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getting previous week highs and lows in pandas dataframe using 30 min data

I have a few set of days where the index is based on 30min data from monday to friday. There might some missing dates (Might be because of holidays). But i would like to find the highest from column high and lowest from column low for ever past week. Like i am calculating today so previous week high and low is marked in the yellow of attached image.
Tried using rolling , resampling but some how not working. Can any one help
enter image description here

You really should add sample data to your question (by that I mean a piece of code/text that can easily be used to create a dataframe for illustrating how the proposed solution works).
Here's a suggestion. With df your dataframe, and column datatime with datetimes (and not strings):
df["week"] = (
df["datetime"].dt.isocalendar().year.astype(str)
+ df["datetime"].dt.isocalendar().week.astype(str)
)
mask = df["high"] == df.groupby("week")["high"].transform("max")
df = df.merge(
df[mask].rename(columns={"low": "high_low"})
.groupby("week").agg({"high_low": "min"}).shift(),
on="week", how="left"
).drop(columns="week")
Add a week column to df (year + week) for grouping along weeks.
Extract the rows with the weekly maximum highs by mask (there could be more than one for a week).
Build a corresponding dataframe with the weekly minimum of the lows corresponding to the weekly maximum highs (column named high_low), shift it once to get the value from the previous week, and .merge it to df.
If column datetime doesn't contain datetimes:
df["datetime"] = pd.to_datetime(df["datetime"])

If I have understood correctly, the solution should be
get the week number from the date
groupby the week number and fetch the max and min number.
groupby the week fetch max date to get max/last date for a week
now merge all the dataframes into one based on date key
Once the steps are done, you could do any formatting as required.

Replacing dates that fall on weekends to next business day in dataframe

I have a dataframe with bunch of dates in them, I would like to check each entry if it is a weekday or weekend, if it is a weekend i would like to increase the date to the next weekday. What is the most pythonic way of doing this ? I was thinking about using a list comprehension and s.th like
days = pd.date_range(start='1/1/2020', end='1/08/2020')
dates = pd.DataFrame(days,columns=['dates'])
dates['dates'] = [day+pd.DateOffset(days=1) if day.weekday() >4 else day for day in dates['dates']]
How could I adjust the code to cover day+pd.DateOffset(days=1) (for Sunday) and day+pd.DateOffset(days=2) (for Saturday) and get an updated column with the shifted dates? Running the code twice with +1 should work but is certainly not pretty

Add hours column to regular list of minutes, group by it, and average the data in Python

I have looked for similar questions, but none seems to be addressing the following challenge. I have a pandas dataframe with a list of minutes and corresponding values, like the following:
minute value
0 454
1 434
2 254
The list is a year-long list, thus counting 60 minutes * 24 hours * 365 days = 525600 observations.
I would like to add a new column called hour, which indeed expresses the hour of the day (assuming minutes 0-59 are 12AM, 60-119 are 1AM, and so forth until the following day, where the sequence restarts).
Then, once the hour column is added, I would like to group observations by it and calculate the average value for every hour of the year, and end up with a dataframe with 24 observations, each expressing the average value of the original data at each hour n.

Using integer and remainder division you can get the hour.
df['hour'] = df['minute']//60%24
If you want other date information it can be useful to use January 1st of some year (not a leap year) as the origin and convert to a datetime. Then you can grab a lot of the date attributes, in this case hour.
df['hour'] = pd.to_datetime(df['minute'], unit='m', origin='2017-01-01').dt.hour
Then for your averages you get the resulting 24 row Series with:
df.groupby('hour')['value'].mean()

Here's a way to do:
# sample df
df = pd.DataFrame({'minute': np.arange(525600), 'value': np.arange(525600)})
# set time format
df['minute'] = pd.to_timedelta(df['minute'], unit='m')
# calculate mean
df_new = df.groupby(pd.Grouper(key='minute', freq='1H'))['value'].mean().reset_index()
Although, you don't need hour column explicity to calculate these value, but if you want to get it, you can do it by:
df_new['hour'] = pd.to_datetime(df_new['minute']).dt.hour

Find if there is any holidays between two dates in a large dataset?

I am working on a dataset that has some 26 million rows and 13 columns including two datetime columns arr_date and dep_date. I am trying to create a new boolean column to check if there is any US holidays between these dates.
I am using apply function to the entire dataframe but the execution time is too slow. The code has been running for more than 48 hours now on Goolge Cloud Platform (24GB ram, 4 core). Is there a faster way to do this?
The dataset looks like this:
Sample data
The code I am using is -
import pandas as pd
import numpy as np
from pandas.tseries.holiday import USFederalHolidayCalendar as calendar
df = pd.read_pickle('dataGT70.pkl')
cal = calendar()
def mark_holiday(df):
df.apply(lambda x: True if (len(cal.holidays(start=x['dep_date'], end=x['arr_date']))>0 and x['num_days']<20) else False, axis=1)
return df
df = mark_holiday(df)

This took me about two minutes to run on a sample dataframe of 30m rows with two columns, start_date and end_date.
The idea is to get a sorted list of all holidays occurring on or after the minimum start date, and then to use bisect_left from the bisect module to determine the next holiday occurring on or after each start date. This holiday is then compared to the end date. If it is less than or equal to the end date, then there must be at least one holiday in the date range between the start and end dates (both inclusive).
from bisect import bisect_left
import pandas as pd
from pandas.tseries.holiday import USFederalHolidayCalendar as calendar
# Create sample dataframe of 10k rows with an interval of 1-19 days.
np.random.seed(0)
n = 10000 # Sample size, e.g. 10k rows.
years = np.random.randint(2010, 2019, n)
months = np.random.randint(1, 13, n)
days = np.random.randint(1, 29, n)
df = pd.DataFrame({'start_date': [pd.Timestamp(*x) for x in zip(years, months, days)],
'interval': np.random.randint(1, 20, n)})
df['end_date'] = df['start_date'] + pd.TimedeltaIndex(df['interval'], unit='d')
df = df.drop('interval', axis=1)
# Get a sorted list of holidays since the fist start date.
hols = calendar().holidays(df['start_date'].min())
# Determine if there is a holiday between the start and end dates (both inclusive).
df['holiday_in_range'] = df['end_date'].ge(
df['start_date'].apply(lambda x: bisect_left(hols, x)).map(lambda x: hols[x]))
>>> df.head(6)
start_date end_date holiday_in_range
0 2015-07-14 2015-07-31 False
1 2010-12-18 2010-12-30 True # 2010-12-24
2 2013-04-06 2013-04-16 False
3 2013-09-12 2013-09-24 False
4 2017-10-28 2017-10-31 False
5 2013-12-14 2013-12-29 True # 2013-12-25
So, for a given start_date timestamp (e.g. 2013-12-14), bisect_right(hols, '2013-12-14') would yield 39, and hols[39] results in 2013-12-25, the next holiday falling on or after the 2013-12-14 start date. The next holiday calculated as df['start_date'].apply(lambda x: bisect_left(hols, x)).map(lambda x: hols[x]). This holiday is then compared to the end_date, and holiday_in_range is thus True if the end_date is greater than or equal to this holiday value, otherwise the holiday must fall after this end_date.

Have you already considered using pandas.merge_asof for this?
I could imagine that map and apply with lambda functions cannot be executed that efficiently.
UPDATE: Ah sorry, I just read, that you only need a boolean if there are any holidays inbetween, this makes it much easier. If thats enough you just need to perform steps 1-5 then group the DataFrame that is the result of step5 by start/end date and use count as the aggregate function to have the number of holidays in the ranges. This result you can join to your original dataset similar to step 8 described below. Then fill the rest of the values with fillna(0). Do something like joined_df['includes_holiday']= joined_df['joined_count_column']>0. After that, you can delete the joined_count_column again from your DataFrame, if you like.
If you use pandas_merge_asof you could work through these steps (step 6 and 7 are only necessary if you need to have all the holidays inbetween start and end in your result DataFrame as well, not just the booleans):
Load your holiday records in a DataFrame and index it on the date. The holidays should be one date per line (storing ranges like for christmas from 24th-26th in one row, would make it much more complex).
Create a copy of your dataframe with just the start, end date columns. UPDATE: every start, end date should only occur once in it. E.g. by using groupby.
Use merge_asof with a reasonable tolerance value (if you join over the start of the period, use direction='forward', if you use the end date, use direction='backward' and how='inner'.
As a result you have a merged DataFrame with your start, end columns and the date column from your holiday dataframe. You get only records, for which a holiday was found with the given tolerance, but later you can merge this data back with your original DataFrame. You will probably now have duplicates of your original records.
Then check the joined holiday for your records with indexers by comparing them with the start and end column and remove the holidays, which are not inbetween.
Sort the dataframe you obtained form step 5 (use something like df.sort_values(['start', 'end', 'holiday'], inplace=True). Now you should insert a number column that numbers the holidays between your periods (the ones you obtained after step 5) form 1 to ... (for each period starting from 1). This is necesary to use unstack in the next step to get the holidays in columns.
Add an index on your dataframe based on period start date, period end date and the count column you inserted in step 6. Use df.unstack(level=-1) on the DataFrame you prepared in steps 1-7. What you now have, is a condensed DataFrame with your original periods with the holidays arranged columnwise.
Now you only have to merge this DataFrame back to your original data using original_df.merge(df_from_step7, left_on=['start', 'end'], right_index=True, how='left')
The result of this is a file with your original data containing the date ranges and for each date range the holidays that lie inbetween the period are stored in a separte columns each behind the data. Loosely speaking the numbering in step 6 assigns the holidays to the columns and has the effect, that the holidays are always assigned from right to left to the columns (you wouldn't have a holiday in column 3 if column 1 is empty).
Step 6. is probably also a bit tricky, but you can do that for example by adding a series filled with a range and then fixing it, so the numbering starts by 0 or 1 in each group by using shift or grouping by start, end with aggregate({'idcol':'min') and joining the result back to subtract it from the value assigned by the range-sequence.
In all, I think it sounds more complicated, than it is and it should be performed quite efficient. Especially if your periods are not that large, because then after step 5, your result set should be much smaller than your original dataframe, but even if that is not the case, it should still be quite efficient, since it can use compiled code.

Python: Date conversion to year-weeknumber, issue at switch of year

I am trying to convert a dataframe column with a date and timestamp to a year-weeknumber format, i.e., 01-05-2017 03:44 = 2017-1. This is pretty easy, however, I am stuck at dates that are in a new year, yet their weeknumber is still the last week of the previous year. The same thing that happens here.
I did the following:
df['WEEK_NUMBER'] = df.date.dt.year.astype(str).str.cat(df.date.dt.week.astype(str), sep='-')
Where df['date'] is a very large column with date and times, ranging over multiple years.
A date which gives a problem is for example:
Timestamp('2017-01-01 02:11:27')
The output for my code will be 2017-52, while it should be 2016-52. Since the data covers multiple years, and weeknumbers and their corresponding dates change every year, I cannot simply subtract a few days.
Does anybody have an idea of how to fix this? Thanks!

Replace df.date.dt.year by this:
(df.date.dt.year- ((df.date.dt.week>50) & (df.date.dt.month==1)))
Basically, it means that you will substract 1 to the year value if the week number is greater than 50 and the month is January.