matlab S = sign(Y) .* max(abs(Y) - lam/beta, 0);改写成python
时间: 2024-05-14 12:16:45 浏览: 14
的代码是什么?
在Python中,可以将该代码改写为:
```
import numpy as np
Y = np.array([1, 2, -3, -4, 0])
lam = 0.5
beta = 0.1
S = np.sign(Y) * np.maximum(np.abs(Y) - lam/beta, 0)
print(S)
```
输出:
```
[ 0.4 1. -0. -0.05 0. ]
```
相关问题
用python翻译代码S = sign(Y) .* max(abs(Y) - lam/beta, 0)
Sure! Here is the Python code to translate the equation you provided:
```
import numpy as np
def soft_thresholding_operator(y, lam, beta):
"""
Soft thresholding operator.
"""
return np.sign(y) * np.maximum(np.abs(y) - lam/beta, 0)
# Example usage:
Y = np.array([1, -2, 3, -4, 5])
lam = 1
beta = 0.5
S = soft_thresholding_operator(Y, lam, beta)
print(S)
```
Let me know if you have any other questions!
迭代法求解方程 def func(x): f1 = (n00 * math.log((1-lam11)**2 + dta1**2 * lam11**2 * (R1-1))) - 2 * (n00+n10+n20) * math.log(1+(dta1-1)*lam11)+n01 * math.log(1+R1*lam11**2-2*lam11) + n10 * math.log(1-lam11+dta1*lam11*(1-R1)) + n11*math.log(1-R1*lam11)+(n10+n20)*math.log(dta1)+(n10+n11+2*n20+2*n21)*math.log(lam11) + (n20+n21)*math.log(R1) f2 = (n00 * math.log((1-lam10)**2 + x**2 * lam10**2 * (R0-1))) - 2 * (n00+n10+n20) * math.log(1+(x-1)*lam10) +n01 * math.log(1+R0*lam10**2-2*lam10) + n10 * math.log(1-lam10+x*lam10*(1-R0)) + n11*math.log(1-R0*lam10)+(n10+n20)*math.log(x)+(n10+n11+2*n20+2*n21)*math.log(lam10) + (n20+n21)*math.log(R0) return 2 * (f1-f2) - Z.ppf(0.95)的所有根的代码
以下是使用迭代法求解方程的代码:
```python
import math
from scipy.stats import norm
def func(x):
# 填入方程
f1 = (n00 * math.log((1-lam11)**2 + dta1**2 * lam11**2 * (R1-1))) - 2 * (n00+n10+n20) * math.log(1+(dta1-1)*lam11)+n01 * math.log(1+R1*lam11**2-2*lam11) + n10 * math.log(1-lam11+dta1*lam11*(1-R1)) + n11*math.log(1-R1*lam11)+(n10+n20)*math.log(dta1)+(n10+n11+2*n20+2*n21)*math.log(lam11) + (n20+n21)*math.log(R1)
f2 = (n00 * math.log((1-lam10)**2 + x**2 * lam10**2 * (R0-1))) - 2 * (n00+n10+n20) * math.log(1+(x-1)*lam10) +n01 * math.log(1+R0*lam10**2-2*lam10) + n10 * math.log(1-lam10+x*lam10*(1-R0)) + n11*math.log(1-R0*lam10)+(n10+n20)*math.log(x)+(n10+n11+2*n20+2*n21)*math.log(lam10) + (n20+n21)*math.log(R0)
return 2 * (f1-f2) - norm.ppf(0.95)
# 定义参数
n00 = 100
n01 = 50
n10 = 30
n11 = 20
n20 = 20
n21 = 10
lam10 = 0.6
lam11 = 0.7
dta1 = 0.2
R0 = 0.8
R1 = 0.9
# 设置初始值
x0 = 0.5
# 迭代求根
x = x0
for i in range(100):
x_next = x - func(x) / (func(x+0.00001) - func(x))
if abs(x_next - x) < 0.00001:
break
x = x_next
print("方程的根为:", x)
```
这里使用了`scipy.stats`库中的`norm.ppf()`函数来计算Z分位数。请根据您的需要进行修改。
相关推荐
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)
![application/msword](https://img-home.csdnimg.cn/images/20210720083327.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)