for x=1:0.5:20
how can store the value for this type of for loop.

6 Kommentare

madhan ravi
madhan ravi am 27 Nov. 2018
Bearbeitet: madhan ravi am 27 Nov. 2018
If you show the calculation we can vectorize it even without a loop if there is a possibility.
wong loong
wong loong am 27 Nov. 2018
for i=1:0.5:20
Base =(BaseValue.inputSingleScan-1.63)*(327/10)*100;
Top = (TopValue.inputSingleScan-1.63)*(327/10)*100;
PI1 = abs(Top - Base);
this is calculation.
madhan ravi
madhan ravi am 27 Nov. 2018
BaseValue.inputSingleScan ? provide the datas
wong loong
wong loong am 27 Nov. 2018
data scan from sensor.
madhan ravi
madhan ravi am 27 Nov. 2018
is it stored or live data?
wong loong
wong loong am 27 Nov. 2018
live data...i want store the data every time loop...then find out the maximun value.

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madhan ravi
madhan ravi am 27 Nov. 2018
Bearbeitet: madhan ravi am 27 Nov. 2018

0 Stimmen

EDITED
x=1:0.5:20;
n=numel(x);
Base=cell(1,n); % PRE-ALLOCATION
Top=cell(1,n);
PI1=cell(1,n);
for i=1:n
Base{i} =(BaseValue.inputSingleScan-1.63)*(327/10)*100;
Top{i} = (TopValue.inputSingleScan-1.63)*(327/10)*100;
PI1{i} = abs([Top{i}] - [Base{i}]);
end
values=[PI1{:}];
max_value=max(values);

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