I want to concatenate 2 DataFrames with different PeriodIndex frequencies, and using for sorting the second level index which is a position.
For example, I have following 2 DataFrames.
import pandas as pd
pr1h = pd.period_range(start='2020-01-01 08:00', end='2020-01-01 11:00', freq='1h')
pr2h = pd.period_range(start='2020-01-01 08:00', end='2020-01-01 11:00', freq='2h')
n_array_1h = [2, 2, 2, 2]
n_array_2h = [0, 1, 0, 1]
index_labels_1h = [pr1h, n_array_1h]
index_labels_2h = [[pr2h[0],pr2h[0],pr2h[1],pr2h[1]], n_array_2h]
values_1h = [[1], [2], [3], [4]]
values_2h = [[10], [20], [30], [40]]
df1h = pd.DataFrame(values_1h, index=index_labels_1h, columns=['Data'])
df1h.index.names=['Period','Position']
df2h = pd.DataFrame(values_2h, index=index_labels_2h, columns=['Data'])
df2h.index.names=['Period','Position']
df1h
Data
Period Position
2020-01-01 08:00 2 1
2020-01-01 09:00 2 2
2020-01-01 10:00 2 3
2020-01-01 11:00 2 4
df2h
Data
Period Position
2020-01-01 08:00 0 10
1 20
2020-01-01 10:00 0 30
1 40
I woudl like to obtain df1h_new, which:
keeps the PeriodIndex from df1h,
keeps data from the block in df2h having a period.start_time immediately lower or equal than current perdiod.start_time in df1h,
keeps data from df1h obviously
So result would be.
df1h_new
Data
Period Position
2020-01-01 08:00 0 10 # |---> data coming from df2h, block with
1 20 # | start_time =< df1h.index[0].start_time
2 1 # ----> data from df1h.index[0]
2020-01-01 09:00 0 10 # |---> data coming from df2h, block with
1 20 # | start_time =< df1h.index[1].start_time
2 2 # ----> data from df1h.index[1]
2020-01-01 10:00 0 30 # and so on...
1 40
2 3
2020-01-01 11:00 0 30
1 40
2 4
Please, what would be the recommended way to achieve that?
I thank you for your help and support! Bests,
One idea is use concat with Series.unstack and change frequency to same by Series.asfreq, then back filling misisng values and reshape back to MultiIndex:
df = (pd.concat([df1h['Data'].unstack(),
df2h['Data'].unstack().asfreq('H')], axis=1)
.bfill()
.stack()
.sort_index()
.to_frame('Data'))
print (df)
Data
Period Position
2020-01-01 08:00 0 10.0
1 20.0
2 1.0
2020-01-01 09:00 0 10.0
1 20.0
2 2.0
2020-01-01 10:00 0 30.0
1 40.0
2 3.0
2020-01-01 11:00 0 30.0
1 40.0
2 4.0
Related
i want to interpolate (Linear interpolation) data. but There is no NA.
Here is my data.with many missing values.
timestamp
id
strength
1383260400000
1
-0.3803901328171995
1383261000000
1
-0.42196042219455937
1383265200000
1
-0.460714706261982
My expected :
timestamp
id
strength
1383260400000
1
-0.3803901328171995
1383261000000
1
-0.42196042219455937
1383261600000
1
Linear interpolated data
1383262200000
1
Linear interpolated data
1383262800000
1
Linear interpolated data
1383263400000
1
Linear interpolated data
1383264000000
1
Linear interpolated data
1383264600000
1
Linear interpolated data
1383265200000
1
-0.460714706261982
timestamp starts 1383260400000, ends 1383343800000
and another id(from 1 to 2025) has same issues.
Idea is create datetimes, convert to DatetimeIndex and in lambda function add missing datetimes by Series.asfreq with interpolate:
df['timestamp'] = pd.to_datetime(df['timestamp'], unit='ms')
f = lambda x: x.asfreq('10Min').interpolate()
df = df.set_index('timestamp').groupby('id')['strength'].apply(f).reset_index()
print (df)
id timestamp strength
0 1 2013-10-31 23:00:00 -0.380390
1 1 2013-10-31 23:10:00 -0.421960
2 1 2013-10-31 23:20:00 -0.427497
3 1 2013-10-31 23:30:00 -0.433033
4 1 2013-10-31 23:40:00 -0.438569
5 1 2013-10-31 23:50:00 -0.444106
6 1 2013-11-01 00:00:00 -0.449642
7 1 2013-11-01 00:10:00 -0.455178
8 1 2013-11-01 00:20:00 -0.460715
Last if need original format of timestamps:
df['timestamp'] = df['timestamp'].astype(np.int64) // 1000000
print (df)
id timestamp strength
0 1 1383260400000 -0.380390
1 1 1383261000000 -0.421960
2 1 1383261600000 -0.427497
3 1 1383262200000 -0.433033
4 1 1383262800000 -0.438569
5 1 1383263400000 -0.444106
6 1 1383264000000 -0.449642
7 1 1383264600000 -0.455178
8 1 1383265200000 -0.460715
EDIT:
#data from question
df =pd.DataFrame({'timestamp': [1383260400000, 1383261000000, 1383265200000],
'id': [1, 1, 1],
'strength':[-0.3803901328171995,-0.4219604221945593,-0.460714706261982]})
print (df)
timestamp id strength
0 1383260400000 1 -0.380390
1 1383261000000 1 -0.421960
2 1383265200000 1 -0.460715
Solution create for each id all datetimes by date_range and create missing values by DataFrame.reindex with MultiIndex, last per id is used interpolate:
df['timestamp'] = pd.to_datetime(df['timestamp'], unit='ms')
r = pd.date_range(pd.to_datetime(1383260400000, unit='ms') ,
pd.to_datetime(1383343800000, unit='ms'),
freq='10Min')
ids = df['id'].unique()
mux = pd.MultiIndex.from_product([r, ids], names=['timestamp','id'])
f = lambda x: x.interpolate()
df = (df.set_index(['timestamp', 'id'])
.reindex(mux)
.groupby('id')['strength']
.transform(f)
.reset_index())
print (df)
timestamp id strength
0 2013-10-31 23:00:00 1 -0.380390
1 2013-10-31 23:10:00 1 -0.421960
2 2013-10-31 23:20:00 1 -0.427497
3 2013-10-31 23:30:00 1 -0.433033
4 2013-10-31 23:40:00 1 -0.438569
.. ... .. ...
135 2013-11-01 21:30:00 1 -0.460715
136 2013-11-01 21:40:00 1 -0.460715
137 2013-11-01 21:50:00 1 -0.460715
138 2013-11-01 22:00:00 1 -0.460715
139 2013-11-01 22:10:00 1 -0.460715
[140 rows x 3 columns]
This is a real use case that I am trying to implement in my work.
Sample data (fake data but similar data structure)
Lap Starttime Endtime
1 10:00:00 10:05:00
format: hh:mm:ss
Desired output
Lap time
1 10:00:00
1 10:01:00
1 10:02:00
1 10:03:00
1 10:04:00
1 10:05:00
so far only trying to think of the logic and techniques required... the codes are not correct
import re
import pandas as pd
df = pd.read_csv('sample.csv')
#1. to determine how many rows to generate. eg. 1000 to 1005 is 6 rows
df['time'] = df['Endtime'] - df['Startime']
#2. add one new row with 1 added minute. eg. 6 rows
for i in No_of_rows:
if df['time'] < df['Endtime']: #if 'time' still before end time, then continue append
df['time'] = df['Startime'] += 1 #not sure how to select Minute part only
else:
continue
pardon my limited coding skills. appreciate all the help from you experts.. thanks!
Try with pd.date_range and explode:
#convert to datetime if needed
df["Starttime"] = pd.to_datetime(df["Starttime"], format="%H:%M:%S")
df["Endtime"] = pd.to_datetime(df["Endtime"], format="%H:%M:%S")
#create list of 1min ranges
df["Range"] = df.apply(lambda x: pd.date_range(x["Starttime"], x["Endtime"], freq="1min"), axis=1)
#explode, drop unneeded columns and keep only time
df = df.drop(["Starttime", "Endtime"], axis=1).explode("Range")
df["Range"] = df["Range"].dt.time
>>> df
Range
Lap
1 10:00:00
1 10:01:00
1 10:02:00
1 10:03:00
1 10:04:00
1 10:05:00
Input df:
df = pd.DataFrame({"Lap": [1],
"Starttime": ["10:00:00"],
"Endtime": ["10:05:00"]}).set_index("Lap")
>>> df
Starttime Endtime
Lap
1 10:00:00 10:05:00
You can convert the times to datetimes, that will arbitrarily prepend the date of today (at whatever date you’re running) but we can then remove that later and it allows for easier manupulation:
>>> bounds = df[['Starttime', 'Endtime']].transform(pd.to_datetime)
>>> bounds
Starttime Endtime
0 2021-09-29 10:00:00 2021-09-29 10:05:00
1 2021-09-29 10:00:00 2021-09-29 10:02:00
Then we can simply use pd.date_range with a 1 minute frequency:
>>> times = bounds.agg(lambda s: pd.date_range(*s, freq='1min'), axis='columns')
>>> times
0 DatetimeIndex(['2021-09-29 10:00:00', '2021-09...
1 DatetimeIndex(['2021-09-29 10:00:00', '2021-09...
dtype: object
Now joining that with the Lap info and using df.explode():
>>> result = df[['Lap']].join(times.rename('time')).explode('time').reset_index(drop=True)
>>> result
Lap time
0 1 2021-09-29 10:00:00
1 1 2021-09-29 10:01:00
2 1 2021-09-29 10:02:00
3 1 2021-09-29 10:03:00
4 1 2021-09-29 10:04:00
5 1 2021-09-29 10:05:00
6 2 2021-09-29 10:00:00
7 2 2021-09-29 10:01:00
8 2 2021-09-29 10:02:00
Finally we wanted to remove the day:
>>> result['time'] = result['time'].dt.time
>>> result
Lap time
0 1 10:00:00
1 1 10:01:00
2 1 10:02:00
3 1 10:03:00
4 1 10:04:00
5 1 10:05:00
6 2 10:00:00
7 2 10:01:00
8 2 10:02:00
The objects in your series are now datetime.time
Here is another way without using apply/agg:
Convert to datetime first:
df["Starttime"] = pd.to_datetime(df["Starttime"], format="%H:%M:%S")
df["Endtime"] = pd.to_datetime(df["Endtime"], format="%H:%M:%S")
Get difference between the end and start times and then using index.repeat, repeat the rows. Then using groupby & cumcount, get pd.to_timedelta in minutes and add to the existing start time:
repeats = df['Endtime'].sub(df['Starttime']).dt.total_seconds()//60
out = df.loc[df.index.repeat(repeats+1),['Lap','Starttime']]
out['Starttime'] = (out['Starttime'].add(
pd.to_timedelta(out.groupby("Lap").cumcount(),'min')).dt.time)
print(out)
Lap Starttime
0 1 10:00:00
0 1 10:01:00
0 1 10:02:00
0 1 10:03:00
0 1 10:04:00
0 1 10:05:00
First time posting, newbie to python.
I have a data frame consisting of 3 columns: ['ID', 'date', 'profit_forecast']
'ID': is product ID
'date': start date
'profit_forecast': a list containing 367 items, each item is a profit forecast for date+n
I am looking to create a new data frame that maps each item in profit_forecast to the ID and corresponding date+n for its position in the list.
Not sure how to start.
Thanks in advance!
If I understand you correctly, the following example data captures the essence of your question:
df = pd.DataFrame({'ID': [1, 2, 3],
'date': pd.date_range('2019-01-01', freq='YS', periods=3),
'profit_forecast': [[1, 2, 3], [4, 5], [6, 7, 8, 9]]})
df
ID date profit_forecast
0 1 2019-01-01 [1, 2, 3]
1 2 2020-01-01 [4, 5]
2 3 2021-01-01 [6, 7, 8, 9]
One solution is to make sure you've upgraded to pandas 0.25, and then to explode the profit_forecast column:
res = df.explode('profit_forecast')
res
ID date profit_forecast
0 1 2019-01-01 1
0 1 2019-01-01 2
0 1 2019-01-01 3
1 2 2020-01-01 4
1 2 2020-01-01 5
2 3 2021-01-01 6
2 3 2021-01-01 7
2 3 2021-01-01 8
2 3 2021-01-01 9
At this point, your question is not clear enough on how you need to increment the dates of each ID. If by "date + n" you mean to add one day to each consecutive date within each ID, then something like this should work:
res['date'] = res['date'] + pd.to_timedelta(res.groupby('ID').cumcount(), 'D')
res
ID date profit_forecast
0 1 2019-01-01 1
0 1 2019-01-02 2
0 1 2019-01-03 3
1 2 2020-01-01 4
1 2 2020-01-02 5
2 3 2021-01-01 6
2 3 2021-01-02 7
2 3 2021-01-03 8
2 3 2021-01-04 9
I have a dataframe and some columns. I want to sum column "Gap" where time is in some time slots.
region. date. time. gap
0 1 2016-01-01 00:00:08 1
1 1 2016-01-01 00:00:48 0
2 1 2016-01-01 00:02:50 1
3 1 2016-01-01 00:00:52 0
4 1 2016-01-01 00:10:01 0
5 1 2016-01-01 00:10:03 1
6 1 2016-01-01 00:10:05 0
7 1 2016-01-01 00:10:08 0
I want to sum gap column. I have time slots in dict like that.
'slot1': '00:00:00', 'slot2': '00:10:00', 'slot3': '00:20:00'
Now after summation, above dataframe should like that.
region. date. time. gap
0 1 2016-01-01 00:10:00/slot1 2
1 1 2016-01-01 00:20:00/slot2 1
I have many regions and 144 time slots from 00:00:00 to 23:59:49. I have tried this.
regres=reg.groupby(['start_region_hash','Date','Time'])['Time'].apply(lambda x: (x >= hoursdict['slot1']) & (x <= hoursdict['slot2'])).sum()
But it doesn't work.
Idea is convert column time to datetimes with floor by 10Min, then convert to strings HH:MM:SS:
d = {'slot1': '00:00:00', 'slot2': '00:10:00', 'slot3': '00:20:00'}
d1 = {v:k for k, v in d.items()}
df['time'] = pd.to_datetime(df['time']).dt.floor('10Min').dt.strftime('%H:%M:%S')
print (df)
region date time gap
0 1 2016-01-01 00:00:00 1
1 1 2016-01-01 00:00:00 0
2 1 2016-01-01 00:00:00 1
3 1 2016-01-01 00:00:00 0
4 1 2016-01-01 00:10:00 0
5 1 2016-01-01 00:10:00 1
6 1 2016-01-01 00:10:00 0
7 1 2016-01-01 00:10:00 0
Aggregate sum and last map values by dictionary with swapped keys with values:
regres = df.groupby(['region','date','time'], as_index=False)['gap'].sum()
regres['time'] = regres['time'] + '/' + regres['time'].map(d1)
print (regres)
region date time gap
0 1 2016-01-01 00:00:00/slot1 2
1 1 2016-01-01 00:10:00/slot2 1
If want display next 10Min slots:
d = {'slot1': '00:00:00', 'slot2': '00:10:00', 'slot3': '00:20:00'}
d1 = {v:k for k, v in d.items()}
times = pd.to_datetime(df['time']).dt.floor('10Min')
df['time'] = times.dt.strftime('%H:%M:%S')
df['time1'] = times.add(pd.Timedelta('10Min')).dt.strftime('%H:%M:%S')
print (df)
region date time gap time1
0 1 2016-01-01 00:00:00 1 00:10:00
1 1 2016-01-01 00:00:00 0 00:10:00
2 1 2016-01-01 00:00:00 1 00:10:00
3 1 2016-01-01 00:00:00 0 00:10:00
4 1 2016-01-01 00:10:00 0 00:20:00
5 1 2016-01-01 00:10:00 1 00:20:00
6 1 2016-01-01 00:10:00 0 00:20:00
7 1 2016-01-01 00:10:00 0 00:20:00
regres = df.groupby(['region','date','time','time1'], as_index=False)['gap'].sum()
regres['time'] = regres.pop('time1') + '/' + regres['time'].map(d1)
print (regres)
region date time gap
0 1 2016-01-01 00:10:00/slot1 2
1 1 2016-01-01 00:20:00/slot2 1
EDIT:
Improvement for floor and convert to strings is use bining by cut or searchsorted:
df['time'] = pd.to_timedelta(df['time'])
bins = pd.timedelta_range('00:00:00', '24:00:00', freq='10Min')
labels = np.array(['{}'.format(str(x)[-8:]) for x in bins])
labels = labels[:-1]
df['time1'] = pd.cut(df['time'], bins=bins, labels=labels)
df['time11'] = labels[np.searchsorted(bins, df['time'].values) - 1]
Just to avoid the complication of the Datetime comparison (unless that is your whole point, in which case ignore my answer), and show the essence of this group by slot window problem, I here assume times are integers.
df = pd.DataFrame({'time':[8, 48, 250, 52, 1001, 1003, 1005, 1008, 2001, 2003, 2056],
'gap': [1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1]})
slots = np.array([0, 1000, 1500])
df['slot'] = df.apply(func = lambda x: slots[np.argmax(slots[x['time']>slots])], axis=1)
df.groupby('slot')[['gap']].sum()
Output
gap
slot
-----------
0 2
1000 1
1500 3
The way to think about approaching this problem is converting your time column to the values you want first, and then doing a groupby sum of the time column.
The below code shows the approach I've used. I used np.select to include in as many conditions and condition options as I want. After I have converted time to the values I wanted, I did a simple groupby sum
None of the fuss of formatting time or converting strings etc is really needed. Simply let pandas dataframe handle it intuitively.
#Just creating the DataFrame using a dictionary here
regdict = {
'time': ['00:00:08','00:00:48','00:02:50','00:00:52','00:10:01','00:10:03','00:10:05','00:10:08'],
'gap': [1,0,1,0,0,1,0,0],}
df = pd.DataFrame(regdict)
import pandas as pd
import numpy as np #This is the library you require for np.select function
#Add in all your conditions and options here
condlist = [df['time']<'00:10:00',df['time']<'00:20:00']
choicelist = ['00:10:00/slot1','00:20:00/slot2']
#Use np.select after you have defined all your conditions and options
answerlist = np.select(condlist, choicelist)
print (answerlist)
['00:10:00/slot1' '00:10:00/slot1' '00:10:00/slot1' '00:10:00/slot1'
'00:20:00/slot2' '00:20:00/slot2' '00:20:00/slot2' '00:20:00/slot2']
#Assign answerlist to df['time']
df['time'] = answerlist
print (df)
time gap
0 00:10:00 1
1 00:10:00 0
2 00:10:00 1
3 00:10:00 0
4 00:20:00 0
5 00:20:00 1
6 00:20:00 0
7 00:20:00 0
df = df.groupby('time', as_index=False)['gap'].sum()
print (df)
time gap
0 00:10:00 2
1 00:20:00 1
If you wish to keep the original time you can instead do df['timeNew'] = answerlist and then filter from there.
df['timeNew'] = answerlist
print (df)
time gap timeNew
0 00:00:08 1 00:10:00/slot1
1 00:00:48 0 00:10:00/slot1
2 00:02:50 1 00:10:00/slot1
3 00:00:52 0 00:10:00/slot1
4 00:10:01 0 00:20:00/slot2
5 00:10:03 1 00:20:00/slot2
6 00:10:05 0 00:20:00/slot2
7 00:10:08 0 00:20:00/slot2
#Use transform function here to retain all prior values
df['aggregate sum of gap'] = df.groupby('timeNew')['gap'].transform(sum)
print (df)
time gap timeNew aggregate sum of gap
0 00:00:08 1 00:10:00/slot1 2
1 00:00:48 0 00:10:00/slot1 2
2 00:02:50 1 00:10:00/slot1 2
3 00:00:52 0 00:10:00/slot1 2
4 00:10:01 0 00:20:00/slot2 1
5 00:10:03 1 00:20:00/slot2 1
6 00:10:05 0 00:20:00/slot2 1
7 00:10:08 0 00:20:00/slot2 1
I have a dataframe with values per day (see df below).
I want to group the "Forecast" field per week but with Monday as the first day of the week.
Currently I can do it via pd.TimeGrouper('W') (see df_final below) but it groups the week starting on Sundays (see df_final below)
import pandas as pd
data = [("W1","G1",1234,pd.to_datetime("2015-07-1"),8),
("W1","G1",1234,pd.to_datetime("2015-07-30"),2),
("W1","G1",1234,pd.to_datetime("2015-07-15"),2),
("W1","G1",1234,pd.to_datetime("2015-07-2"),4),
("W1","G2",2345,pd.to_datetime("2015-07-5"),5),
("W1","G2",2345,pd.to_datetime("2015-07-7"),1),
("W1","G2",2345,pd.to_datetime("2015-07-9"),1),
("W1","G2",2345,pd.to_datetime("2015-07-11"),3)]
labels = ["Site","Type","Product","Date","Forecast"]
df = pd.DataFrame(data,columns=labels).set_index(["Site","Type","Product","Date"])
df
Forecast
Site Type Product Date
W1 G1 1234 2015-07-01 8
2015-07-30 2
2015-07-15 2
2015-07-02 4
G2 2345 2015-07-05 5
2015-07-07 1
2015-07-09 1
2015-07-11 3
df_final = (df
.reset_index()
.set_index("Date")
.groupby(["Site","Product",pd.TimeGrouper('W')])["Forecast"].sum()
.astype(int)
.reset_index())
df_final["DayOfWeek"] = df_final["Date"].dt.dayofweek
df_final
Site Product Date Forecast DayOfWeek
0 W1 1234 2015-07-05 12 6
1 W1 1234 2015-07-19 2 6
2 W1 1234 2015-08-02 2 6
3 W1 2345 2015-07-05 5 6
4 W1 2345 2015-07-12 5 6
Use W-MON instead W, check anchored offsets:
df_final = (df
.reset_index()
.set_index("Date")
.groupby(["Site","Product",pd.Grouper(freq='W-MON')])["Forecast"].sum()
.astype(int)
.reset_index())
df_final["DayOfWeek"] = df_final["Date"].dt.dayofweek
print (df_final)
Site Product Date Forecast DayOfWeek
0 W1 1234 2015-07-06 12 0
1 W1 1234 2015-07-20 2 0
2 W1 1234 2015-08-03 2 0
3 W1 2345 2015-07-06 5 0
4 W1 2345 2015-07-13 5 0
I have three solutions to this problem as described below. First, I should state that the ex-accepted answer is incorrect. Here is why:
# let's create an example df of length 9, 2020-03-08 is a Sunday
s = pd.DataFrame({'dt':pd.date_range('2020-03-08', periods=9, freq='D'),
'counts':0})
> s
dt
counts
0
2020-03-08 00:00:00
0
1
2020-03-09 00:00:00
0
2
2020-03-10 00:00:00
0
3
2020-03-11 00:00:00
0
4
2020-03-12 00:00:00
0
5
2020-03-13 00:00:00
0
6
2020-03-14 00:00:00
0
7
2020-03-15 00:00:00
0
8
2020-03-16 00:00:00
0
These nine days span three Monday-to-Sunday weeks. The weeks of March 2nd, 9th, and 16th. Let's try the accepted answer:
# the accepted answer
> s.groupby(pd.Grouper(key='dt',freq='W-Mon')).count()
dt
counts
2020-03-09 00:00:00
2
2020-03-16 00:00:00
7
This is wrong because the OP wants to have "Monday as the first day of the week" (not as the last day of the week) in the resulting dataframe. Let's see what we get when we try with freq='W'
> s.groupby(pd.Grouper(key='dt', freq='W')).count()
dt
counts
2020-03-08 00:00:00
1
2020-03-15 00:00:00
7
2020-03-22 00:00:00
1
This grouper actually grouped as we wanted (Monday to Sunday) but labeled the 'dt' with the END of the week, rather than the start. So, to get what we want, we can move the index by 6 days like:
w = s.groupby(pd.Grouper(key='dt', freq='W')).count()
w.index -= pd.Timedelta(days=6)
or alternatively we can do:
s.groupby(pd.Grouper(key='dt',freq='W-Mon',label='left',closed='left')).count()
a third solution, arguably the most readable one, is converting dt to period first, then grouping, and finally (if needed) converting back to timestamp:
s.groupby(s.dt.dt.to_period('W'))['counts'].count().to_timestamp()
# a variant of this solution is: s.set_index('dt').to_period('W').groupby(pd.Grouper(freq='W')).count().to_timestamp()
all of these solutions return what the OP asked for:
dt
counts
2020-03-02 00:00:00
1
2020-03-09 00:00:00
7
2020-03-16 00:00:00
1
Explanation: when freq is provided to pd.Grouper, both closed and label kwargs default to right. Setting freq to W (short for W-Sun) works because we want our week to end on Sunday (Sunday included, and g.closed == 'right' handles this). Unfortunately, the pd.Grouper docstring does not show the default values but you can see them like this:
g = pd.Grouper(key='dt', freq='W')
print(g.closed, g.label)
> right right