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How to retrieve a certain day of the month for each row based on a dataframe value?

I am trying to replace some hardcoded SQL queries related to timezone changes with a more dynamic/data-driven Python script. I have a dataset that looks like this spreadsheet below. WEEK_START/DAY/MONTH is the week, day, and month when daylight savings time begins (for example Canberra starts the first Sunday of April while Vienna is the last Sunday of March). The end variables are in the same format and display when it ends.
Dataset
Here is the issue. I have seen solutions for specific use cases such as this, finding the last Sunday of the month:
current_year=today.year
current_month=today.month
current_day=today.day
month = calendar.monthcalendar(current_year, current_month)
day_of_month = max(month[-1][calendar.SUNDAY], month[-2][calendar.SUNDAY])
print(day_of_month)
31
This tells me that the last day of this month is the 31st. I can adjust the attributes for one given month/scenario, but how would I make a column for each and every row (city) to retrieve each? That is, several cities that change times on different dates? I thought if I could set attributes in day_of_month in an apply function it would work but when I do something like weekday='SUNDAY' it returns an error because of course the string 'SUNDAY' is not the same as SUNDAY the attribute of calendar. My SQL queries are grouped by cities that change on the same day but ideally anyone would be able to edit the CSV that loads the above dataset as needed and then each day the script would run once to see if today is between the start and end of daylight savings. We might have new cities to add in the future. I'm confident in doing that bit but quite lost on how to retrieve the dates for a given year.
My alternate, less resilient, option is to look at the distinct list of potential dates (last Sunday of March, first Sunday of April, etc.), write code to retrieve each one upfront (as in the above snippet above), and assign the dates in that way. I say that this is less resilient because if a city is added that does not fit in an existing group for time changes, the code would need to be altered as well.
So stackoverflow, is there a way to do this in a data driven way in pandas through an apply or something similar? Thanks in advance.

Basically I think you have most of what you need. Just map the WEEK_START / WEEK_END column {-1, 1} to last or first day of month, put it all in a function and apply it to each row. EX:
import calendar
import operator
import pandas as pd
def get_date(year: int, month: int, dayname: str, first=-1) -> pd.Timestamp:
"""
get the first or last day "dayname" in given month and year.
returns last by default.
"""
daysinmonth = calendar.monthcalendar(year, month)
getday = operator.attrgetter(dayname.upper())
if first == 1:
day = daysinmonth[0][getday(calendar)]
else:
day = max(daysinmonth[-1][getday(calendar)], daysinmonth[-2][getday(calendar)])
return pd.Timestamp(year, month, day)
year = 2021 # we need a year...
df['date_start'] = df.apply(lambda row: get_date(year,
row['MONTH_START'],
row['DAY_START'],
row['WEEK_START']), # selects first or last
axis=1) # to each row
df['date_end'] = df.apply(lambda row: get_date(year,
row['MONTH_END'],
row['DAY_END'],
row['WEEK_END']),
axis=1)
giving you for the sample data
df[['CITY', 'date_start', 'date_end']]
CITY date_start date_end
0 Canberra 2021-04-04 2021-10-03
1 Melbourne 2021-04-04 2021-10-03
2 Sydney 2021-04-04 2021-10-03
3 Kitzbuhel 2021-03-28 2021-10-31
4 Vienna 2021-03-28 2021-10-31
5 Antwerp 2021-03-28 2021-10-31
6 Brussels 2021-03-28 2021-10-31
7 Louvain-la-Neuve 2021-03-28 2021-10-31
Once you start working with time zones and DST transitions, Q: Is there a way to infer in Python if a date is the actual day in which the DST (Daylight Saving Time) change is made? might also be interesting to you.

Keeping the first day of the month in a timestamp object

I have a DataFrame with a timestamp column containing all the days of the year.
I would like to keep only the first day of the month, any idea of how should I do this?

see this article for an example on how you can ask a good question on Stack Overflow and provide a minimum reproducible example:
https://stackoverflow.com/help/how-to-ask
With that in mind, you can access the day attribute of a datetime object as follows:
from datetime import datetime
dt = datetime.today()
dt.day
Output:
2021-07-11 09:37:23.122548
11
You could then use masking to select rows in your dataframe that have a value of 1 for day as below:
df = df[df['date_column'].dt.day == 1]
You'll just need to replace 'date_column' with whatever your date column is called.
We've got a tutorial for a complete introduction to pandas on our website, feel free to take a look if you'd like to learn more!
https://www.learndatasci.com/tutorials/python-pandas-tutorial-complete-introduction-for-beginners/

Subtract datetime now from a dataframe column without Saturday and Sunday

Currently my script is subtracting my current time with the times that i have in a Dataframe column called "Creation", generating a new column with the days of the difference. I get the difference days with this code:
df['Creation']= pandas.to_datetime(df["Creation"],dayfirst="True")
#Generates new column with the days.
df['Difference'] = df.to_datetime('now') - df['Creation']
What i want to now is for it to give me the days like hes giving me but dont count the Saturdays and Sundays. How can i do that ?

you can make use of numpy's busday_count, Ex:
import pandas as pd
import numpy as np
# some dummy data
df = pd.DataFrame({'Creation': ['2021-03-29', '2021-03-30']})
# make sure we have datetime
df['Creation'] = pd.to_datetime(df['Creation'])
# set now to a fixed date
now = pd.Timestamp('2021-04-05')
# difference in business days, excluding weekends
# need to cast to datetime64[D] dtype so that np.busday_count works
df['busday_diff'] = np.busday_count(df['Creation'].values.astype('datetime64[D]'),
np.repeat(now, df['Creation'].size).astype('datetime64[D]'))
df['busday_diff'] # since I didn't define holidays, potential Easter holiday is excluded:
0 5
1 4
Name: busday_diff, dtype: int64
If you need the output to be of dtype timedelta, you can easily cast to that via
df['busday_diff'] = pd.to_timedelta(df['busday_diff'], unit='d')
df['busday_diff']
0 5 days
1 4 days
Name: busday_diff, dtype: timedelta64[ns]
Note: np.busday_count also allows you to set a custom weekmask (exclude days other than Saturday and Sunday) or a list of holidays. See the docs I linked on top.
Related: Calculate difference between two dates excluding weekends in python?, how to use (np.busday_count) with pandas.core.series.Series

How to aggregate irregularly sampled data for Time Series Analysis

I am trying to forecast daily profit using time series analysis, but daily profit is not only recorded unevenly, but some of the data is missing.
Raw Data:
Date
Revenue
2020/1/19
10$
2020/1/20
7$
2020/1/25
14$
2020/1/29
18$
2020/2/1
12$
2020/2/2
17$
2020/2/9
28$
The above table is an example of what kind of data I have. Profit is not recorded daily, so date between 2020/1/20 and 2020/1/24 does not exist. Not only that, say the profit recorded during the period between 2020/2/3 and 2020/3/8 went missing in the database. I would like to recover this missing data and use time series analysis to predict the profit after 2020/2/9 ~.
My approach was to first aggregate the profit every 6 days since I have to recover the profit between 2020/2/3 and 2020/3/8. So my cleaned data will look something like this
Date
Revenue
2020/1/16 ~ 2020/1/21
17$
2020/1/22 ~ 2020/1/27
14$
2020/1/28 ~ 2020/2/2
47$
2020/2/3 ~ 2020/2/8
? (to predict)
After applying this to a time series model, I would like to further predict the profit after 2020/2/9 ~.
This is my general idea, but as a beginner at Python, using pandas library, I have trouble executing my ideas. Could you please help me how to aggregate the profit every 6 days and have the data look like the above table?

Easiest way is using pandas resample function.
Provided you have an index of type Datetime resampling to aggregate profits at every 6 days would be as simple as your_dataframe.resample('6D').sum()
You can do all sorts of resampling (end of month, end of quarter, begining of week, every hour, minute, second, ...). Check the full documentation if you're interested: https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.resample.html?highlight=resample#pandas.DataFrame.resample

I suggest using a combination of .rolling, pd.date_range, and .reindex
Say your DataFrame is df, with proper datetime indexing:
df = pd.DataFrame([['2020/1/19',10],
['2020/1/20',7],
['2020/1/25',14],
['2020/1/29',18],
['2020/2/1',12],
['2020/2/2',17],
['2020/2/9',28]],columns=['Date','Revenue'])
df['Date'] = pd.to_datetime(df['Date'])
df.set_index('Date',inplace=True)
The first step is to 'fill in' the missing days with dummy, zero revenue. We can use pd.date_range to get an index with evenly spaced dates from 2020/1/16 to 2020/2/8, and then .reindex to bring this into the main df DataFrame:
evenly_spaced_idx = pd.date_range(start='2020/1/16',end='2020/2/8',freq='1d')
df = df.reindex(evenly_spaced_idx, fill_value=0)
Now we can take a rolling sum for each 6 day period. We're not interested in every day's six day revenue total, only every 6th days, though:
summary_df = df.rolling('6d').sum().iloc[5::6, :]
The last thing with summary_df is just to format it the way you'd like so that it clearly states the date range which each row refers to.
summary_df['Start Date'] = summary_df.index-pd.Timedelta('6d')
summary_df['End Date'] = summary_df.index
summary_df.reset_index(drop=True,inplace=True)

You can use resample for this.
Make sure to have the "Date" column as datetime type.
>>> df = pd.DataFrame([["2020/1/19" ,10],
... ["2020/1/20" ,7],
... ["2020/1/25" ,14],
... ["2020/1/29" ,18],
... ["2020/2/1" ,12],
... ["2020/2/2" ,17],
... ["2020/2/9" ,28]], columns=['Date', 'Revenue'])
>>> df['Date'] = pd.to_datetime(df.Date)
For pandas < 1.1.0
>>> df.set_index('Date').resample('6D', base=3).sum()
Revenue
Date
2020-01-16 17
2020-01-22 14
2020-01-28 47
2020-02-03 0
2020-02-09 28
For pandas >= 1.1.0
>>> df.set_index('Date').resample('6D', origin='2020-01-16').sum()
Revenue
Date
2020-01-16 17
2020-01-22 14
2020-01-28 47
2020-02-03 0
2020-02-09 28

How to use resample count to screen original data

Say I have a dataset at daily scale, but not all days have valid data. In other words, some days are missing in the data. I want to compute the summer season mean from the dataset, and want to remove the month which has less than 20 days of valid data.
How do I achieve this (in pythonic fashion)?
Say my dataframe (df) is like this:
DATE VAR
1900-01-01 123
1900-01-02 456
1900-01-10 789
...
I know how to compute the count:
df_count = df.resample('MS').count()
I also know how to compute the summer season mean:
df_summer = df.resample('Q-NOV').mean()

You can based on df_count to filter out the month which have less than 20 days of valid data. After that compute the summer season mean using your formula.
df_count = df.resample('MS').count()
relevant_month = df_count[df_count > 10].index
df_summer = df[df.index.isin(relevant_month)].resample('Q-NOV').mean()
I suppose you store the month in index. If the month or time is stored in a different column, change df.index.isin(relevant_month) to df.columnName.isin(relevant_month).
I also don't know the format of your time column (date or datetime) so you might need to modify the code to change this part df.index.isin(relevant_month) accordingly. It is just the general idea.