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How to use IF statements to find data greater than or equal to a certain amount

4 vues (au cours des 30 derniers jours)
If you have a large amount of data, tens of millions of rows of data, how can you use IF statements in MATLAB to find specific data points that spike above a certain threshold?
Lets say for example you have a load of capacitance data against time. And you wan to find where the data spikes above 1 Farad. An example data table:
Time, Capacitance
1 uS, 0.3
2 uS, 0.5
3 uS, 0.8
4 uS, 1
5 uS, 1.3
6 uS, 1.4
7 uS, 1.2
8 uS, 0.9
9 uS, 0.5
You want to sift through this table and create a new variable that only includes data where the capacitance exceeds 1 Farad

Réponse acceptée

Karim
Karim le 19 Sep 2022
Modifié(e) : Karim le 19 Sep 2022
see below for a demonstration
Time = [1 2 3 4 5 6 7 8 9]';
Capacitance = [0.3 0.5 0.8 1 1.3 1.4 1.2 0.9 0.5]';
MyTable = table(Time,Capacitance)
MyTable = 9×2 table
Time Capacitance ____ ___________ 1 0.3 2 0.5 3 0.8 4 1 5 1.3 6 1.4 7 1.2 8 0.9 9 0.5
% delete variables, just to show that we work with the data from the table
clearvars Time Capacitance
% set a treshold value
Treshold = 1;
% use some logice to find the values larger then the treshold (if you want
% to include it use >= instead of >
KeepMe = MyTable.Capacitance > Treshold;
% extract a part of the table using the logical array
NewTable = MyTable( KeepMe , : )
NewTable = 3×2 table
Time Capacitance ____ ___________ 5 1.3 6 1.4 7 1.2
  6 commentaires
Voss
Voss le 19 Sep 2022
Modifié(e) : Voss le 19 Sep 2022
@AluminiumMan: You left off the second subscript, like the error message suggests:
A=B.C(C.Capacitance>Threshold,:);
% ^^ you need this part
Notice that @Karim's answer has it:
NewTable = MyTable( KeepMe , : )
Here's a demo with your mat file and variables, as given:
S = load('CapacitanceData.mat');
B = struct('C',S.CapacitanceData.RecordedCapacitanceData);
C = B.C;
Threshold = 1;
A=B.C(C.Capacitance>Threshold,:)
A = 3×2 table
Time Capacitance ____ ___________ 3 1.3 4 1.6 7 1.3
AluminiumMan
AluminiumMan le 20 Sep 2022
Its working. Thank you for your help

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Plus de réponses (1)

Chunru
Chunru le 19 Sep 2022
c = randn(20,1) % random data
c = 20×1
-0.1315 1.3920 0.3866 0.1342 0.7914 -0.5643 2.9497 -0.1842 -0.3139 0.3103
c1 = c(c>1)
c1 = 2×1
1.3920 2.9497
  5 commentaires
AluminiumMan
AluminiumMan le 20 Sep 2022
Its working. Thank you for your help

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