Question

我有以下数据：

,dateTime,magnitude,occurrence,dateTime_s
1,2017-11-20 08:00:09.052260,12861,1,2017-11-20 08:00:09.000000
2,2017-11-20 08:00:09.052270,12868.12,1,2017-11-20 08:00:09.000000
3,2017-11-20 08:00:09.052282,12868.12,1,2017-11-20 08:00:09.000000
4,2017-11-20 08:00:09.052291,12867.5,2,2017-11-20 08:00:09.000000
5,2017-11-20 08:00:09.052315,12867.5,4,2017-11-20 08:00:09.000000
6,2017-11-20 08:00:09.052315,12867,1,2017-11-20 08:00:09.000000
7,2017-11-20 08:00:09.052315,12865.5,1,2017-11-20 08:00:09.000000
8,2017-11-20 08:00:09.052315,12865.89,1,2017-11-20 08:00:09.000000
9,2017-11-20 08:00:12.064744,12867.5,1,2017-11-20 08:00:12.000000
10,2017-11-20 08:00:12.131555,12868.5,2,2017-11-20 08:00:12.000000
11,2017-11-20 08:00:12.333511,12868.5,4,2017-11-20 08:00:12.000000
12,2017-11-20 08:00:12.333511,12869.95,2,2017-11-20 08:00:12.000000
13,2017-11-20 08:00:12.341516,12869.5,1,2017-11-20 08:00:12.000000
14,2017-11-20 08:00:12.343538,12868.5,1,2017-11-20 08:00:12.000000
15,2017-11-20 08:00:12.343538,12868.17,5,2017-11-20 08:00:12.000000
16,2017-11-20 08:00:12.343538,12867.5,2,2017-11-20 08:00:12.000000
17,2017-11-20 08:00:14.148704,12882.5,1,2017-11-20 08:00:14.000000
18,2017-11-20 08:00:14.148748,12882.5,1,2017-11-20 08:00:14.000000
19,2017-11-20 08:00:14.218977,12883.66,1,2017-11-20 08:00:14.000000
20,2017-11-20 08:00:14.218977,12883.5,1,2017-11-20 08:00:14.000000
21,2017-11-20 08:00:14.385283,12882.09,1,2017-11-20 08:00:14.000000
22,2017-11-20 08:00:14.388518,12881.5,1,2017-11-20 08:00:14.000000
23,2017-11-20 08:00:14.577002,12882.5,1,2017-11-20 08:00:14.000000

我正在使用以下代码按时间汇总它（因为它是毫秒，而我需要几秒钟。

import pandas as pd
import numpy as np

df = pd.read_csv('C:/Users/Data/test.csv')
print(df.head(30))

groups = df.groupby('dateTime_s')
df_grouped = (groups.agg({
            'magnitude': np.mean,
            'occurrence': np.sum,
            }))
print(df_grouped.head())

结果很好：

                               magnitude  occurrence
dateTime_s                                          
2017-11-20 08:00:09.000000  12866.328750          12
2017-11-20 08:00:12.000000  12868.515000          18
2017-11-20 08:00:14.000000  12882.607143           7

但是对于我的研究，我需要添加最常见的幅度和它的发生。如何组合（在当前组内）并以最大频率计算幅度并显示幅度和频率？

我正在寻找这样的结果：

                    groupby magnitude   
    dateTime_s      magnitude   occurrence  max sum
2017-11-20  08:00:09.000000     12866.32875 12  12867.5 6
2017-11-20  08:00:12.000000     12868.515   18  12868.5 7
2017-11-20  08:00:14.000000     12882.607143    7   12882.5 3

Answer 1

我认为您需要sum occurrence个magnitude值的自定义函数，其值为groups = df.groupby('dateTime_s') df_grouped = (groups.agg({ 'magnitude': np.mean, 'occurrence': np.sum, })) #print (df_grouped) def f(x): a = x['magnitude'].value_counts().index[0] b = x.loc[x['magnitude'] == a, 'occurrence'].sum() return pd.Series([a,b],['max magn','freq oc']) df_grouped1 = groups.apply(f) #print (df_grouped1) df = pd.concat([df_grouped, df_grouped1], axis=1) print (df) magnitude occurrence max magn freq oc dateTime_s 2017-11-20 08:00:09 12866.328750 12 12867.5 6.0 2017-11-20 08:00:12 12868.515000 18 12868.5 7.0 2017-11-20 08:00:14 12882.607143 7 12882.5 3.0个值{/ p>

groups = df.groupby('dateTime_s')

def f(x):
    a = x['magnitude'].value_counts().index[0]
    b = x.loc[x['magnitude'] == a, 'occurrence'].sum()
    c = x['magnitude'].mean()
    d = x['occurrence'].sum()
    return pd.Series([a,b,c,d],['max magn','freq oc', 'mean', 'sum'])

df_grouped1 = groups.apply(f)
print (df_grouped1)

                     max magn  freq oc          mean   sum
dateTime_s                                                
2017-11-20 08:00:09   12867.5      6.0  12866.328750  12.0
2017-11-20 08:00:12   12868.5      7.0  12868.515000  18.0
2017-11-20 08:00:14   12882.5      3.0  12882.607143   7.0

或仅限自定义功能：

var MessageSchema = new Schema({
    client_id: Number,
    messages: [{ 
        client_id: Number, 
        messages: [{
            type: { type: String },
            data: String
        }] 
    }]
})

pandas如何在groupby中进行分组

1 个答案: