Decision Based on Multiple Values

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Dear All
I have sensor measurement data arranged in the format shown below:
The first three values in each row represents the sensing pair and the last value (highlighted in yellow) shows the position of the object in the sensing area. The decision of the position can only be made by considering the first 3 values. 1 1 1 is the minimum value and 8 8 8 is the maximum value and there are lots of combinations as shown in the figure. If I use simple if else statement I have to use a lot of them which to be honest is not a good approach.
My question is if there is a way to do this smartly without using a lots of if and else?
Thank you
  4 Comments
Walter Roberson
Walter Roberson on 7 Dec 2020
line 5 compared to line 8
6 5 4 2 line 5
6 5 4 4 line 8
M Sattar
M Sattar on 7 Dec 2020
I am sorry this is a mistake

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Accepted Answer

Walter Roberson
Walter Roberson on 7 Dec 2020
Edited: Walter Roberson on 7 Dec 2020
Decision trees can construct an appropriate output structure from data

More Answers (1)

Harry Laing
Harry Laing on 7 Dec 2020
Unless there is some mathematical formula that you use to determine the position (which we don't know), then I'm afraid a long list of is/else statements seems likely.
Alternatively, you could create a 'master' reference array, that contains all the exisiting combinations and the resulting position, and then just compare the sensor data to determine the resulting positon.
E.g:
% Assuming you have an array of all sensor values / positons called MASTER_array
% Assuming your sensor data is called SENSOR_data
[NumRows_data,NumCols_data] = size(SENSOR_data);
[NumRows_master,NumCols_master] = size(MASTER_array);
object_position = NaN(NumRows_data,1); % Initialise 'position' vector for each row in sensor data
for i = 1:NumRows_data
for j = 1:NumRows_master
% test to compare each row of master array with current sensor array
if MASTER_array(j,1:3)==SENSOR_data(i,1:3)
% if sensor matches master, use 4th col in master for position
object_position(i)=MASTER_array(j,4);
end
end
end

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