使用 pandas 根据日期时间分配时间范围

Question

我需要根据输入的时间向大师求出时间段。 cust_id 开始时间 0 1 2000-01-01 09:00:03 1 2 2000-01-01 18:01:03

我需要的输出是 cust_id 开始时间 0 1 2000-01-01 09:00:03 早上 1 2 2000-01-01 18:01:03 晚上

用于创建主时间表详细信息的代码 mastdf={'timeframe':['morning','latemorning','midnoon','evening'],'start_time':['8:00:00', '11:00:00','13:00:00','17:00:00'],'end_time':['10:59:59','13:59:59','16 :59:59','7:59:59']}在此处输入代码

创建输入数据框的代码 inputdf={'cust_id':[1,2],'starttime':['2000-01-01 09:00:03', '2000-01-01 18:01:03']}

Answer 1

使用 cut for binning but first convert values to timedeltas by to_timedelta，创建带有添加端点 24H 的容器，00:00:00 到 8:00:00 之间的时间范围由 fillna 列的最后一个值使用 fillna =17=]:

mastdf={'timeframe':['morning','latemorning','midnoon','evening'],
        'start_time':['8:00:00','11:00:00','13:00:00','17:00:00'],
        'end_time':['10:59:59','13:59:59','16:59:59','7:59:59']}
mastdf = pd.DataFrame(mastdf)
print (mastdf)
     timeframe start_time  end_time
0      morning    8:00:00  10:59:59
1  latemorning   11:00:00  13:59:59
2      midnoon   13:00:00  16:59:59
3      evening   17:00:00   7:59:59

inputdf={'cust_id':[1,2],'starttime':['2000-01-01 09:00:03', '2000-01-01 18:01:03']}
inputdf = pd.DataFrame(inputdf)
inputdf['starttime'] = pd.to_datetime(inputdf['starttime'])

start =  pd.to_timedelta(mastdf['start_time']).tolist() + [pd.Timedelta(24, unit='h')]
s = pd.to_timedelta(inputdf['starttime'].dt.strftime('%H:%M:%S'))
last = mastdf['timeframe'].iat[-1]
inputdf['timeframe'] = pd.cut(s, 
                              bins=start, 
                              labels=mastdf['timeframe'], right=False).fillna(last)
print (inputdf)
   cust_id           starttime timeframe
0        1 2000-01-01 09:00:03   morning
1        2 2000-01-01 18:01:03   evening

使用 pandas 根据日期时间分配时间范围

Allocating the timeframe based on the datetime using pandas

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