如何在 Python 中将 Days Duration 划分为更小的部分答案

【问题标题】：How to divide Days Duration into further smaller sections in Python如何在 Python 中将 Days Duration 划分为更小的部分
【发布时间】：2021-08-31 23:05:43
【问题描述】：

'''

import pandas as pd

import pymysql as sql

import pygsheets

import datetime

#%%

today =datetime.date.today()
month=today.month
year =today.year
d =today.day

#%%

date =datetime.datetime.now().date() - pd.to_timedelta("90day")

date_filter =datetime.datetime.strftime(date,'%Y-%m-%d')

#%%

db = sql.connect(host = 'ffff', 
                 user = 'aadsd', 
                 password ='#######',
                 database = 'ewewewe')

cursor=db.cursor()
cursor.execute(f"""select * from table1 t1
left join table2  t2 on t1.id=t2.id
where t2.name ='xct' and date(addtime(t1.created_on,"05:30:00"))>{date}; """)
'''

我是 Python 新手，我已将最后一个单元格中的 {date} 定义为

日期 = datetime.datetime.now().date() - pd.to_timedelta("90day") ,

我想在另一个单元格中将 90 天部分拆分为前 45 天和下一个 45 天的概念，因为目前由于大量数据库记录，我面临“超出最大语句执行时间”错误的问题

【问题讨论】：

标签： python pandas spyder

【解决方案1】：

您可以使用它来将日期拆分为任意大小

import datetime
#%%

now = datetime.datetime.now()

date_filter =datetime.datetime.strftime(date,'%Y-%m-%d')

#%%
bin_size = 5
for i in range(90,0,-bin_size):
    end_date = now.strftime('%Y-%m-%d')
    start_date = now-datetime.timedelta(days=bin_size)
    now = start_date
    start_date = start_date.strftime('%Y-%m-%d')
    
    print(f"""select * from table1 t1
left join table2  t2 on t1.id=t2.id
where t2.name ='xct' and date(addtime(t1.created_on,"05:30:00")) between {start_date} and {end_date}; """)

这样的结果

select * from table1 t1
left join table2  t2 on t1.id=t2.id
where t2.name ='xct' and date(addtime(t1.created_on,"05:30:00")) between 2021-06-11 and 2021-06-16; 
select * from table1 t1
left join table2  t2 on t1.id=t2.id
where t2.name ='xct' and date(addtime(t1.created_on,"05:30:00")) between 2021-06-06 and 2021-06-11; 
select * from table1 t1
left join table2  t2 on t1.id=t2.id
where t2.name ='xct' and date(addtime(t1.created_on,"05:30:00")) between 2021-06-01 and 2021-06-06;
...

如果您希望 sql 查询更快。如果可能，请将 date(addtime(t1.created_on,"05:30:00")) 替换为 t1.created_on。
使用 python time shift 即更改 start_date 和 end_date 确实减少时间会更好。

【讨论】：