组合下列代码: if (Ef>=0)&&(Ef<1/3) theta=0.7; r=2+t/NGen; end if (Ef>=1/3)&&(Ef<=2/3) theta=0.8; r=1+t/NGen; end if Ef>2/3 theta=2; r=0.8; end
时间: 2023-11-29 21:04:03 浏览: 29
```matlab
if (Ef >= 0) && (Ef < 1/3)
theta = 0.7;
r = 2 + t/NGen;
end
if (Ef >= 1/3) && (Ef <= 2/3)
theta = 0.8;
r = 1 + t/NGen;
end
if (Ef > 2/3)
theta = 2;
r = 0.8;
end
```
相关问题
if (rho >= 0 && rho < max_rho) { hough_space[rho][theta]++; }这两个是什么作用
这段代码是Hough变换的核心部分,用于将图像中的像素点转换为Hough空间中的直线参数。在这里,每个像素点都被转换为一条直线,并且这些直线的参数被累加在Hough空间中。具体来说,这两个语句的作用如下:
- `if (rho >= 0 && rho < max_rho)`:判断当前计算得到的rho值是否在Hough空间的有效范围内,如果是则继续执行下一步,否则跳过该像素点。
- `hough_space[rho][theta]++`:将当前像素点对应的直线在Hough空间中的计数器加1,表示该直线在该点上的投票数增加了1。
通过这种方式,最终在Hough空间中统计出一些得票数较高的直线,这些直线就是原图像中的直线。
ra1=0; r2=0.018; r1=0.0178; miu0=0.003065; b=0.015; while theta > pi/9 theta = theta-pi/9; end if theta == 0 ra1= 1/(miu0*b/(pi/2)*log(r2/r1)+0.5*miu0/(pi/90)*log(r2/r1)); 将下面这段代码做成循环函数ra1=0; r2=0.018; r1=0.0178; miu0=0.003065; b=0.015; while theta > pi/9 theta = theta-pi/9; end if theta == 0 ra1= 1/(miu0*b/(pi/2)*log(r2/r1)+0.5*miu0/(pi/90)*log(r2/r1)); else if (theta>0)&(theta<=pi/90) ra1 = 1/(2*miu0*b/(pi/2)*log(r2/r1)+miu0/(pi/90-theta)*log(r2/r1)); else if (theta>pi/90)&(theta<=pi/18) ra1 = 1/(miu0*b/log(r2/r1)*(theta-pi/90)); else if (theta>pi/18)&(theta<=pi/10) ra1 = 1/(miu0*b/log(r2/r1)*(2*pi/45+1/18*pi-theta)); else if (theta>pi/10)&(theta<=pi/9) ra1= 1/(2*miu0*b/(pi/2)*log(r2/r1)+miu0/(theta-pi/10)*log(r2/r1)); end end end end end
你的要求是将上述代码转换为循环函数。下面是转换后的代码:
```python
def calculate_ra1(theta):
ra1 = 0
r2 = 0.018
r1 = 0.0178
miu0 = 0.003065
b = 0.015
while theta > pi/9:
theta = theta - pi/9
if theta == 0:
ra1 = 1 / (miu0*b/(pi/2)*log(r2/r1) + 0.5*miu0/(pi/90)*log(r2/r1))
elif (theta > 0) and (theta <= pi/90):
ra1 = 1 / (2*miu0*b/(pi/2)*log(r2/r1) + miu0/(pi/90-theta)*log(r2/r1))
elif (theta > pi/90) and (theta <= pi/18):
ra1 = 1 / (miu0*b/log(r2/r1)*(theta-pi/90))
elif (theta > pi/18) and (theta <= pi/10):
ra1 = 1 / (miu0*b/log(r2/r1)*(2*pi/45+1/18*pi-theta))
elif (theta > pi/10) and (theta <= pi/9):
ra1 = 1 / (2*miu0*b/(pi/2)*log(r2/r1) + miu0/(theta-pi/10)*log(r2/r1))
return ra1
```
你可以将 `theta` 的值传递给 `calculate_ra1` 函数,它将返回相应的 `ra1` 值。请注意,这只是一个转换,我没有对代码逻辑进行任何更改。