如何通过调整权重来优化皮尔逊相关系数？

Question

我想调整权重w以优化皮尔逊相关系数的 r 平方。

import numpy as np
from scipy import stats
x1_raw=np.array([277, 115, 196])
x2_raw=np.array([263, 118, 191])
x3_raw=np.array([270, 114, 191])
w=np.array([w1, w2, w3])
x1=np.prod([w,x1_raw], axis=0).sum()
x2=np.prod([w,x2_raw], axis=0).sum()
x3=np.prod([w,x3_raw], axis=0).sum()
x=np.array([x1, x2, x3])
y=np.array([71.86, 71.14, 70.76])
slope, intercept, r_value, p_value, std_err = stats.linregress(x,y)
r_squared = r_value**2

那么调整[w1, w2, w3]以最大化r_squared的代码是什么？

---

谢谢你@mathew冈瑟

我从print(res)得到的结果是：

final_simplex: (array([[ 0.41998763,  2.66314965,  3.34462572],
  [ 0.4199877 ,  2.66314968,  3.34462654],
  [ 0.41998749,  2.66314983,  3.34462649],
  [ 0.41998765,  2.66314917,  3.34462607]]), array([-1., -1., -1., -1.]))
      fun: -0.99999999999999822
  message: 'Optimization terminated successfully.'
     nfev: 130
      nit: 65
   status: 0
  success: True
        x: array([ 0.41998763,  2.66314965,  3.34462572])

我可以理解x: array([ 0.41998763, 2.66314965, 3.34462572])是w; nfev是函数评估的数量; nit是迭代次数

但是以下参数是什么？

array([[ 0.41998763,  2.66314965,  3.34462572],
  [ 0.4199877 ,  2.66314968,  3.34462654],
  [ 0.41998749,  2.66314983,  3.34462649],
  [ 0.41998765,  2.66314917,  3.34462607]])
array([-1., -1., -1., -1.]))
status: 0

Answer 1

我敢打赌有一些封闭形式的解决方案，但如果被黑客入侵的代码就足够了，请参见下文

(此解决方案基于 scipy.optimize 包https://docs.scipy.org/doc/scipy/reference/tutorial/optimize.html(

(通过返回 -1 倍r_squared将最小值转换为最大化(

import numpy as np
from scipy import stats
from scipy import optimize
import IPython
def get_linregress(*args):
    #IPython.embed()
    w1,w2,w3 = args[0]
    x1_raw=np.array([277, 115, 196])
    x2_raw=np.array([263, 118, 191])
    x3_raw=np.array([270, 114, 191])
    w=np.array([w1, w2, w3])
    #w=np.array([1, 1, 1])
    x1=np.prod([w,x1_raw], axis=0).sum()
    x2=np.prod([w,x2_raw], axis=0).sum()
    x3=np.prod([w,x3_raw], axis=0).sum()
    x=np.array([x1, x2, x3])
    y=np.array([71.86, 71.14, 70.76])
    slope, intercept, r_value, p_value, std_err = stats.linregress(x,y) r_squared = r_value**2
    return -1*r_squared
res = optimize.minimize(get_linregress, [1,2,3], method='Nelder-Mead', tol=1e-6)
res.x