time_diff = pd.Timedelta('0 days 00:00:00') # 将时间差值转换为小时 hours = time_diff.total_seconds() / 3600 1. 2. 3. 4. 5. 6.
days:天数,可以为正数、负数或零。 使用timedelta可以很容易地计算日期时间之间的差值,例如: ``` import datetime # 计算1天后的日期 one_day = datetime.timedelta(days=1) print(one_day) 输出结果如下: ``` 1 day, 0:00:00 0:05:00 14 days, 0:00:00 ``` # 计算两个日期的差值 d1 = datetim...
delta1 = datetime.timedelta(days=1) print(delta1) # 输出: 1 day, 0:00:00 # 创建一个表示2小时的timedelta对象 delta2 = datetime.timedelta(hours=2) print(delta2) # 输出: 2:00:00 # 创建一个表示30分钟的timedelta对象 delta3 = datetime.timedelta(minutes=30) print(delta3) # 输出: 0:30...
0 days 06:00:00 1. Shell 数据偏移 例如- 周,天,小时,分钟,秒,毫秒,微秒,纳秒的数据偏移也可用于构建。 import pandas as pd timediff = pd.Timedelta(days=2) print(timediff) 1. 2. 3. 4. Python 执行上面救命代码,得到以下结果 - 2 days 00:00:00 1. Shell 运算操作 可以在Series/DataFrames...
delta = timedelta(days=1) print(delta) #输出:1 day, 0:00:00 #创建一个表示两天两小时三分钟四秒五毫秒的timedelta对象 delta = timedelta(days=2, hours=2, minutes=3, seconds=4, milliseconds=5) print(delta) #输出:2 days, 02:03:04.000005 三、timedelta对象的基本操作 一旦我们有了一个timedelt...
1 #class datetime.datetime.timedelta(days=0, seconds=0, microseconds=0,milliseconds=0, minutes=0, hours=0, weeks=0) 生成时间差 1 2 3 4 5 6 7 #生成时间差 today = datetime.date.today() td = datetime.timedelta(10) # 10 days, 0:00:00 ,<class 'datetime.timedelta'> td = datetime...
rng =timedelta_range('1 days','10 days') result = rng - delta expected =timedelta_range('0 days 22:00:00','9 days 22:00:00') tm.assert_index_equal(result, expected) rng -= delta tm.assert_index_equal(rng, expected)# intrng =timedelta_range('1 days 09:00:00', freq='H', ...
timedelta 浏览14提问于2019-05-28得票数 0 回答已采纳 2回答 Python:查找第一组连续数字(包括重复的数字) {5072: Timedelta('0 days 00:00:00'), 5085: Timedelta('0 days 00:00:00'), 5107: Timedelta('0 days 00:00:00'), 5126: Timedelta('1 days 00:00:00'), 5169: Timedelta('1 days...
2013-10-13 00:00:00 >>> print tt1-tt 2 days, 0:00:00 >>> print (tt1-tt).days 2 当有人问你昨天是几号,是很容易就得到答案的,但当要计算出100天前是几号,就不那么容易得出了。而Python中datetime的timedelta则可以轻松完成计算
TimedeltaIndex(['1 days 06:05:01.000030', '0 days 00:00:00.000015', NaT], dtype='timedelta64[ns]', freq=None) Converting numbers by specifying the `unit` keyword argument: >>> pd.to_timedelta(np.arange(5), unit='s') TimedeltaIndex(['00:00:00', '00:00:01', '00:00:02', ...