How can I re-write this code to fit any size matrix? Help!
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I have written a code to calculate the end grade(sum) while dropping the lowest grade value between a specific number of columns in a matrix. This code only works for a 10*5 matrix I need to create a code that works for any size matrix. My current code will be written below. Thanks in advance.
function updatedGrades = computeTotal(sectionGrades) % Modify xxx,yyy to variable names of your choice. % Comment what this function is about here.
% Put the body of your function here. Don't forget to assign output variable! e.g. xxx = something
grades2=sectionGrades([1], [3:8]) grades3=sectionGrades([2], [3:8]) grades4=sectionGrades([3], [3:8]) grades5=sectionGrades([4], [3:8])
grades2a=sum(grades2)-min(grades2) grades3b=sum(grades3)-min(grades3) grades4c=sum(grades4)-min(grades4) grades5d=sum(grades5)-min(grades5)
grades23=grades2a+sum(sectionGrades([1], [9:10])) grades34=grades3b+sum(sectionGrades([2], [9:10])) grades45=grades4c+sum(sectionGrades([3], [9:10])) grades56=grades5d+sum(sectionGrades([4], [9:10]))
finalGrades=[grades23;grades34;grades45;grades56]
sectionGrades([1:4],[11])=finalGrades
updatedGrades=sectionGrades
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Respuesta aceptada
KL
el 25 de Nov. de 2017
Editada: KL
el 25 de Nov. de 2017
You're complicating a simple one line calculation way too much. Here's how to do it effectively,
sectionGrades(:,end+1) = sum(sectionGrades(:,3:8),2)-min(sectionGrades(:,3:8),1,2)...
+sum(sectionGrades(:,9:10),2);
Now it does't matter how many rows you add to the matrix, the last column should be calculated accordingly.
3 comentarios
KL
el 25 de Nov. de 2017
Sorry, I missed one argument in min command, fixed it now and it should work. If you want to assign it to 'updatedGrades', simply assign it!
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