How to avoid the multiple use of 'isnan' to filter inconsitent datasets?

Question

I have two corresponded (has a relation and has the same dimension) dataset:

Time
Salinity

Some data in salinity dataset is NaN.

I can remove the NaN value by:

 Salinity_new=Salinity(~isnan(Salinity))

But it will not correspond to the Time dataset any more.

How can I remove the corresponded time in also?

Thanks

`nanElements = isnan(Salinity);`? If you do not have enough memory for this, you should consider another solution. — patrik, Mar 25 '14 at 10:38

score 2 · Answer 1 · answered Mar 25 '14 at 16:57

2

Another solution is the following:

indexes = find(isnan(Salinity)==1);
Salinity(indexes) = []; 
Time(indexes) = []

In this way you eliminate the non-numerical value from your vectors.

answered Mar 25 '14 at 16:57

roschach

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Robert Seifert · Accepted Answer · 2014-03-25T12:36:34.450

1

The comments of Diavakar and patrik are correct. To sum it up and get this question answered, some further remarks.

mask = isfinite(Salinity) 
[Time,Salinity] = deal( Time(mask), Salinity(mask) )

isfinite is the same as ~isnan - but with one computation step less, its about 50-90% faster. By introducing a mask you're avoiding the double use of isfinite. deal just saves you some space.

edited Mar 25 '14 at 12:36

answered Mar 25 '14 at 11:35

Robert Seifert

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How to avoid the multiple use of 'isnan' to filter inconsitent datasets?

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