优化更改变量以获得多列的最大 Pearson 相关系数

Question

修改：

如果我有一个包含 5 列 Col1 & Col2 & Col3 & Col4 & Col5 的 pandas DataFrame，我需要获得最大 Pearson 相关系数介于 (Col2,Col3 ) & (Col2,Col4) & (Col2,Col5) 通过考虑Col1中的值

Col2的修改值由下式得到：

df['Col1']=np.power((df['Col1']),B)
df['Col2']=df['Col2']*df['Col1']

其中B 是变化变量（单个值），以获得最大 Pearson 相关系数（Col2,Col3）和（Col2,Col4 的新值）和（ Col2,Col5) 的新值。

更新：

上表包含我上面提到的5列，说明了(Col2,Col3) & (Col2,Col4) & (Col2,Col5)之间的系数之间的相关性桌子下方。

我需要根据上面提到的两个等式更改Col2 的值，其中更改值为B。

所以问题是如何获得B 的最佳值，它给出的新相关系数大于或等于其对应物（旧）？

更新 2：

Col1,Col2,Col3,Col4,Col5

2,0.051361397,2618,1453,1099

4,0.053507779,306,153,150

2,0.041236151,39,54,34

6,0.094526419,2755,2209,1947

4,0.079773397,2313,1261,1022

4,0.083891415,3528,2502,2029

6,0.090737243,3594,2781,2508

2,0.069552772,370,234,246

2,0.052401789,690,402,280

2,0.039930675,1218,846,631

4,0.065952096,1706,523,453

2,0.053064126,314,197,123

6,0.076847486,4019,1675,1452

2,0.044881545,604,402,356

2,0.073102611,2214,1263,1050

0,0.046998526,938,648,572

Answer 1

不是很优雅，但很有效；随意使它更通用：

import pandas as pd
from scipy.optimize import minimize


def minimize_me(b, df):

    # we want to maximize, so we have to multiply by -1
    return -1 * df['Col3'].corr(df['Col2'] * df['Col1'] ** b )

# read your dataframe from somehwere, e.g. csv
df = pd.read_clipboard(sep=',')

# B is greater than 0 for now
bnds = [(0, None)]

res = minimize(minimize_me, (1), args=(df,), bounds=bnds)

if res.success:
    # that's the optimal B
    print(res.x[0])

    # that's the highest correlation you can get
    print(-1 * res.fun)
else:
    print("Sorry, the optimization was not successful. Try with another initial"
          " guess or optimization method")

这将打印：

0.9020784246026575 # your B
0.7614993786787415 # highest correlation for corr(col2, col3)

我现在从clipboard 读取，将其替换为您的.csv 文件。然后，您还应该避免对列进行硬编码；上面的代码仅用于演示目的，以便您了解如何设置优化问题本身。

如果您对总和感兴趣，可以使用（其余代码未修改）：

def minimize_me(b, df):

    col_mod = df['Col2'] * df['Col1'] ** b

    # we want to maximize, so we have to multiply by -1
    return -1 * (df['Col3'].corr(col_mod) +
                 df['Col4'].corr(col_mod) +
                 df['Col5'].corr(col_mod))

这将打印：

1.0452394748131613
2.3428368479642137