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python时间日期相加减的实现示例

datetime relativedelta class dateutil 300    来源:    2024-10-17

前言

时间日期相加减,单独看没啥意思,但是比如要预测从当前时间开始,未来12个月的股票变化情况,你只拿到了股票的价格的序列值一共12个,需要把它扑到未来12个月,中间可能跨年。

安装插件

python-dateutil

pip install python-dateutil

一、python时间处理月份加减

import datetime
from dateutil.relativedelta import relativedelta
 
datetime_now = datetime.datetime.now()
datetime_three_month_ago = datetime_now - relativedelta(months=3)
print(datetime_three_month_ago)

二、python日期相减(秒、小时、天、月、年)

1.例子

代码如下(示例):

import datetime
today=datetime.datetime.now()
datetime1=today-datetime.timedelta(seconds=10)#减10秒
datetime2=today-datetime.timedelta(minutes=10)#减10分钟
datetime3=today-datetime.timedelta(hours=1)#减1小时
datetime4=today-datetime.timedelta(days=7)#减1天
datetime5=today-datetime.timedelta(weeks=1)#减1周
datetime6=today.strftime('%Y-%m-%d')#将时间格式化为字符串
print('today:',today)
print('second:',datetime1)
print('minute:',datetime2)
print('hour:',datetime3)
print('day:',datetime4)
print('week:',datetime5)
print('时间格式化为字符串:',datetime6)
  
  
import dateutil.relativedelta
datetime7= datetime.datetime.strptime(datetime6, '%Y-%m-%d')#将字符串格式为时间
datetime8 = datetime7 - dateutil.relativedelta.relativedelta(seconds=10)#减10秒
datetime9 = datetime7 - dateutil.relativedelta.relativedelta(minutes=10)#减10分钟
datetime10 = datetime7 - dateutil.relativedelta.relativedelta(hours=1)#减减1小时
datetime11 = datetime7 - dateutil.relativedelta.relativedelta(days=1)#减1天
datetime12 = datetime7 - dateutil.relativedelta.relativedelta(months=1)#减一个月
datetime13 = datetime7 - dateutil.relativedelta.relativedelta(years=1)#减一年
print('字符串格式为时间:',datetime7)
print('second2:',datetime8)
print('minute2:',datetime9)
print('hour2:',datetime10)
print('day2:',datetime11)
print('month2:',datetime12)
print('year:',datetime13)

2.案例:

处理未来12个月

def get_dynamic_year(i):
    today=datetime.datetime.now()
    tmp = today + dateutil.relativedelta.relativedelta(months=i)
    return tmp.year

def get_dynamic_month(i):
    today=datetime.datetime.now()
    tmp = today + dateutil.relativedelta.relativedelta(months=i)
    return tmp.month

pd_insert["year"] = [get_dynamic_year(i) for i in range(null,13)]
pd_insert["month"] = [ get_dynamic_month(i) for i in range(null,13) ]

处理未来7天的日期

    """
    pandas实在是太变态了,用row.name的方式获取行索引
    把每一天都加上
    """
    data_date = today + datetime.timedelta(days=i+1)
    return data_date.strftime("%Y-%m-%d")
pd_insert['forecast_date'] = today.strftime("%Y-%m-%d")
pd_insert['data_date'] = pd_insert.apply(
         lambda x: get_date(x.name, x['forecast_date']), axis=1)

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