Python 需要以历元格式获取日期范围
需要以历元格式获取当前月份的范围(从开始到月底,两个变量,以毫秒为单位的历元格式)。怎么做?也许有人已经这样做了,并且可以提供帮助?来自熊猫文档:。将datetime转换为int64,然后除以转换单位Python 需要以历元格式获取日期范围,python,pandas,datetime,time,Python,Pandas,Datetime,Time,需要以历元格式获取当前月份的范围(从开始到月底,两个变量,以毫秒为单位的历元格式)。怎么做?也许有人已经这样做了,并且可以提供帮助?来自熊猫文档:。将datetime转换为int64,然后除以转换单位 import pandas as pd def to_epoch(**kwargs): """returns: NumPy ndarray.""" stamps = pd.date_range(**kwargs) return stamps.view('int64') /
import pandas as pd
def to_epoch(**kwargs):
"""returns: NumPy ndarray."""
stamps = pd.date_range(**kwargs)
return stamps.view('int64') // pd.Timedelta(1, unit='s')
to_epoch(start='2017-04-01', end='2017-04-30')
array([1491004800, 1491091200, 1491177600, 1491264000, 1491350400,
1491436800, 1491523200, 1491609600, 1491696000, 1491782400,
1491868800, 1491955200, 1492041600, 1492128000, 1492214400,
1492300800, 1492387200, 1492473600, 1492560000, 1492646400,
1492732800, 1492819200, 1492905600, 1492992000, 1493078400,
1493164800, 1493251200, 1493337600, 1493424000, 1493510400])
不是熊猫,但一般来说:
from datetime import datetime
import calendar
epoch = datetime.utcfromtimestamp(0)
def unix_time(dt):
return (dt - epoch).total_seconds() * 1000.0
today = datetime.now()
lastDay = calendar.monthrange(today.year, today.month)[1]
firstDayOfMonth = today.replace(day=1)
lastDayofMonth = today.replace(day=lastDay)
firstEpoch = unix_time(firstDayOfMonth)
lastEpoch = unix_time(lastDayOfMonth)
或: