(Time series - time interval) Based on the daily online time and offline time, calculate the user's longest continuous login time

The table is the login log of a hypothetical APP user. The goal is to count the highest continuous login time (number of days).
data = pd.read_excel('./time.xlsx',enconding='gb18030')

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Merge tables, output unique times, and perform positive time sorting
df1 = data[['id','s_time']].rename(columns={
    
     's_time': 'time'})
df2 = data[['id','e_time']].rename(columns={
    
     'e_time': 'time'})
data_1 = pd.concat([df1,df2]).sort_values(by='time',ascending=True)
data_1

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Output a unique time list, and a continuous time list (time_list)
def time_interval(li):
    list_e,start_time = [],None
    for x,y in zip(li,li[1:]):
        if start_time is None:
            start_time = x
        if x==li[-1]:
            list_e.append((start_time,y))
            print(start_time,y)
            start_time = None
        if  (y - x)/ np.timedelta64(1, 'D')> 1:
            list_e.append((start_time,x))
            start_time = None
    else:    
        list_e.append((start_time,y))
    return list_e
df_unique = data_1.groupby(['id'])['time'].agg({
    
    'unique'})
df_unique['time_list'] = df_unique['unique'].apply(lambda x:time_interval(x))
df_unique

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Perform time processing on time_list, output the number of consecutive login days within the time period, and then output the maximum number of days.
def days_time(x):
    time_intervals = [(end - start)/ np.timedelta64(1, 'D') for start, end in x]
    return time_interv
df_unique['days'] = df_unique['time_list'].apply(lambda x:days_time(x))
df_unique['max_day'] = df_unique['days'].apply(lambda x:max(x))
df_unique

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Origin blog.csdn.net/weixin_43502706/article/details/132169491