从python中的数据帧字典中跨数据帧捕获变量值

时间:2018-11-01 08:34:50

标签: python pandas dataframe unique

我在字典中有3个数据框,其中键是月份标识符,值是数据框:

enter image description here

下面是数据帧和键的快照:

enter image description here

现在,对于每个唯一变量,我想捕获其在所有月份/数据框中的相关强度。 如果变量在df中具有相关值,则应将其捕获,否则该值将为0。类似于excel中的VLOOKUP。

最终数据框如下所示:

enter image description here

这对我来说用python实现似乎很复杂,所以有人可以帮我吗?

以下是用于生成示例数据并创建数据帧的字典的代码:

import pandas as pd
import numpy as np

df1 = pd.DataFrame([{'Variable_Name':'Pending_Disconnect','correlation': 0.553395448},
                    {'Variable_Name':'status_Active','correlation': 0.539464806},
                    {'Variable_Name':'days_active','correlation':0.414774231},
                    {'Variable_Name':'days_pend_disco','correlation':0.392915837},
                    {'Variable_Name':'prop_tenure','correlation':0.074321692},
                    {'Variable_Name':'abs_change_3m','correlation':0.062267386}
                    ])


df2 = pd.DataFrame([{'Variable_Name':'Pending_Change','correlation': 0.043461995},
                    {'Variable_Name':'status_Active','correlation': 0.038057697},
                    {'Variable_Name':'ethnic','correlation':0.037503202},
                    {'Variable_Name':'days_active','correlation':0.037227245},
                    {'Variable_Name':'archetype_grp','correlation':0.035761434},
                    {'Variable_Name':'age_nan','correlation':0.035761434}
                    ])


df3 = pd.DataFrame([{'Variable_Name':'active_frq_N','correlation':0.025697016},
                    {'Variable_Name':'active_frq_Y','correlation': 0.025697016},
                    {'Variable_Name':'ethnic','correlation':0.025195149},
                    {'Variable_Name':'ecgroup','correlation':0.023192408},
                    {'Variable_Name':'age','correlation':0.023121305},
                    {'Variable_Name':'archetype_nan','correlation':0.023121305}
                    ])

dfs = [df1,df2,df3]
months = ['Jan - Feb 2018','Jan - Mar 2018','Jan - Apr 2018']

sample_dict = dict(zip(months,dfs))

3 个答案:

答案 0 :(得分:1)

pd.concat用作:

df1.set_index('Variable_Name',inplace=True)
df2.set_index('Variable_Name',inplace=True)
df3.set_index('Variable_Name',inplace=True)

df = pd.concat([df1,df2,df3], axis=1, sort=False).fillna(0)
df.reset_index(inplace=True)
df.columns = ['Variable_Name','Jan - Feb 2018','Jan - Mar 2018','Jan - Apr 2018']

print(df)
         Variable_Name  Jan - Feb 2018  Jan - Mar 2018  Jan - Apr 2018
0   Pending_Disconnect        0.553395        0.000000        0.000000
1        status_Active        0.539465        0.038058        0.000000
2          days_active        0.414774        0.037227        0.000000
3      days_pend_disco        0.392916        0.000000        0.000000
4          prop_tenure        0.074322        0.000000        0.000000
5        abs_change_3m        0.062267        0.000000        0.000000
6       Pending_Change        0.000000        0.043462        0.000000
7               ethnic        0.000000        0.037503        0.025195
8        archetype_grp        0.000000        0.035761        0.000000
9              age_nan        0.000000        0.035761        0.000000
10        active_frq_N        0.000000        0.000000        0.025697
11        active_frq_Y        0.000000        0.000000        0.025697
12             ecgroup        0.000000        0.000000        0.023192
13                 age        0.000000        0.000000        0.023121
14       archetype_nan        0.000000        0.000000        0.023121

答案 1 :(得分:1)

您可以替换数据框的列名,然后使用pd.concat来连接数据框。

for key, df in sample_dict.items():
    df.rename(columns={'correlation':'correlation '+ key}, inplace=True)
pd.concat(dfs)

编辑:您还可以省略字典,并从数据框列表中执行此操作。

for i, df in enumerate(dfs):
    df.rename(columns={'correlation':'correlation '+ months[i]}, inplace=True)
pd.concat(dfs)  

答案 2 :(得分:0)

我的最终代码是@onno和@Sandeep Kadapa的代码的组合:

final_df = pd.DataFrame()
for key, df in sample_dict.items():
        df = sample_dict[key]
        df = df.iloc[:,0:2]
        df.rename(columns={'correlation':'correlation '+ key}, inplace=True)
        final_df = pd.concat([final_df,df],axis = 1,sort = False).fillna(0)

非常感谢您的快速恢复。