finding clustered NaNs but leaving lonely NaNs alone - matlab

Question

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?

Answer 1

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 :).

Answer 2

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));

Answer 3

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

Answer 4

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.