标题：TB转金字塔

1楼

lc1227 发表于：2013/9/3 10:51:49

SumX = N * ( N - 1 ) * 1/2;

SumY = Summation( Close, N) ;

for i = 0 to N-1

{

SumXY = SumXY + i * Close[i] ;

SumXX = SumXX + i * i ;

SumYY = SumYY + Close[i] * Close[i] ;

}

R2 = (SumXY - N*(SumX/N)*(SumY/N)) / Sqrt((SumXX - SumX*SumX/N)*(SumYY - SumY*SumY/N));

2楼

jinzhe 发表于：2013/9/3 10:55:31

解释下代码

3楼

lc1227 发表于：2013/9/3 10:59:06

例如 n=30

SumX = N * ( N - 1 ) * 1/2;

SumY = Summation( Close, N) ;//求和

for i = 0 to N-1

{

SumXY = SumXY + i * Close[i] ; //close[i]就是ref(c,i)

SumXX = SumXX + i * i ;

SumYY = SumYY + Close[i] * Close[i] ;

}

4楼

jinzhe 发表于：2013/9/3 11:07:19

sumxy

sumxx

sumyy的初始值是？

5楼

lc1227 发表于：2013/9/3 11:09:10

初始值都是0

6楼

jinzhe 发表于：2013/9/3 13:20:26

SumX = N * ( N - 1 ) * 1/2;

SumY = Summation( Close, N) ;

for i = 0 to N-1

{

SumXY = SumXY + i * Close[i] ;

SumXX = SumXX + i * i ;

SumYY = SumYY + Close[i] * Close[i] ;

}

R2 = (SumXY - N*(SumX/N)*(SumY/N)) / Sqrt((SumXX - SumX*SumX/N)*(SumYY - SumY*SumY/N));

7楼

lc1227 发表于：2013/9/3 14:21:26

你这是回答么？

8楼

jinzhe 发表于：2013/9/3 14:29:30

不好意思啊，复制错了，抱歉

n:=30;
sumx:=n*(n-1)/2;
sumy:=sum(c,n);
variable:sumxx[n]=0,sumyy[30]=0,sumxy[30]=0;
r1:=ref(c,1);
for i=2 to n do begin
    sumxy[i]:=sumxy[i-1]+i*r1;
    sumxx[i]:=sumxx[i-1]+i*i;
    sumyy[i]:=sumyy[i]+r1*r1;
end

r2:=(sumxy-n*(sumx/n))/sqrt((sumxx-sumx*sumx/n)*(sumyy-sumy*sumy/n));

9楼

lc1227 发表于：2013/9/3 16:24:35

十分感谢，哈哈

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