pandas resample uneven hourly data into 1D or 24h bins - pandas

Question

I have weekly hourly FX data which I need to resample into '1D' or '24hr' bins Monday through Thursday 12:00pm and at 21:00 on Friday, totaling 5 days per week:
Date rate
2020-01-02 00:00:00 0.673355
2020-01-02 01:00:00 0.67311
2020-01-02 02:00:00 0.672925
2020-01-02 03:00:00 0.67224
2020-01-02 04:00:00 0.67198
2020-01-02 05:00:00 0.67223
2020-01-02 06:00:00 0.671895
2020-01-02 07:00:00 0.672175
2020-01-02 08:00:00 0.672085
2020-01-02 09:00:00 0.67087
2020-01-02 10:00:00 0.6705800000000001
2020-01-02 11:00:00 0.66884
2020-01-02 12:00:00 0.66946
2020-01-02 13:00:00 0.6701600000000001
2020-01-02 14:00:00 0.67056
2020-01-02 15:00:00 0.67124
2020-01-02 16:00:00 0.6691699999999999
2020-01-02 17:00:00 0.66883
2020-01-02 18:00:00 0.66892
2020-01-02 19:00:00 0.669345
2020-01-02 20:00:00 0.66959
2020-01-02 21:00:00 0.670175
2020-01-02 22:00:00 0.6696300000000001
2020-01-02 23:00:00 0.6698350000000001
2020-01-03 00:00:00 0.66957
So the number of hours in each some days of the week is uneven, ie "Monday" = 00:00:00 Monday through 12:00:00 Monday, "Tuesday" (and also Weds, Thu) = i.e. 13:00:00 Monday though 12:00:00 Tuesday, and Friday = 13:00:00 through 21:00:00
In trying to find a solution I see that base is now deprecated, and offset/origin methods aren't working as expected, likely due to uneven number of rows per day:
df.rate.resample('24h', offset=12).ohlc()
I've spent hours attempting to find a solution
How can one simply bin into ohlc() columns all data rows between each 12:00:00 timestamp?
the desired output would look something like this:
Out[69]:
open high low close
2020-01-02 00:00:00.0000000 0.673355 0.673355 0.673355 0.673355
2020-01-03 00:00:00.0000000 0.673110 0.673110 0.668830 0.669570
2020-01-04 00:00:00.0000000 0.668280 0.668280 0.664950 0.666395
2020-01-05 00:00:00.0000000 0.666425 0.666425 0.666425 0.666425

Answer 1

Is this what you are looking for, using both origin and offset as parameters:
df.resample('24h', origin='start_day', offset='13h').ohlc()
For your example, this gives me:
open high low close
datetime
2020-01-01 13:00:00 0.673355 0.673355 0.66884 0.66946
2020-01-02 13:00:00 0.670160 0.671240 0.66883 0.66957

Answer 2

Since the period lengths are unequal, IMO it is necessary to craft the mapping wheel yourself. Speaking precisely, the 1.5-day length on Monday makes it impossible for freq='D' to do the mapping correctly at once.
The hand-crafted code is also able to guard against records outside the well-defined periods.
Data
A slightly different timestamp is used to demonstrate the correctness of the code. The days are from Mon. to Fri.
import pandas as pd
import numpy as np
from datetime import datetime
import io
from pandas import Timestamp, Timedelta
df = pd.read_csv(io.StringIO("""
rate
Date
2020-01-06 00:00:00 0.673355
2020-01-06 23:00:00 0.673110
2020-01-07 00:00:00 0.672925
2020-01-07 12:00:00 0.672240
2020-01-07 13:00:00 0.671980
2020-01-07 23:00:00 0.672230
2020-01-08 00:00:00 0.671895
2020-01-08 12:00:00 0.672175
2020-01-08 23:00:00 0.672085
2020-01-09 00:00:00 0.670870
2020-01-09 12:00:00 0.670580
2020-01-09 23:00:00 0.668840
2020-01-10 00:00:00 0.669460
2020-01-10 12:00:00 0.670160
2020-01-10 21:00:00 0.670560
2020-01-10 22:00:00 0.671240
2020-01-10 23:00:00 0.669170
"""), sep=r"\s{2,}", engine="python")
df.set_index(pd.to_datetime(df.index), inplace=True)
Code
def find_day(ts: Timestamp):
"""Find the trading day with irregular length"""
wd = ts.isoweekday()
if wd == 1:
return ts.date()
elif wd in (2, 3, 4):
return ts.date() - Timedelta("1D") if ts.hour <= 12 else ts.date()
elif wd == 5:
if ts.hour <= 12:
return ts.date() - Timedelta("1D")
elif 13 <= ts.hour <= 21:
return ts.date()
# out of range or nulls
return None
# map the timestamps, and set as new index
df.set_index(pd.DatetimeIndex(df.index.map(find_day)), inplace=True)
# drop invalid values and collect ohlc
ans = df["rate"][df.index.notnull()].resample("D").ohlc()
Result
print(ans)
open high low close
Date
2020-01-06 0.673355 0.673355 0.672240 0.672240
2020-01-07 0.671980 0.672230 0.671895 0.672175
2020-01-08 0.672085 0.672085 0.670580 0.670580
2020-01-09 0.668840 0.670160 0.668840 0.670160
2020-01-10 0.670560 0.670560 0.670560 0.670560

Answer 3

I ended up using a combination of grouby and datetime day of the week identification to arrive at my specific solution
# get idxs of time to rebal (12:00:00)-------------------------------------
df['idx'] = range(len(df)) # get row index
days = [] # identify each row by day of week
for i in range(len(df.index)):
days.append(df.index[i].date().weekday())
df['day'] = days
dtChgIdx = [] # stores "12:00:00" rows
justDates = df.index.date.tolist() # gets just dates
res = [] # removes duplicate dates
[res.append(x) for x in justDates if x not in res]
justDates = res
grouped_dates = df.groupby(df.index.date) # group entire df by dates
for i in range(len(grouped_dates)):
tempDf = grouped_dates.get_group(justDates[i]) # look at each grouped dates
if tempDf['day'][0] == 6:
continue # skip Sundays
times = [] # gets just the time portion of index
for y in range(len(tempDf.index)):
times.append(str(tempDf.index[y])[-8:])
tempDf['time'] = times # add time column to df
tempDf['dayCls'] = np.where(tempDf['time'] == '12:00:00',1,0) # idx "12:00:00" row
dtChgIdx.append(tempDf.loc[tempDf['dayCls'] == 1, 'idx'][0]) # idx value