I have a simple example dataset below:
a =
1 1 1 NaN NaN
1 1 1 NaN NaN
1 1 1 1 NaN
1 1 1 1 1
1 1 1 1 1
I want to work out the average cumulative value per row. However, cumsum gives the following output:
cumsum(a)
1 1 1 NaN NaN
2 2 2 NaN NaN
3 3 3 1 NaN
4 4 4 2 1
5 5 5 3 2
Then calculating a row mean gives:
nanmean(a,2)
1
2
2.5
3
4
I want to be able to account for the fact that different columns start later i.e. the row mean values for rows (3:5) are reduced with respect to their true values due to low numbers in columns (4:5).
I want to achieve this by replacing the last NaN above the first numeric element in each column in the matrix (a) with the mean of the other columns in that row in the cumulative matrix.This would need to be done iteratively to reflect the changing values in the cumulative matrix. So the new matrix would first look as follows:
(a)
1 1 1 NaN NaN
1 1 1 *2* NaN
1 1 1 1 NaN
1 1 1 1 1
1 1 1 1 1
which would lead to:
cumsum(a)
1 1 1 NaN NaN
2 2 2 2 NaN
3 3 3 3 NaN
4 4 4 4 1
5 5 5 5 2
and then iteratively, (a) would equal:
(a)
1 1 1 NaN NaN
1 1 1 2 NaN
1 1 1 1 *3*
1 1 1 1 1
1 1 1 1 1
which would lead to:
cumsum(a)
1 1 1 NaN NaN
2 2 2 2 NaN
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
which would give the desired row means values as:
nanmean(a,2)
1
2
3
4
5
There may be a way to further vectorise this. However, I think that because each row depends on the previous values, you have to update the matrix row-by-row as follows:
% Cycle through each row in matrix
for i = 1:length(a)
if i > 1
% This makes elements equal to the sum of themselves and above element
% Equivalent outcome to cumsum
a(i,:) = a(i,:) + a(i-1,:);
end
% Replace all NaN values in the row with the average of the non-NaN values
a(i,isnan(a(i,:))) = mean(a(i,~isnan(a(i,:))));
end
This replicates your input and output examples. It doesn't replicate all your iterative steps, it in fact uses many less steps, only 5 (number of rows) for entire operation.
Edit: equally,
for i = 1:length(a)
% Replace all NaN values in the row with the average of the non-NaN values
a(i,isnan(a(i,:))) = mean(a(i,~isnan(a(i,:))));
end
a = cumsum(a);
Related
I am trying to create a function that would replace repeated values in each row of a matrix with NaNs. My code works except that I can't seem to get it to apply to every row. It applies the code to each row, but will only accept "n" as the number of columns, not the number of rows, so if there are more rows than columns then it won't do every row. For example, if I have 8 rows and 6 columns it will only let me put in 6 for n, not 8. Yet it applies the code to the rows but only the first 6 rows. Here is a working example:
X = [1 2 3 4 4 4; 1 2 3 4 5 5; 1 2 2 2 2 2; 1 2 3 3 3 3;
1 1 1 1 1 1; 1 2 3 4 4 4; 1 2 3 3 3 3; 1 2 2 2 2 2];
n=6
nanfill = nan(1,n);
for i=1:n
temp = [unique(X(i,:)), nanfill];
X(i,:) = temp(1:n);
end
It gives me this for my original matrix:
1 2 3 4 4 4
1 2 3 4 5 5
1 2 2 2 2 2
1 2 3 3 3 3
1 1 1 1 1 1
1 2 3 4 4 4
1 2 3 3 3 3
1 2 2 2 2 2
Then this is the output I get for the new matrix after running my code:
1 2 3 NaN NaN NaN
1 2 3 4 NaN NaN
1 2 NaN NaN NaN NaN
1 2 3 NaN NaN NaN
1 NaN NaN NaN NaN NaN
1 2 3 4 NaN NaN
1 2 3 3 3 3
1 2 2 2 2 2
If I attempt to put in the number of rows, 8, for "n" then it gives me the following error:
Unable to perform assignment because the size of the left side is 1-by-6 and the size
of the right side is 1-by-8.
If the values in each row are always non-decreasing, as in your example, you can use the following. It computes the difference between each entry and the one to its left, and writes NaN if that difference is zero:
X([false(size(x,1),1) ~diff(X,[],2)]) = NaN;
That is because your matrix's rows have elements in them. You can't assign a vector that is 8 elements long into a shorter one. Better would be to check the length of the rows so that they will definitely match:
X = [1 2 3 4 4 4; 1 2 3 4 5 5; 1 2 2 2 2 2; 1 2 3 3 3 3;
1 1 1 1 1 1; 1 2 3 4 4 4; 1 2 3 3 3 3; 1 2 2 2 2 2];
n=size(X,2) % <- # of columns (length of one row)
nanfill = nan(1,n);
for i=1:n
temp = [unique(X(i,:)), nanfill];
X(i,:) = temp(1:n);
end
I have a simple example matrix as follows: (The actual matrix I'm working on is 674x11 and is not simply all '1' elements).
a =
1 1 1 NaN NaN
1 1 1 NaN NaN
1 1 1 1 NaN
1 1 1 1 1
1 1 1 1 1
I want to create a cumulative matrix which accounts for the fact that numeric elements start in each column at different rows. I want to achieve this by replacing the NaN value above the first numeric element in each column with the mean of that row.
So instead of:
cumsum(a)=
1 1 1 NaN NaN
2 2 2 NaN NaN
3 3 3 1 NaN
4 4 4 2 1
5 5 5 3 2
what I want to achieve is:
cumsum(a) =
1 1 1 NaN NaN
2 2 2 2 NaN
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
where element (2,4) is the mean of a(2,1:3) and element (3,5) is the mean of a(3,1:4).
You can compute the mean of each row (ignoring the NaN values) by using nanmean. We can then use find to identify the row in which each NaN is and replace the values with the mean of that row. Then we can follow that up with the cumsum operation
% Get the rows of each NaN value
bool = isnan(a);
[row,col] = find(bool);
% Compute the mean value of each row
rowmeans = nanmean(a, 2);
% Replace the NaN values with their row means
a(bool) = rowmeans(row);
% Perform the cumulative sum
result = cumsum(a);
If you want to leave the initial NaN values as NaN values afterwards, then you can follow it up with
result(bool) = NaN;
I have the following matrix: first column are the values of 1 to 5, second column is 1 to 20, and the third column are random values.
1 1 2545
1 2 0
1 3 0
1 4 0
2 5 0
2 6 0
2 7 231
2 8 54587
3 9 41
3 10 1111
3 11 0
3 12 1213
4 13 0
4 14 0
4 15 0
4 16 0
5 17 898
5 18 6887
5 19 522
5 20 23
What I am trying to do is to get the sum in groups of four when all values are different of zero. As an example, in the matrix the output I want is:
1 NaN
2 NaN
3 NaN
4 NaN
5 8330
Assuming that the first column delineates what values in the third column belong to what group, the easiest would be to change all values that are zero to NaN, then use accumarray to sum all of the values that belong to each group. This is crucial because as soon as you sum over a matrix / array and any value is NaN, the result will be NaN. This is nice because if you sum over each group, you will get a NaN result if at least one of the values in the group was equal to 0 before the change.
I'm going to assume that your matrix is stored in X like so:
X = [1 1 2545
1 2 0
1 3 0
1 4 0
2 5 0
2 6 0
2 7 231
2 8 54587
3 9 41
3 10 1111
3 11 0
3 12 1213
4 13 0
4 14 0
4 15 0
4 16 0
5 17 898
5 18 6887
5 19 522
5 20 23 ];
Make a copy of the third column, and let's do some magic:
col = X(:,3);
col(col == 0) = NaN;
out = accumarray(X(:,1), col);
We thus get:
out =
NaN
NaN
NaN
NaN
8330
The nice thing about this approach is that the group ID for each value in your matrix doesn't have to be in order as you have placed in your post.
If however your matrix is guaranteed to have the order where each group consists of consecutive 4-tuples of elements, you can do the same thing with the NaN assignment, but you can avoid using accumarray and reshape the third column into a matrix of four rows then sum over each row individually:
col = X(:,3);
col(col == 0) = NaN;
out = sum(reshape(col, 4, []), 1);
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 have an incomplete dataset,
N = [NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN]'
I wish to identify a cluster of Nans, that is, if the subsequent number of them exceeds 2. how do i do that?
You could do something like this:
aux = diff([0; isnan(N); 0]);
clusters = [find(aux == 1) find(aux == -1) - 1];
Then clusters will be a Nx2 matrix, where N is the number of NaN clusters (all of them), and each row gives you the start and end index of the cluster.
In this example, that would be:
clusters =
1 1
5 5
8 9
15 19
It means you have 4 NaN clusters, and cluster one ranges from index 1 to index 1, cluster two ranges from 5 to 5, cluster three ranges from 8 to 9 and cluster four ranges from 15 to 19.
If you want only the clusters with at least K NaNs, you could do it like this (for example, with K = 2):
K = 2;
clusters(clusters(:,2) - clusters(:,1) + 1 >= K, :)
That would give you this:
ans =
8 9
15 19
That is, clusters 8-9 and 15-19 have 2 or more NaNs.
Explanation:
Finding the clusters
isnan(N) gives you a logical vector containing the NaNs as ones:
N --------> NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
isnan(N) -> 1 0 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1
We want to know where each sequence of ones start, so we use diff, which calculates each value minus the previous one, and gives us this:
aux = diff(isnan(N));
N ----> NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
aux --> -1 0 0 1 -1 0 1 0 -1 0 0 0 0 1 0 0 0 0
Where a 1 indicates the group start and a -1 indicates a group end. But it misses the first group start and the last group end, because the first 1 element is absent (it doesn't have a previous on N because it is the first) and the last -1 is absent too (because there is nothing after the last 1 on N). A common fix is to add a zero before and after the array, which gives us this:
aux = diff([0; isnan(N); 0]);
N ----> NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
aux --> 1 -1 0 0 1 -1 0 1 0 -1 0 0 0 0 1 0 0 0 0 -1
Notice two things:
If the diff at index i is 1, N(i) is the start of the NaN block.
If the diff at index i is -1, N(i - 1) is the end of the NaN block.
To get the start and end, we use find to get the indexes where aux == 1 and aux == -1. Hence, we call find twice, and concatenate both calls using [ and ]:
aux = diff([0; isnan(N); 0]);
clusters = [find(aux == 1) find(aux == -1) - 1];
Filtering the clusters whick have K or more elements
The last step is to find clusters which have K or more elements. To do that, we first take the cluster matrix and subtract the first column from the first, and add 1, like this:
clusters(:,2) - clusters(:,1) + 1
ans =
1
1
2
5
It means clusters 1 and 2 have 1 NaN, cluster 3 have 3 NaNs and cluster 4 have 5 NaNs. If we ask which values are greather than or equal K, we get this:
clusters(:,2) - clusters(:,1) + 1 >= K
ans =
0
0
1
1
It's a logical array. We can use that to index only the 1 (true) rows of the cluster matrix, like this:
clusters(clusters(:,2) - clusters(:,1) + 1 >= K, :)
ans =
8 9
15 19
It's like asking: give us only the clusters where the rows match the ones on this logical vector, and give us all columns (denoted by the :).
Here is a modular solution:
% the number of NaN you consider as a cluster
num = 3;
% moving average filter
Z = filter(ones(num,1),1,isnan(N));
x = arrayfun(#(x) find(Z == num) - num + x, 1:num,'uni',0)
y = unique(cell2mat(x))
(UPDATE: faster version below)
gives for num = 1:
y = 1 5 8 9 15 16 17 18 19
for num = 2:
y = 8 9 15 16 17 18 19
for num = 3, num = 4 and num = 5:
y = 15 16 17 18 19
and finally for num = 6 ... and more
y = Empty matrix: 1-by-0
Explanation
isnan(N) returns a logical array with ones at the positions of NaN.
Z = filter(ones(num,1),1,isnan(N)); is a implementation for a moving average filter with a filter window of ones(num,1) = [1 1 1] (for num = 3). So the filter of size 3 glides of the array and just reaches the value num = 3 when there are 3 NaN in a row.
So it basicall looks like:
%// N isnan(N) Z
NaN 1 1
1 0 1
2 0 1
3 0 0
NaN 1 1
5 0 1
6 0 1
NaN 1 1
NaN 1 2
7 0 2
8 0 1
10 0 0
12 0 0
20 0 0
NaN 1 1
NaN 1 2
NaN 1 3
NaN 1 3
NaN 1 3
Now it is easy to find all elements which are 3: find(Z == num) - but you also need all 2 right before: find(Z == num) - num + 2 and all 1 right before: find(Z == num) - num + 1. Instead of a loop arrayfun is used, which is basically the same. As result you get a matrix with a lot of indices, lot of them mulitple, but you just need the unique ones. I hope everything is clear now.
Actually it would be much faster to get find out of arrayfun, which can then even be substituted by bsxfun and you can get rid of cell2mat, which leads to the following form:
Faster:
Z = find( filter(ones(num,1),1,isnan(N)) == num ) - num;
y = unique( bsxfun(#plus, Z,1:num) );
or faster the obligatory fancy one-liner:
y = unique(bsxfun(#plus,find(filter(ones(num,1),1,isnan(N))==num)-num,1:num));
STRFIND Approach
I. Fancy One Liner:
%%// Given input N
N = [NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN]
out = [strfind(num2str(isnan([ 0 N 0]),'%1d'),'011');strfind(num2str(isnan([ 0 N 0]),'%1d'),'110')]'
Output
out =
8 9
15 19
II. Detailed one with explanation:
Basically you are trying to do sliding window checks, for which there is no direct method when working with double arrays, but after converting to strings, one can use strfind. This trick is used here.
I would suggest following the comments used in the code and the output numbers to understand it. Please note that for this particular case a cluster means a group of two or more consecutive NaNs
Code
%%// Given input N
N = [NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN]
%%// Set the locations where NaNs are present and then
%%// append at the start and end with zeros
N2 = isnan([ 0 N 0])
%%// Find the start indices of all NaN clusters
start_ind = strfind(num2str(N2,'%1d'),'011')
%%// Find the stop indices of all NaN clusters
stop_ind = [strfind(num2str(N2,'%1d'),'110')]
%%// Put start and stop indices into a Mx2 matrix
out = [start_ind' stop_ind']
Output
N =
NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
N2 =
0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0
start_ind =
8 15
stop_ind =
9 19
out =
8 9
15 19
This uses diff, as Rafael Monteiro's answer, but seems to be simpler:
ind = diff([0; isnan(N(:))]);
result = find(ind(1:end-1)==1 & ind(2:end)==0);
In your example, this gives [8 15].
How it works: ind takes the values:
1 where a run of (one or more) NaN values starts;
0 where there is no change between NaN and numeric with respect to the previous value;
-1 where a run of (one o more) numeric values starts.
The second line selects the positions at which a run of NaN starts and such that the next position is also NaN. Thus it gives the start of each run of more than one NaN, as desired.