Python Pandas - 如何以秒频率对 DateTimeIndex 执行地板操作

要以秒频率对DateTimeIndex执行地板操作，请使用方法。对于秒频率，使用值为“S”的freq参数DateTimeIndex.floor()

首先，导入所需的库-

import pandas as pd

创建一个日期时间索引，周期为7，频率为S，即秒-

datetimeindex = pd.date_range(2021-10-18 07:20:32.261811624, periods=5, tz=Australia/Adelaide, freq=40S)

显示日期时间索引-

print("DateTimeIndex...\n", datetimeindex)

以秒频率对DateTimeIndex日期进行地板操作。对于秒频率，我们使用了“S”-

print("\nPerforming floor operation with seconds frequency...\n", datetimeindex.floor(freq=S))

示例

以下是代码-

import pandas as pd #DatetimeIndexwithperiod7andfrequencyasSi.e.seconds #timezoneisAustralia/Adelaide datetimeindex = pd.date_range(2021-10-18 07:20:32.261811624, periods=5, tz=Australia/Adelaide, freq=40S) #displayDateTimeIndex print("DateTimeIndex...\n", datetimeindex) #displayDateTimeIndex frequency print("DateTimeIndex frequency...\n", datetimeindex.freq) #gettingthesecond res = datetimeindex.second #displayonlythesecond print("\nThe seconds from DateTimeIndex...\n", res) #FlooroperationonDateTimeIndexdatewithsecondsfrequency # For seconds frequency, we have used S print("\nPerforming floor operation with seconds frequency...\n", datetimeindex.floor(freq=S))输出结果

这将产生以下代码-

DateTimeIndex... DatetimeIndex([2021-10-18 07:20:32.261811624+10:30, 2021-10-18 07:21:12.261811624+10:30, 2021-10-18 07:21:52.261811624+10:30, 2021-10-18 07:22:32.261811624+10:30, 2021-10-18 07:23:12.261811624+10:30], dtype=datetime64[ns, Australia/Adelaide], freq=40S) DateTimeIndex frequency... <40 * Seconds> The seconds from DateTimeIndex... Int64Index([32, 12, 52, 32, 12], dtype=int64) Performing floor operation with seconds frequency... DatetimeIndex([2021-10-18 07:20:32+10:30, 2021-10-18 07:21:12+10:30, 2021-10-18 07:21:52+10:30, 2021-10-18 07:22:32+10:30, 2021-10-18 07:23:12+10:30], dtype=datetime64[ns, Australia/Adelaide], freq=None)