python - How to get minimum of each group for each day based on hour criteria

Question

I have given two dataframes below for you to test

df = pd.DataFrame({
    'subject_id':[1,1,1,1,1,1,1,1,1,1,1],
    'time_1' :['2173-04-03 12:35:00','2173-04-03 17:00:00','2173-04-03 
         20:00:00','2173-04-04 11:00:00','2173-04-04 11:30:00','2173-04-04 
       12:00:00','2173-04-05 16:00:00','2173-04-05 22:00:00','2173-04-06 
       04:00:00','2173-04-06 04:30:00','2173-04-06 06:30:00'],
  'val' :[5,5,5,10,5,10,5,8,3,8,10]
 })


df1 = pd.DataFrame({
 'subject_id':[1,1,1,1,1,1,1,1,1,1,1],
 'time_1' :['2173-04-03 12:35:00','2173-04-03 12:50:00','2173-04-03 
           12:59:00','2173-04-03 13:14:00','2173-04-03 13:37:00','2173-04-04 
           11:30:00','2173-04-05 16:00:00','2173-04-05 22:00:00','2173-04-06 
           04:00:00','2173-04-06 04:30:00','2173-04-06 08:00:00'],
 'val' :[5,5,5,5,10,5,5,8,3,4,6]
 })

what I would like to do is

1) Find all values (from val column) which have been same for more than 1 hour in each day for each subject_id and get the minimum of it

Please note that values can also be captured at every 15 min duration as well, so you might have to consider 5 records to see > 1 hr condition). See sample screenshot below

2) If there are no values which were same for more than 1 hour in a day, then just get the minimum of that day for that subject_id

The below screenshot for one subject will help you understand and the code I tried is given below

This is what I tried

df['time_1'] = pd.to_datetime(df['time_1'])
df['time_2'] = df['time_1'].shift(-1)
df['tdiff'] = (df['time_2'] - df['time_1']).dt.total_seconds() / 3600
df['reading_day'] = pd.DatetimeIndex(df['time_1']).day

# don't know how to apply if else condition here to check for 1 hr criteria
t1 = df.groupby(['subject_id','reading_start_day','tdiff])['val'].min() 

As I have to apply this to million records, any elegant and efficient solution would be helpful

Answer 1

df = pd.DataFrame({
 'subject_id':[1,1,1,1,1,1,1,1,1,1],
 'time_1' :['2173-04-03 12:35:00','2173-04-03 17:00:00','2173-04-03 20:00:00','2173-04-04 11:00:00','2173-04-04 11:30:00','2173-04-04 12:00:00','2173-04-04 16:00:00','2173-04-04 22:00:00','2173-04-05 04:00:00','2173-04-05 06:30:00'],
  'val' :[5,5,5,10,5,10,5,8,8,10]
 })

# Separate Date and time
df['time_1']=pd.to_datetime(df['time_1'])
df['new_date'] = [d.date() for d in df['time_1']]
df['new_time'] = [d.time() for d in df['time_1']]


# find time diff in group with the first element to check > 1 hr
df['shift_val'] = df['val'].shift()
df1=df.assign(time_diff=df.groupby(['subject_id','new_date']).time_1.apply(lambda x: x - x.iloc[0]))

# Verify if time diff > 1 and value is not changed
df2=df1.loc[(df1['time_diff']/ np.timedelta64(1, 'h') >= 1) & (df1.val == df1.groupby('new_date').first().val[0])]
df3=df1.loc[(df1['time_diff']/ np.timedelta64(1, 'h') <= 1) & (df1.val == df1.shift_val)]

# Get the minimum within the group
df4=df2.append(df3).groupby(['new_date'], sort=False).min()

# drop unwanted columns
df4.drop(['new_time','shift_val','time_diff'],axis=1, inplace=True)

df4

Output

          subject_id    time_1     val
new_date            
2173-04-03  1   2173-04-03 17:00:00 5
2173-04-04  1   2173-04-04 16:00:00 5
2173-04-05  1   2173-04-05 04:00:00 8