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i have written a while loop which displays 'yes' or 'no' if the difference is less than 0.5 for a given problem. how can i write this output into an excel sheet.

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this is the code i have used. it gives the desired result in the command window but i need it to write the output into an excel sheet. how can i incorporate that?
data=readtable("DATA.xlsx");
X=table2array(data);
i=1;
j=2;
while i<57
M=max(X(:,i));
n = mean(X(M-50:end,j),'omitnan');
m= M-50;
while m<M+1
y=X(m,j);
z=((y-n)/n)*100;
disp(z)
if (0.5>z) && (-0.5<z)
disp("YES")
else
disp("NO")
end
m = m+1;
end
i=i+2;
j=j+2;
end
  3 Kommentare
Aditi
Aditi am 1 Nov. 2023
yes but the output matrix is only showing the last value.
it is my first time using matlab so if you could help with making a simpler code for this problem.
Siddharth Bhutiya
Siddharth Bhutiya am 1 Nov. 2023
Can you attach the DATA.xlsx? As KSSV mentioned there might be ways to simplify the code a bit.

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Akzeptierte Antwort

Arun
Arun am 1 Nov. 2023
Hey Aditi,
One possible way to do so is:
data=readtable("DATA.xlsx");
X=table2array(data);
i=1;
j=2;
while i<57
M=max(X(:,i));
n = mean(X(M-50:end,j),'omitnan');
m= M-50;
counterToPrint=1; %increment it for each row
% Specify the file name and sheet name
filename = 'example.xlsx';
sheet = 'Sheet1';
while m<M+1
y=X(m,j)
z=((y-n)/n)*100;
disp(z)
%location to write
location = ['A' num2str(counterToPrint)]; % Use the column accordingly
if (0.5>z) && (-0.5<z)
xlswrite(filename,{"Yes"}, sheet, location); %write to the sheet
else
xlswrite(filename,{"No"}, sheet, location); %write to the sheet
end
counterToPrint = counterToPrint + 1;
m = m+1;
end
i=i+2;
j=j+2;
end
Hope this helps.
  17 Kommentare
Walter Roberson
Walter Roberson am 2 Nov. 2023
As a matter of efficiency, but not a change in functionality:
data = readtable('checking3.xlsx');
display = data(data.Var2=="no",:)
This will give the same result as your find() version, just a little faster.
It should give you just the rows that match "no" in the second column.... assuming the xlsx does not have any header.

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Weitere Antworten (1)

Walter Roberson
Walter Roberson am 1 Nov. 2023
data=readtable("DATA.xlsx");
X=table2array(data);
i=1;
j=2;
decisions = strings(0);
decision_count = 0;
while i<57
M=max(X(:,i));
n = mean(X(M-50:end,j),'omitnan');
m= M-50;
while m<M+1
y=X(m,j);
z=((y-n)/n)*100;
disp(z)
decision_count = decision_count + 1;
if (0.5>z) && (-0.5<z)
decisions(decision_count) = "YES";
disp("YES")
else
decisions(decisions_count) = "NO";
disp("NO")
end
m = m+1;
end
i=i+2;
j=j+2;
end
writematrix(decisions, FILENAME)

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R2023b

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