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修正案：

如果我有一个包含 5 列& Col1& Col2& Col3&Col4的pandas DataFrame ，Col5我需要通过考虑Col2Col3Col2Col4Col2Col5Col1

Col2下式得到的修改值：

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

其中 B是变化的变量（单个值），以获得（新值Col2, Col3）&（新值Col2, Col4）和（新值Col2, ）之间的最大 Pearson 相关系数Col5。

更新：

在此处输入图像描述

Col2上表包含我上面提到的 5 列，( , Col3) & ( Col2, Col4) & ( Col2, )之间的系数之间的相关性Col5如下表所示。

我需要Col2根据两个提到的方程式更改 的值，其中更改值为B。

所以问题是如何获得最好的值B，使新的相关系数大于或等于其对应物（旧）？

在此处输入图像描述

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

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1 回答 1

2

不是非常优雅，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 中读取clipboard，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
于 2019-04-28T14:27:20.963 回答