How to count "gaps" consisting of designated values in a vector?

4 vues (au cours des 30 derniers jours)
Al_G
Al_G le 18 Déc 2018
Commenté : Matt J le 8 Juin 2021
I have a very long vector of data that includes gaps of "invalid" values where data that's missing has been replaced with a designated missing data value (e.g. -1) or with "NaN".
invalid_values = [-1 NaN];
sample_data_vector = [22 23 22 24 -1 -1 -1 25 20 24 NaN Nan NaN 25 24 -1 -1 22 20 NaN 23];
I want to summarize info on the length of the gaps of invalid data (and no need to distinguish between whether the value is Nan or -1 or some other number). It might be helpful to know that my goal is to compare gap lengths over subsequent trials to determine whether gaps are getting longer/shorter and more/less frequent.
So for the above sample vector I'd get a matrix output that summarizes the following info about the gaps:
Length: 1 Number of occurences: 1
Length: 2 Number of occurences: 1
Length: 3 Number of occurences: 2
How would I write code to do this?

Réponse acceptée

Matt J
Matt J le 18 Déc 2018
Modifié(e) : Matt J le 18 Déc 2018
One way, assuming you have the Image Processing Toolbox,
bw=ismember(sample_data_vector, invalid_values)|isnan(sample_data_vector);
S=regionprops(bw,'Area');
gaplengths=[S.Area];
H = histcounts(gaplengths,1:max(gaplengths)+1) %output histogram
  5 commentaires
Al_G
Al_G le 8 Juin 2021
Revisiting this a few years later in an attempt to improve my code. Is there a straightforward way to obtain the indices of where the gaps start and end?
The ideal output would look like:
gap.length = [3, 3, 2, 1];
gap.start_indices = [5, 11, 16, 20];
gap.end_indices = [7, 13, 17, 20];
Matt J
Matt J le 8 Juin 2021
That will just be,
gap.start_indices = find( diff([0,bw])>0 );
gap.end_indices = find( diff([bw,0])<0 );
gap.length = gap.end_indices-gap.start_indices+1;

Connectez-vous pour commenter.

Plus de réponses (0)

Tags

Produits


Version

R2015b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by