mat = isnan(X1) | X1==0 %// binary array of matches
match_blk_id = find(sum(match_mat)>=cl) %// blocks that satisfy requirements
match_colstart = (match_blk_id-1).*3+1 %// start column indices that satisfy
all_col_ind = bsxfun(#plus,match_colstart,[0:2]') %//'columns indices to be removed
X(:,all_col_ind)=[] %// final output after removing to be removed columns
Or if you prefer "compact" codes -
X1 = X(1:rl,1:3:end);
X(:,bsxfun(#plus,(find(sum(isnan(X1) | X1==0)>=cl)-1).*3+1,[0:2]'))=[];
Approach #2
X1 = X(1:rl,1:3:end)
match_mat = isnan(X1) | X1==0 %// binary array of matches
X(:,repmat(sum(match_mat)>=cl,[3 1]))=[] %// Find matching blocks, replicate to
%// next two columns and remove them from X
Note: If X is not a multiple of 3, use this before using the codes - X = [X zeros(size(X,1) ,3 - mod(size(X,2),3))].
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)