如何根据 pandas 中的 3 分钟间隔对 ID 进行分组？

Question

I have a dataframe that looks like this:

   ID     time      city        transport
0  1      10:20:00  London      car
1  20     08:50:20  Berlin      air plane
2  44     21:10:00  Paris       train
3  32     10:24:00  Rome        car
4  56     08:53:10  Berlin      air plane
5  90     21:8:00   Paris       train
.
.
.
1009 446  10:21:24  London     car

I want to group these data so that same value in 'city' and 'transport' but with time difference of +3min or -3min should have the same 'ID'.

I already tried pd.Grouper() like this but didn't work:

df['time'] = pd.to_datetime(df['time'])
df['ID'] = df.groupby([pd.Grouper(key= 'time',freq ='3min'),'city','transport'])['ID'].transform('first')

The output is the first dataframe I had without any changes. One reason could be that by using.datetime the date will be added as well to "time" and because my data is very big the date will differ and groupby doesn't work. I couldn't figure it out how to add time intervall (+3min or -3min) while using groupby and without adding DATE to 'time' column.

What I'm expecting is this:

   ID     time      city        transport
0  1      10:20:00  London      car
1  20     08:50:20  Berlin      air plane
2  44     21:10:00  Paris       train
3  32     10:24:00  Rome        car
4  20     08:53:10  Berlin      air plane
5  44     21:8:00   Paris       train
.
.
.
1009 1  10:21:24  London     car

it has been a while that I'm struggling with this question and I really appreciate any help. Thanks in advance

Answer 1

Exploring pd.Grouper()

1. found it useful to insert start time so that it's more obvious how buckets are being generated
2. you requirement +/- 3mins, most closely is a 6min bucket. Mostly matches your requirement but +/- 3 mins of what?
3. have done something that just shows what has been grouped and shows time bucket

setup

df = pd.read_csv(io.StringIO("""   ID     time      city        transport
0  1      10:20:00  London      car
1  20     08:50:20  Berlin      air plane
2  44     21:10:00  Paris       train
3  32     10:24:00  Rome        car
4  56     08:53:10  Berlin      air plane
5  90     21:08:00   Paris       train
6  33  05:08:22  Paris  train"""), sep="\s\s+", engine="python")

# force in origin so grouper generates bucket every Xmins from midnight with no seconds...
df = pd.concat([pd.DataFrame({"time":[pd.Timedelta(0)],"dummy":[True]}), df]).assign(dummy=lambda dfa: dfa.dummy.fillna(False))
df = df.assign(td=pd.to_timedelta(df.time))

analysis

### DEBUGGER ### - see whats being grouped...
df.groupby([pd.Grouper(key="td", freq="6min"), "city","transport"]).agg(lambda x: list(x) if len(x)>0 else np.nan).dropna()

• see that two time buckets will group >1 ID

	time	dummy	ID
(Timedelta('0 days 05:06:00'), 'Paris', 'train')	['05:08:22']	[False]	[33.0]
(Timedelta('0 days 08:48:00'), 'Berlin', 'air plane')	['08:50:20', '08:53:10']	[False, False]	[20.0, 56.0]
(Timedelta('0 days 10:18:00'), 'London', 'car')	['10:20:00']	[False]	[1.0]
(Timedelta('0 days 10:24:00'), 'Rome', 'car')	['10:24:00']	[False]	[32.0]
(Timedelta('0 days 21:06:00'), 'Paris', 'train')	['21:10:00', '21:08:00']	[False, False]	[44.0, 90.0]

solution

# finally +/- double the window.  NB this is not +/- but rows that group the same
(df.assign(ID=lambda dfa: dfa
           .groupby([pd.Grouper(key= 'td',freq ='6min'),'city','transport'])['ID']
           .transform('first'))
 # cleanup... NB needs changing if dummy row is not inserted
 .query("not dummy")
 .drop(columns=["td","dummy"])
 .assign(ID=lambda dfa: dfa.ID.astype(int))
)

time	ID	city	transport
10:20:00	1	London	car
08:50:20	20	Berlin	air plane
21:10:00	44	Paris	train
10:24:00	32	Rome	car
08:53:10	20	Berlin	air plane
21:08:00	44	Paris	train
05:08:22	33	Paris	train

Answer 2

def convert(seconds): 
    seconds = seconds % (24 * 3600) 
    hour = seconds // 3600
    seconds %= 3600
    minutes = seconds // 60
    seconds %= 60
    return hour,minutes,seconds

def get_sec(h,m,s):
     """Get Seconds from time."""
    if h==np.empty:
        h=0
    if m==np.empty:
        m=0
    if s==np.empty:
        s=0
    return int(h) * 3600 + int(m) * 60 + int(s)    

 df['time']=df['time'].apply(lambda x:       datetime.strptime(x,'%H:%M:%S') if isinstance(x,str) else x )

 df=df.sort_values(by=["time"])
 print(df)

 prev_hour=np.empty
 prev_minute=np.empty
 prev_second=np.empty
 for key,item in df.iterrows():
    curr_hour=item.time.hour
    curr_minute=item.time.minute
    curr_second=item.time.second
    curr_id=item.id
    curr_seconds=get_sec(curr_hour, curr_minute ,curr_second)
    prev_seconds=get_sec(prev_hour, prev_minute,prev_second)
    diff_seconds=curr_seconds-prev_seconds
hour,minute,second=convert(diff_seconds)
    if (hour==0) & (minute <=3):
        df.loc[key,'id']=prev_id
    prev_hour=item.time.hour
    prev_minute=item.time.minute
    prev_second=item.time.second
    prev_id=item.id

print(df)


output:
   id                time    city  transport
1  20 1900-01-01 08:50:20  Berlin  air plane
4  20 1900-01-01 08:53:10  Berlin  air plane
0   1 1900-01-01 10:20:00  London        car
3  32 1900-01-01 10:24:00    Rome        car
5  90 1900-01-01 21:08:00   Paris      train
2  90 1900-01-01 21:10:00   Paris      train