Question

我在熊猫中有以下数据框

 code    tank     var     nozzle_1    nozzle_2     nozzle_3
 123     1        23.34   12.23       54.56        12.22
 123     1        22.32   11.32       7.89         3.45
 123     1        21.22   19.93       5.54         5.66
 123     1        21.34   12.23       54.56        22.22
 123     1        32.32   13.32       4.89         32.45
 123     1        32.22   29.93       23.54        23.66
 123     2        23.34   12.23       54.56        12.22
 123     2        22.32   11.32       7.89         3.45
 123     2        21.22   19.93       5.54         5.66
 123     2        21.34   12.23       54.56        22.22
 123     2        32.32   13.32       4.89         32.45
 123     2        32.22   29.93       23.54        23.66

我想找到与tank分组的var_1列相关的喷嘴_1，喷嘴_2，喷嘴_3的相关性，并每3行获取相关性

我想要的数据框是

 code    tank     nozzle_1    nozzle_2     nozzle_3
 123     1        0.20        0.30         0.23 
 123     1        0.12        0.08         0.12
 123     2        0.14        0.12         0.01
 123     2        0.15        0.04         0.13

我正在跟踪熊猫

cols= df.columns[df.columns.str.contains(pat='nozzle_\d+$', regex=True)] 
cols= np.array(cols)


def corrVar(df, cols):
        for col in cols_to_scale:
            for i in range(0, df.shape[0], 3):
                df[col] = df.groupby('tank')[col, 'var'].corr()
        return df

test = corrVar（df，cols）

但是，它没有给我想要的结果。我们如何在熊猫中做到这一点？

Answer 1

对此没有简单的解决方案，所以这是我的细分：

我们获取以nozzle开头的列的列索引
我们获得列var的列索引

GroupBy

code, tank，并为每个nozzle列计算数据框的上半部的相关性
执行相同的操作，但下半场
Concat的两半互相重叠，作为final数据框

cols_idx = [df.columns.get_loc(c) for c in df.filter(like='nozzle').columns]
var_idx = df.columns.get_loc('var')

df1 = pd.concat([
    df.groupby(['code','tank']).apply(lambda x: x.iloc[:len(x)//2, var_idx].corr(x.iloc[:len(x)//2, idx])) for idx in cols_idx
], axis=1).reset_index()

df2 = pd.concat([
    df.groupby(['code','tank']).apply(lambda x: x.iloc[len(x)//2:, var_idx].corr(x.iloc[len(x)//2:, idx])) for idx in cols_idx
], axis=1).reset_index()


df_final = pd.concat([df1,df2]).sort_values('tank').reset_index(drop=True)

输出

   code  tank         0         1         2
0   123     1 -0.826376  0.876202  0.703793
1   123     1  0.540176 -0.931286  0.614626
2   123     2 -0.826376  0.876202  0.703793
3   123     2  0.540176 -0.931286  0.614626

如果您想正确地重命名列

Answer 2

import pandas as pd


data = [
        [ 123, 1, 23.34, 12.23, 54.56, 12.22 ],
        [ 123, 1, 22.32, 11.32, 7.89,  3.45  ],
        [ 123, 1, 21.22, 19.93, 5.54,  5.66  ],
        [ 123, 1, 21.34, 12.23, 54.56, 22.22 ],
        [ 123, 1, 32.32, 13.32, 4.89,  32.45 ],
        [ 123, 1, 32.22, 29.93, 23.54, 23.66 ],
        [ 123, 2, 23.34, 12.23, 54.56, 12.22 ],
        [ 123, 2, 22.32, 11.32, 7.89,  3.45  ],
        [ 123, 2, 21.22, 19.93, 5.54,  5.66  ],
        [ 123, 2, 21.34, 12.23, 54.56, 22.22 ],
        [ 123, 2, 32.32, 13.32, 4.89,  32.45 ],
        [ 123, 2, 32.22, 29.93, 23.54, 23.66 ]
]

columns = ['code', 'tank',  'var',  'nozzle_1', 'nozzle_2',  'nozzle_3']

df = pd.DataFrame(data=data, columns=columns)

print(df[['tank', 'var',  'nozzle_1', 'nozzle_2',  'nozzle_3']].groupby(['tank']).corr())


# ------------------------------------------------------
# RESULT:
#                     var  nozzle_1  nozzle_2  nozzle_3
# tank                                                 
# 1    var       1.000000  0.501164 -0.309435  0.761017
#      nozzle_1  0.501164  1.000000 -0.214982  0.168518
#      nozzle_2 -0.309435 -0.214982  1.000000  0.107815
#      nozzle_3  0.761017  0.168518  0.107815  1.000000
# 2    var       1.000000  0.501164 -0.309435  0.761017
#      nozzle_1  0.501164  1.000000 -0.214982  0.168518
#      nozzle_2 -0.309435 -0.214982  1.000000  0.107815
#      nozzle_3  0.761017  0.168518  0.107815  1.000000

如何为大熊猫中的n行计算与groupby的相关性

2 个答案: