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Is there any general way to remove NaNs from a matrix?
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Closed 6 years ago.
New to MATLAB, any help with this would be appreciated.
I have a dataset that is 1000 elements in 1 column, and most of the elements are numbers but some are NaN's. Is there a way I can, 1. Find them, and 2. Remove them and put them in a variable (or just remove them)?
Would I have to do this the reverse way and find and remove the non-NaN's (the numbers) and store them in a variable?
Use logical indexing to extract the elements that are not NaN and then store them anywhere you like. Here's how it works. If x is your column vector containing NaN, y = ~isnan(x) will give a logical vector y such that |y|=|x| and y(i) is 1 iff x(1) is not NaN. You can use this logical vector to extract non NaN elements:
x = [1 2 3 NaN 5 6 NaN NaN 9];
y = ~isnan(x); % now y is [1 1 1 0 1 1 0 0 1]
x = x(y) % now x is [1 2 3 5 6 9]
Logical indexing is powerful and efficient. You could also say:
x = [1 2 3 NaN 5 6 NaN NaN 9];
x(isnan(x)) = -1 % now x is [1 2 3 -1 5 6 -1 -1 9]
For more information on logical indexing see the official Matlab documentation here.
A option equivalent to #sadeghmir's answer:
x = [1 2 3 NaN 5 6 NaN NaN 9];
x(x==NaN)=[];
>x
1 2 3 5 6 9
Related
I have a 3x3x3 matrix which contains a particular set of elements that I would like to extract. However, I would like for the elements to be ordered in a matrix after selecting them. An example is:
a(1,:,:)=1*[1 2 3; 4 5 6; 7 8 9];
a(2,:,:)=2*[1 2 3; 4 5 6; 7 8 9];
a(3,:,:)=3*[1 2 3; 4 5 6; 7 8 9];
a(a>1.0)
The condition a(a>1.0) gives me a vector of elements, but is there a way to order them in a matrix following their original ordering?
What would you like to do to the elements that don't satisfy your criteria?
You could do something like a(a<=1) = nan;
Suppose 2-D matrix for simplicity:
a = [1 2 3; ...
4 5 6];
Let's take only even values and keep them in their original shape:
a(mod(a, 2) == 0)
You might want:
2
4 6
However, in the world of matrix, is there such a matrix which has empty space? Yes, a sparse matrix. But, you must note that a sparse matrix is filled with 0 not just missing.
So, my suggestion is to replace other values with NaN
b = a; % just make a duplicate
b(mod(b, 2) == 1) = nan
b =
nan 2 nan
4 nan 6
There is likely a quick little trick for this problem, but I cannot find it. I would like code to produce the matrix in the following image:
Here's a way using logical indexing:
n = 4;
A = nan(n);
A(~eye(n)) = 1:n^2-n; %// Only replace values *not* on diagonal
A = A.'
A =
NaN 1 2 3
4 NaN 5 6
7 8 NaN 9
10 11 12 NaN
Here's one way using triu and tril:
n = 4;
A = reshape(1:n*(n-1),n-1,n).';
z = zeros(n,1);
A = [tril(A,-1) z]+[z triu(A)]+diag(NaN(n,1));
which, in this case for a 4-by-4 matrix, returns
A =
NaN 1 2 3
4 NaN 5 6
7 8 NaN 9
10 11 12 NaN
Here's another way just using reshape:
n = 4;
A = [reshape(1:n*(n-1),n,n-1);NaN(1,n-1)];
A = reshape([NaN;A(:)],n,n).'
Here's another way:
n = 4; %// matrix size
x = 1-eye(n);
x(:) = cumsum(x(:));
x = x.' + diag(NaN(1,n));
You can start from a nan matrix, find the linear indices of the diagonal, then fill up the rest of the elements with an incrementing range:
n=4;
A=nan(n);
inds=setdiff(1:n^2,sub2ind([n,n],1:n,1:n));
A(inds)=1:numel(inds);
A=A.'; %' transpose to get the matrix we need
The transpose in the end is necessary, as linear indexing goes column-first, but your specifics need a row-first assignment of matrix elements.
Result:
>> A
A =
NaN 1 2 3
4 NaN 5 6
7 8 NaN 9
10 11 12 NaN
I have a 3x3 matrix, populated with NaN and values of a variable:
NaN 7 NaN
5 NaN 0
NaN NaN 4
matrix = [NaN 7 NaN; 5 NaN 0; NaN NaN 4]
I would like to get the row and column numbers of non-NaN cells and put them in a matrix together with the value of the variable. That is, I would like to obtain the following matrix:
row col value
1 2 7
2 1 5
2 3 0
3 3 4
want = [1 2 7; 2 1 5; 2 3 0; 3 3 4]
Any help would be highly appreciated.
This can be done without loops:
[jj, ii, kk] = find((~isnan(matrix).*(reshape(1:numel(matrix), size(matrix)))).');
result = [ii jj matrix(kk)];
The trick is to multiply ~isnan(matrix) by a matrix of indices so that the third output of find gives the linear index of non-NaN entries. The transpose is needed to have the same order as in the question.
The following should work!
[p,q]=find(~isnan(matrix)) % Loops through matrix to find indices
want = zeros(numel(p),3) % three columns you need with same number of rows as p
for i=1:numel(p)
want[i,:] = [p(i) q(i) matrix(p(i), matrix(i))]
end
Should give you the correct result which is:
2 1 5
1 2 7
2 3 0
3 3 4
If you don't mind the ordering of the rows, you can use a simplified version of Luis Mendo's answer:
[row, col] = find(~isnan(matrix));
result = [row(:), col(:), matrix(~isnan(matrix))];
Which will result in:
2 1 5
1 2 7
2 3 0
3 3 4
I would like to get the minimum nonzero values per row in a sparse matrix. Solutions I found for dense matrices suggested masking out the zero values by setting them to NaN or Inf. However, this obviously doesn't work for sparse matrices.
Ideally, I should get a column vector of all the row-wise minima, as I would get with
minValues = min( A, [], 2);
Except, obviously, using min leaves me with an all-zeros column vector due to the sparsity. Is there a solution using find?
This is perfect for accumarray. Consider the following sparse matrix,
vals = [3 1 1 9 7 4 10 1]; % got this from randi(10,1,8)
S = sparse([1 3 4 4 5 5 7 9],[2 2 3 6 7 8 8 11],vals);
Get the minimum value for each row, assuming 0 for empty elements:
[ii,jj] = find(S);
rowMinVals = accumarray(ii,nonzeros(S),[],#min)
Note that rows 4 and 5 of rowMinVals, which are the only two rows of S with multiple nonzero values are equal to the min of the row:
rowMinVals =
3
0
1
1 % min([1 9]
4 % min([7 4]
0
10
0
1
If the last row(s) of your sparse matrix do not contain any non-zeros, but you want your min row value output to reflect that you have numRows, for example, change theaccumarray command as follows,
rowMinVals = accumarray(ii,nonzeros(S),[numRows 1],#min).
Also, perhaps you also want to avoid including the default 0 in the output. One way to handle that is to set the fillval input argument to NaN:
rowMinVals = accumarray(ii,nonzeros(S),[numRows 1],#min,NaN)
rowMinVals =
3
NaN
1
1
4
NaN
10
NaN
1
NaN
NaN
NaN
Or you can keep using a sparse matrix with the fifth input argument, issparse:
>> rowMinVals = accumarray(ii,nonzeros(S),[],#min,[],true)
rowMinVals =
(1,1) 3
(3,1) 1
(4,1) 1
(5,1) 4
(7,1) 10
(9,1) 1
There is one NaN element per row, I want to remove it.
A=[NaN 1 2;
3 NaN 4;
NaN 5 6];
The desired output is:
[1 2;
3 4;
5 6]
A = [NaN 1 2 ; 3 NaN 4; NaN 5 6]
sz = size(A);
B = reshape(A', size(A,1)*size(A,2), 1);
B(isnan(B)) = [];
B = reshape(B, sz(2)-1, sz(1))'
I thought it could be done in one line, but I was wrong. See solution below:
Given (added row helps me debug my indexing below):
>> A = [NaN 1 2 ; 3 NaN 4; NaN 5 6; 7 8 NaN]
A =
NaN 1 2
3 NaN 4
NaN 5 6
7 8 NaN
Then:
>> Atrans = A';
>> B = reshape( Atrans(~isnan(Atrans)) ,[],size(Atrans,2))'
B =
1 2
3 4
5 6
7 8
Incidentally, the Matlab idiom of performing a simple logical check on an array within an logical indexing operation is very common and incredibly useful. The archetypical example is:
>> x(x>0) %This returns a 1D column vector of all values of x
%which are greater than 0, regardless of the initial
%size of x. Multidimensional inputs are unwrapped
%column-first
Everything else above is size and dimension handling.
Here it is - please note that the code is not robust. It assumes that indeed in every row there is a NaN element.
While it is not a vectorized solution, it has other advantages - like a clear code.
for i=1:size(A,1)
x = A(i,:);
x(isnan(x)) = [];
B(i,:) = x;
end
B
B =
1 2
3 4
5 6