如何根据小时标准获得每天每个组的最小值

Question

我在下面给出了两个数据框供您测试

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]
 })

我想做的是

1) 在each day for each subject_id 中查找所有已为same for more than 1 hour 的值（来自val 列）并获取minimum of it

请注意，值也可以在 every 15 min duration 处捕获，因此您可能需要考虑 5 条记录才能看到 > 1 hr 条件）。请参阅下面的示例屏幕截图

2) 如果一天内没有same for more than 1 hour 的值，则只需获取minimum of that day for that subject_id

以下一个主题的屏幕截图将帮助您理解，我尝试过的代码如下

这是我尝试过的

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() 

由于我必须将其应用于数百万条记录，因此任何优雅高效的解决方案都会有所帮助

Answer 1

我想出了一个像下面这样的方法，它正在工作。欢迎任何建议

s=pd.to_timedelta(24,unit='h')-(df.time_1-df.time_1.dt.normalize())
df['tdiff'] = df.groupby(df.time_1.dt.date).time_1.diff().shift(-1).fillna(s)
df['t_d'] = df['tdiff'].dt.total_seconds()/3600
df['hr'] = df['time_1'].dt.hour
df['date'] = df['time_1'].dt.date
df['day'] = pd.DatetimeIndex(df['time_1']).day

# here I get the freq and cumsum of each val for each day and each hour. Since sort = 'False', timeorder is retained as is

temp_1 = pd.DataFrame(df.groupby(['subject_id','date','hr','val'], sort=False)['t_d'].agg({'cumduration':sum,'freq':'count'}).reset_index())

# here i remove the `hour` component and sum the value duration in same day but different hours (for example `5` was in 12th hour and 13th hour. we sum them)

temp_2 = pd.DataFrame(temp_1.groupby(['subject_id','date','val'], sort=False)['cumduration'].agg({'sum_of_cumduration':sum,'freq':'count'}).reset_index())

# Later, I create a mask for `> 1` hr criteria  

mask = temp_2.groupby(['subject_id','date'])['sum_of_cumduration'].apply(lambda x: x > 1)
output_1 = pd.DataFrame(temp_2[mask].groupby(['subject_id','date'])['val'].min()).reset_index()

 # I check for `< 1 ` hr records here 

output_2 = pd.DataFrame(temp_2[~mask].groupby(['subject_id','date'])['val'].min()).reset_index()

 # I finally check for `subject_id` and `date` and then append
output = output_1.append(output_2[~output_2['subject_id'].isin(output_1['subject_id'])])

output

Answer 2

试试这个

from datetime import timedelta

df1['time_1']= pd.to_datetime(df1['time_1'])
df1['date'] = df1['time_1'].dt.date
df1['t_d'] = df1.groupby(['date'])['time_1'].diff().shift(-1)
mask= df1['t_d']>pd.Timedelta(1,'h')
dfa=df1[mask]
dfb=df1[~mask].groupby('date').first().reset_index()
df_f = dfa.merge(dfb, how='outer')
df_f.drop_duplicates(subset='date', keep='first', inplace=True)
df_f.drop(['date','t_d'], axis=1, inplace=True)
df_f.sort_values('time_1')

Answer 3

试试这个。

from datetime import timedelta

def f(x):
    dif = (x.iloc[0]-x.iloc[-1])//timedelta(minutes=1)
    return dif
df1['time_1']= pd.to_datetime(df1['time_1'])
df1['flag']= df1.val.diff().ne(0).cumsum()
df1['t_d']=df1.groupby('flag')['time_1'].transform(f)
df1['date'] = df1['time_1'].dt.date
mask= df1['t_d'].ne(0)
dfa=df1[mask].groupby(['flag','date']).first().reset_index()
dfb=df1[~mask].groupby('date').first().reset_index().dropna(how='any')
df_f = dfa.merge(dfb, how='outer')
df_f.drop_duplicates(subset='date', keep='first', inplace=True)
df_f.drop(['flag','date','t_d'], axis=1, inplace=True)
df_f

输出。

 subject_id     time_1         val
0   1   2173-04-03 12:35:00     5
1   1   2173-04-04 11:30:00     5
2   1   2173-04-05 16:00:00     5
5   1   2173-04-06 04:00:00     3

Answer 4

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

输出

          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