Best way to reduce to a simple loop?

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Oscar Zampi
Oscar Zampi el 20 de Feb. de 2021
Respondida: Jan el 21 de Feb. de 2021
Hi, I have the following section of code that I would like to simplify down as much as possible so that it can do the same thing, but am unsure how to do so.
Any help would be greatly apopreciated!
if A(1,3)<A(2,3) && A(1,3)<A(3,3) && A(1,3)<A(4,3) && A(1,3)<A(5,3)
E1i = A(1,1);
E1j = A(1,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(1,:) = [];
else if A(2,3)<A(1,3) && A(2,3)<A(3,3) && A(2,3)<A(4,3) && A(2,3)<A(5,3)
E1i = A(2,1);
E1j = A(2,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(2,:) = [];
else if A(3,3)<A(1,3) && A(3,3)<A(2,3) && A(3,3)<A(4,3) && A(3,3)<A(5,3)
E1i = A(3,1);
E1j = A(3,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(3,:) = [];
else if A(4,3)<A(1,3) && A(4,3)<A(2,3) && A(4,3)<A(3,3) && A(4,3)<A(5,3)
E1i = A(4,1);
E1j = A(4,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(4,:) = [];
else A(5,3)<A(1,3) && A(5,3)<A(2,3) && A(5,3)<A(3,3) && A(5,3)<A(4,3)
E1i = A(5,1);
E1j = A(5,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(5,:) = [];
end
end
end
end
if A(1,3)<A(2,3) && A(1,3)<A(3,3) && A(1,3)<A(4,3)
E1i = A(1,1);
E1j = A(1,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(1,:) = [];
else if A(2,3)<A(1,3) && A(2,3)<A(3,3) && A(2,3)<A(4,3)
E1i = A(2,1);
E1j = A(2,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(2,:) = [];
else if A(3,3)<A(1,3) && A(3,3)<A(2,3) && A(3,3)<A(4,3)
E1i = A(3,1);
E1j = A(3,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(3,:) = [];
else if A(4,3)<A(1,3) && A(4,3)<A(2,3) && A(4,3)<A(3,3)
E1i = A(4,1);
E1j = A(4,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(4,:) = [];
end
end
end
end
if A(1,3)<A(2,3) && A(1,3)<A(3,3)
E1i = A(1,1);
E1j = A(1,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(1,:) = [];
else if A(2,3)<A(1,3) && A(2,3)<A(3,3)
E1i = A(2,1);
E1j = A(2,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(2,:) = [];
else if A(3,3)<A(1,3) && A(3,3)<A(2,3)
E1i = A(3,1);
E1j = A(3,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(3,:) = [];
end
end
end
  1 comentario
Adam Danz
Adam Danz el 20 de Feb. de 2021
😨
Before the loops are optimized, there are some serious problems to address with the existing code.
  1. The indentation make is very difficult to understand how these conditions are nested. From the Matlab editor, select all of the text (ctrl+a) and then smart-indent (ctrl+i).
  2. Look at all of the orange tick marks along the right side of the editor. They indicate warnings. Not all warnings are serious but the ones listed here are serious.
  3. "else if" is must different from "elseif". Please review the documentation: doc elseif
  4. Edit your question to clean up the code so that it represents what you want it to do and fix the problems listed above.

Iniciar sesión para comentar.

Respuestas (1)

Jan
Jan el 21 de Feb. de 2021
The first block can be simplified to:
i2 = 2; % This is 2 in the first iteration, and 1 afterwards
index = NaN;
for i1 = 1:5
if A(i1,3)<A(i2,3) && A(i1,3)<A(3,3) && A(i1,3)<A(4,3) && A(i1,3)<A(5,3)
index = i1:
break;
end
i2 = 1;
end
E1i = A(index,1);
E1j = A(index,2);
E1y = 913000 - E1i;
E1x = E1j + 207000;
plot(E1x, E1y,'rx')
A(index, :) = [];
Avoid repeated code by moving it outside the loop.

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