如何在MATLAB中分离if和else [英] how to separate if and else in matlab

问题描述

让我们考虑以下代码

function averageentropy=calculate(f,y)
count1=0;
count0=0;
n=length(f);
n1=0;
n0=0;
 entrop1=0;
  entrop2=0;
bigp=sum(f)/n;
for i=1:n 
     if f(i)==1 && y(i)==1
         count1=count1+1;
     end
end
 for i=1:n
     if  f(i)==0 && y(i)==1
          count0=count0+1;
     end
 end
for i=1:n
     if f(i)==1
          n1=n1+1;
     end
end
 for i=1:n 
     if f(i)==0
         n0=n0+1;
     end
 end
 smallpplus=count1/n1;
 smallpminus=count0/n0;
 if smallpplus==0
     entrop1=0;
 else
 entrop1=-smallpplus*log2(smallpplus)-(1- smallpplus)*log2(1- smallpplus);
end
  if smallpminus==0
     entrop2=0;
  else
  entrop2=-smallpminus*log2(smallpminus)-(1- smallpminus)*log2(1- smallpminus);

 end 
  averageentropy=bigp*entrop1+(1-bigp)*entrop2
end

当我运行此代码时，我得到0.4056，而在Excel中相同的过程返回给我大约0.91，这意味着if和else情况下存在一些错误，因为如果我删除if and else，我将得到答案相同，那是什么问题呢?我正在使用if和else来避免log(0)，但是有一些问题.

when I am running this code, I am getting 0.4056, while in Excel the same procedure returns me approximately 0.91, which means that there is some error in if and else case, because if I remove if and else, I am getting the same answer to, so what is problem? I am using if and else to avoid log(0), but there is some problem.

推荐答案

如果f，y是实值而不是整数，我建议使用

If f,y are real valued and not integers I would recommend using for example

abs(f(i)-1)< = eps

abs(f(i)-1) <= eps

其中eps是一个小数字"(例如eps = 1e-5) 代替

where eps is a 'small number' (e.g eps = 1e-5) instead of

f(i)== 1

f(i)==1

附带说明，您可以编写代码的这一部分:

On a side note, you could write this part of your code:

for i=1:n 
     if f(i)==1 && y(i)==1
         count1=count1+1;
     end
end
 for i=1:n
     if  f(i)==0 && y(i)==1
          count0=count0+1;
     end
 end
for i=1:n
     if f(i)==1
          n1=n1+1;
     end
end
 for i=1:n 
     if f(i)==0
         n0=n0+1;
     end
 end

通过利用Matlab的矢量化表达功能，更紧凑，更高效(我宁愿为了可读性而牺牲一些额外的行):

more compactly and efficiently by taking advantage of Matlab's vectorized expression capabilities (I preferred to sacrifice some extra lines for the sake of readability):

indf1 = f == 1;
indf0 = ~indf1 ;
indy1 = y == 1;
indy0 = ~indy1 ;
count1 = sum(indf1 & indy1) ;
count0 = sum(indf0 & indy1) ;
n1 = sum(indf1);
n0 = sum(indf0);

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