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)