任务

1.按 “时间” 对订单经行 数据拆分

import pandas as pd
import numpy as np
from sqlalchemy import create_engine
engine = create_engine('mysql+pymysql://root:981221@localhost/testdb?charset=utf8mb4')
detail = pd.read_sql_table('meal_order_detail1',con=engine)
detail['place_order_time'] = pd.to_datetime(detail['place_order_time'])
detail['date'] = [i.date() for i in detail['place_order_time']]
detailGroup = detail[['date','counts','amounts']].groupby(by = 'date')
print('订单前 5 组的数目:\n',detailGroup.size().head())

订单前 5 组的数目:date
2016-08-01    217
2016-08-02    138
2016-08-03    157
2016-08-04    144
2016-08-05    193
dtype: int64

2.agg 聚合数据

• 计算售价的 “平均数” 和 “中位数”

dayMean = detailGroup.agg({'amounts':np.mean})
print('订单前 5 组的售价的均值：\n',dayMean.head())

订单前 5 组的售价的均值：amounts
date                 
2016-08-01  43.161290
2016-08-02  44.384058
2016-08-03  43.885350
2016-08-04  52.423611
2016-08-05  44.927461

dayMedian = detailGroup.agg({'amounts':np.median})
print('订单前 5  组的售价的中位数：\n',dayMedian.head())

订单前 5  组的售价的中位数：amounts
date               
2016-08-01     33.0
2016-08-02     35.0
2016-08-03     38.0
2016-08-04     39.0
2016-08-05     37.0

3. apply 聚合数据 统计总数

• counts 销量的总数统计

daySaleSum = detailGroup.apply(np.sum)
print('订单前 5 组的销量的总数:\n',daySaleSum['counts'].head())

订单前 5 组的销量的总数:date
2016-08-01    233.0
2016-08-02    151.0
2016-08-03    192.0
2016-08-04    169.0
2016-08-05    224.0
Name: counts, dtype: float64