Finding the length distribution of zeros in data?

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Pappis
Pappis am 2 Sep. 2025 um 13:05
Bearbeitet: Matt J am 4 Sep. 2025 um 15:44
I have a data set of several measurements, which give me the values at given time intervals (e.g. once a minute). Every now and then, the value is missed, e.g. due to a random fault, and the missing value is marked as "0" (zero).
I have experience handling the data, but got a new request to debug the faults more closely. I need to find the length distribution of these zeros, i.e. how long the faults are typically? Are they single-value misses, or longer?
So, how do I evaluate the lengths for each of these faults, and/or their length distribution? I prefer answers without the Statictical toolbox, but if needed I have access to that, too.

Akzeptierte Antwort

Stephen23
Stephen23 am 2 Sep. 2025 um 13:16
Bearbeitet: Stephen23 am 2 Sep. 2025 um 13:19
Fake data:
V = rand(1,123);
V(randi(123,1,23)) = 0;
V(5:7) = 0
V = 1×123
0.7107 0.4742 0.0842 0.5311 0 0 0 0.0510 0.3399 0.4221 0.8725 0.0811 0.3725 0.0362 0.9471 0.0934 0.2950 0.4100 0.0774 0.6726 0.3011 0.4228 0.9944 0.1799 0 0.5158 0.2221 0.3628 0.6641 0.0212
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D = diff([false;V(:)==0;false]);
L = find(D<0) - find(D>0);
histogram(L)

Weitere Antworten (1)

Matt J
Matt J am 3 Sep. 2025 um 12:13
Bearbeitet: Matt J am 4 Sep. 2025 um 15:44
You can use this File Exchange submission,
V = rand(1,123);
V(randi(123,1,23)) = 0;
V(5:7) = 0;
[~,~,zeroLengths] = groupLims(groupTrue(~V),1);
histogram(zeroLengths)
  1 Kommentar
Pappis
Pappis am 4 Sep. 2025 um 7:29
Verschoben: Rik am 4 Sep. 2025 um 7:55
I like Stephen's elegant solution very much. It's also easier to understand while reading the code. Thank you for your quick reply!

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