当scipy.stats.spearmanr（df1，df2）用于两个数据框时，您可以将输出值解栈以获得2个numpy相关性和p值数组。如何将它们放回与原始数据框具有相同列名的数据框中？

correlation, p_value = scipy.stats.spearmanr(df1, df2)

生成2个numpy数组，但希望获得具有相关性的 Dataframe ，并且对于x列，p_value与每列相邻，如下所示：

不明显...这里有一些代码来获取随机数据（由ouroboros1建议）

import scipy.stats
import numpy as np
import pandas as pd

# Create array of 5 rows and 3 columns,

# filled with random values from 1 to 10

data = np.random.randint(1,10,size=(5,3))

# Create Dataframe with data

df1 = pd.DataFrame(data, columns=['Col_1','Col_2','Col_3'])

# Set target column in new data frame

df2 = df1['Col_3'].to_frame().copy()
df1.drop(['Col_3'], axis=1, inplace=True)

# Obtain correlation coefficient for each

# value in column   against value in every other column

correlation, p_value = scipy.stats.spearmanr(df1, df2)

（编辑）我想我找到了一条路，但必须有一条更短的路：

# Concat frames that contain features with frame that

# contains target

correlation_frame = pd.concat([df1, df2], axis=1)
col_cor_list = list(correlation_frame)

# Create new frame containing all correlation coefficients

cor = pd.DataFrame(correlation, columns=col_list)
cor
    Col_1     Col_2        Col_3
0   1.000000    -0.883883   -0.229416
1   -0.883883   1.000000    0.081111
2   -0.229416   0.081111    1.000000

# Get a dataframe for p-values

col_p_list = ['p_value' + str(x) for x in
              range(1,len(p_value)+1)]
p_frame = pd.DataFrame(p_value, columns=col_p_list)
p_frame

    p_value1        p_value2    p_value3
0   1.404265e-24    0.046662    0.710482
1   4.666188e-02    0.000000    0.896840
2   7.104817e-01    0.896840    0.000000

# Combine values by alternating column names

# so p-values are placed correctly

alter_names = (list(sum(zip(col_cor_list, col_p_list), ())))
final = cor.join(p_frame)
results = final[alter_names]
results

   Col_1            p_value1       Col_2    p_value2      Col_3 p_value3
0   1.000000    1.404265e-24    -0.883883   0.046662 -0.229416  0.710482
1   -0.883883   4.666188e-02    1.000000    0.000000    0.081111    0.896840
2   -0.229416   7.104817e-01    0.081111    0.896840    1.000000    0.000000