python 时间日期处理汇集 转：python 时间日期处理汇集

‍#计算精确时间差

#-----------------------------

# High Resolution Timers

t1=time.clock()

# Do Stuff Here

t2=time.clock()

print t2-t1

# 2.27236813618

# Accuracy will depend on platform and OS,

# but time.clock() uses the most accurate timer it can

time.clock(); time.clock()

# 174485.51365466841

# 174485.55702610247

#-----------------------------

# Also useful;

import timeit

code='[x for x in range(10) if x % 2 == 0]'

eval(code)

# [0, 2, 4, 6, 8]

t=timeit.Timer(code)

print"10,000 repeats of that code takes:",t.timeit(10000),"seconds"

print"1,000,000 repeats of that code takes:",t.timeit(),"seconds"

# 10,000 repeats of that code takes: 0.128238644856 seconds

# 1,000,000 repeats of that code takes: 12.5396490336 seconds

#-----------------------------

import timeit

code='import random; l = random.sample(xrange(10000000), 1000); l.sort()'

t=timeit.Timer(code)

print"Create a list of a thousand random numbers. Sort the list. Repeated a thousand times."

print"Average Time:",t.timeit(1000) /1000

# Time taken: 5.24391507859

‍#time , datetime , string 类型互相转换

# string -> time

> time.strptime(publishDate,"%Y-%m-%d %H:%M:%S")

# time -> string

>time.strftime("%y-%m-%d",t)

#string ->datetime

datetime.strptime(date_string,format)

#datetime->string

datetime.strftime("%Y-%m-%d %H:%M:%S")

strptime formating

‍#两日期相减 ‍时间差：

d1 = datetime.datetime(2005, 2, 16)

d2 = datetime.datetime(2004, 12, 31)

print (d1 - d2).days

starttime = datetime.datetime.now()

endtime = datetime.datetime.now()

print (endtime - starttime).seconds

‍#计算当前时间向后10天的时间。

#如果是小时 days 换成 hours

d1 = datetime.datetime.now()

d3 = d1 + datetime.timedelta(days =10)

print str(d3)

print d3.ctime()

import datetime, calendar

#昨天

def getYesterday():

today=datetime.date.today()

oneday=datetime.timedelta(days=1)

yesterday=today-oneday

return yesterday

#今天

def getToday():

return datetime.date.today()

#获取给定参数的前几天的日期，返回一个list

def getDaysByNum(num):

today=datetime.date.today()

oneday=datetime.timedelta(days=1)

li=[]

for i in range(0,num):

#今天减一天，一天一天减

today=today-oneday

#把日期转换成字符串

#result=datetostr(today)

li.append(datetostr(today))

return li

#将字符串转换成datetime类型

def strtodatetime(datestr,format):

return datetime.datetime.strptime(datestr,format)

#时间转换成字符串,格式为2008-08-02

def datetostr(date):

return str(date)[0:10]

#两个日期相隔多少天，例：2008-10-03和2008-10-01是相隔两天

def datediff(beginDate,endDate):

format="%Y-%m-%d";

bd=strtodatetime(beginDate,format)

ed=strtodatetime(endDate,format)

oneday=datetime.timedelta(days=1)

count=0

while bd!=ed:

ed=ed-oneday

count+=1

return count

#获取两个时间段的所有时间,返回list

def getDays(beginDate,endDate):

format="%Y-%m-%d";

bd=strtodatetime(beginDate,format)

ed=strtodatetime(endDate,format)

oneday=datetime.timedelta(days=1)

num=datediff(beginDate,endDate)+1

li=[]

for i in range(0,num):

li.append(datetostr(ed))

ed=ed-oneday

return li

#获取当前年份 是一个字符串

def getYear():

return str(datetime.date.today())[0:4]

#获取当前月份 是一个字符串

def getMonth():

return str(datetime.date.today())[5:7]

#获取当前天 是一个字符串

def getDay():

return str(datetime.date.today())[8:10]

def getNow():

return datetime.datetime.now()

print getToday()

print getYesterday()

print getDaysByNum(3)

print getDays('2008-10-01','2008-10-05')

print '2008-10-04 00:00:00'[0:10]

print str(getYear())+getMonth()+getDay()

print getNow()

‍#计算精确时间差

#-----------------------------

# High Resolution Timers

t1=time.clock()

# Do Stuff Here

t2=time.clock()

print t2-t1

# 2.27236813618

# Accuracy will depend on platform and OS,

# but time.clock() uses the most accurate timer it can

time.clock(); time.clock()

# 174485.51365466841

# 174485.55702610247

#-----------------------------

# Also useful;

import timeit

code='[x for x in range(10) if x % 2 == 0]'

eval(code)

# [0, 2, 4, 6, 8]

t=timeit.Timer(code)

print"10,000 repeats of that code takes:",t.timeit(10000),"seconds"

print"1,000,000 repeats of that code takes:",t.timeit(),"seconds"

# 10,000 repeats of that code takes: 0.128238644856 seconds

# 1,000,000 repeats of that code takes: 12.5396490336 seconds

#-----------------------------

import timeit

code='import random; l = random.sample(xrange(10000000), 1000); l.sort()'

t=timeit.Timer(code)

print"Create a list of a thousand random numbers. Sort the list. Repeated a thousand times."

print"Average Time:",t.timeit(1000) /1000

# Time taken: 5.24391507859

‍#time , datetime , string 类型互相转换

# string -> time

> time.strptime(publishDate,"%Y-%m-%d %H:%M:%S")

# time -> string

>time.strftime("%y-%m-%d",t)

#string ->datetime

datetime.strptime(date_string,format)

#datetime->string

datetime.strftime("%Y-%m-%d %H:%M:%S")

strptime formating

‍#两日期相减 ‍时间差：

d1 = datetime.datetime(2005, 2, 16)

d2 = datetime.datetime(2004, 12, 31)

print (d1 - d2).days

starttime = datetime.datetime.now()

endtime = datetime.datetime.now()

print (endtime - starttime).seconds

‍#计算当前时间向后10天的时间。

#如果是小时 days 换成 hours

d1 = datetime.datetime.now()

d3 = d1 + datetime.timedelta(days =10)

print str(d3)

print d3.ctime()

import datetime, calendar

#昨天

def getYesterday():

today=datetime.date.today()

oneday=datetime.timedelta(days=1)

yesterday=today-oneday

return yesterday

#今天

def getToday():

return datetime.date.today()

#获取给定参数的前几天的日期，返回一个list

def getDaysByNum(num):

today=datetime.date.today()

oneday=datetime.timedelta(days=1)

li=[]

for i in range(0,num):

#今天减一天，一天一天减

today=today-oneday

#把日期转换成字符串

#result=datetostr(today)

li.append(datetostr(today))

return li

#将字符串转换成datetime类型

def strtodatetime(datestr,format):

return datetime.datetime.strptime(datestr,format)

#时间转换成字符串,格式为2008-08-02

def datetostr(date):

return str(date)[0:10]

#两个日期相隔多少天，例：2008-10-03和2008-10-01是相隔两天

def datediff(beginDate,endDate):

format="%Y-%m-%d";

bd=strtodatetime(beginDate,format)

ed=strtodatetime(endDate,format)

oneday=datetime.timedelta(days=1)

count=0

while bd!=ed:

ed=ed-oneday

count+=1

return count

#获取两个时间段的所有时间,返回list

def getDays(beginDate,endDate):

format="%Y-%m-%d";

bd=strtodatetime(beginDate,format)

ed=strtodatetime(endDate,format)

oneday=datetime.timedelta(days=1)

num=datediff(beginDate,endDate)+1

li=[]

for i in range(0,num):

li.append(datetostr(ed))

ed=ed-oneday

return li

#获取当前年份 是一个字符串

def getYear():

return str(datetime.date.today())[0:4]

#获取当前月份 是一个字符串

def getMonth():

return str(datetime.date.today())[5:7]

#获取当前天 是一个字符串

def getDay():

return str(datetime.date.today())[8:10]

def getNow():

return datetime.datetime.now()

print getToday()

print getYesterday()

print getDaysByNum(3)

print getDays('2008-10-01','2008-10-05')

print '2008-10-04 00:00:00'[0:10]

print str(getYear())+getMonth()+getDay()

print getNow()