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How to calculate number of days between 2 months in Python

I have a requirement where I have to find number of days between 2 months where 1st month value is constant and 2nd month value is present in a data frame.
I have to subtract 24th Feb with values present in the Data Frame.
past_2_month = date.today()
def to_integer(dt_time):
return 1*dt_time.month
past_2_month = to_integer(past_2_month)
past_2_month_num = past_2_month-2
day = 24
date_2 = dt.date(year, past_2_month_num, day)
date_2
Output of above code: datetime.date(2022, 2, 24)
Other values present in the Data frame is below:
dict_1 = {'Col1' : ['2017-05-01', np.NaN, '2017-11-01', np.NaN, '2016-10-01']}
a = pd.DataFrame(dict_1)
How to subtract this 2 values so that I can get difference in days between these 2 values?

If need number of days between datetime column and 2 months shifted values use offsets.DateOffset and convert timedeltas to days by Series.dt.days:
a['Col1'] = pd.to_datetime(a['Col1'])
a['new'] = (a['Col1'] - (a['Col1'] - pd.DateOffset(months=2))).dt.days
print (a)
Col1 new
0 2017-05-01 61.0
1 NaT NaN
2 2017-11-01 61.0
3 NaT NaN
4 2016-10-01 61.0
If need difference by another datetime solution is simplier - subtract and convert values to days:
a['Col1'] = pd.to_datetime(a['Col1'])
a['new'] = (pd.to_datetime('2022-02-24') - a['Col1']).dt.days
print (a)
Col1 new
0 2017-05-01 1760.0
1 NaT NaN
2 2017-11-01 1576.0
3 NaT NaN
4 2016-10-01 1972.0

Calculating values from time series in pandas multi-indexed pivot tables

I've got a dataframe in pandas that stores the Id of a person, the quality of interaction, and the date of the interaction. A person can have multiple interactions across multiple dates, so to help visualise and plot this I converted it into a pivot table grouping first by Id then by date to analyse the pattern over time.
e.g.
import pandas as pd
df = pd.DataFrame({'Id':['A4G8','A4G8','A4G8','P9N3','P9N3','P9N3','P9N3','C7R5','L4U7'],
'Date':['2016-1-1','2016-1-15','2016-1-30','2017-2-12','2017-2-28','2017-3-10','2019-1-1','2018-6-1','2019-8-6'],
'Quality':[2,3,6,1,5,10,10,2,2]})
pt = df.pivot_table(values='Quality', index=['Id','Date'])
print(pt)
Leads to this:
Id
Date
Quality
A4G8
2016-1-1
2
2016-1-15
4
2016-1-30
6
P9N3
2017-2-12
1
2017-2-28
5
2017-3-10
10
2019-1-1
10
C7R5
2018-6-1
2
L4U7
2019-8-6
2
However, I'd also like to...
Measure the time from the first interaction for each interaction per Id
Measure the time from the previous interaction with the same Id
So I'd get a table similar to the one below
Id
Date
Quality
Time From First
Time To Prev
A4G8
2016-1-1
2
0 days
NA days
2016-1-15
4
14 days
14 days
2016-1-30
6
29 days
14 days
P9N3
2017-2-12
1
0 days
NA days
2017-2-28
5
15 days
15 days
2017-3-10
10
24 days
9 days
The Id column is a string type, and I've converted the date column into datetime, and the Quality column into an integer.
The column is rather large (>10,000 unique ids) so for performance reasons I'm trying to avoid using for loops. I'm guessing the solution is somehow using pd.eval but I'm stuck as to how to apply it correctly.
Apologies I'm a python, pandas, & stack overflow) noob and I haven't found the answer anywhere yet so even some pointers on where to look would be great :-).
Many thanks in advance

Convert Dates to datetimes and then substract minimal datetimes per groups by GroupBy.transformb subtracted by column Date and for second new column use DataFrameGroupBy.diff:
df['Date'] = pd.to_datetime(df['Date'])
df['Time From First'] = df['Date'].sub(df.groupby('Id')['Date'].transform('min'))
df['Time To Prev'] = df.groupby('Id')['Date'].diff()
print (df)
Id Date Quality Time From First Time To Prev
0 A4G8 2016-01-01 2 0 days NaT
1 A4G8 2016-01-15 3 14 days 14 days
2 A4G8 2016-01-30 6 29 days 15 days
3 P9N3 2017-02-12 1 0 days NaT
4 P9N3 2017-02-28 5 16 days 16 days
5 P9N3 2017-03-10 10 26 days 10 days
6 P9N3 2019-01-01 10 688 days 662 days
7 C7R5 2018-06-01 2 0 days NaT
8 L4U7 2019-08-06 2 0 days NaT

df["Date"] = pd.to_datetime(df.Date)
df = df.merge(
df.groupby(["Id"]).Date.first(),
on="Id",
how="left",
suffixes=["", "_first"]
)
df["Time From First"] = df.Date-df.Date_first
df['Time To Prev'] = df.groupby('Id').Date.diff()
df.set_index(["Id", "Date"], inplace=True)
df
output:

Is there a Pandas function to highlight a week's 10 lowest values in a time series?

Rookie here so please excuse my question format:
I got an event time series dataset for two months (columns for "date/time" and "# of events", each row representing an hour).
I would like to highlight the 10 hours with the lowest numbers of events for each week. Is there a specific Pandas function for that? Thanks!

Let's say you have a dataframe df with column col as well as a datetime column.
You can simply sort the column with
import pandas as pd
df = pd.DataFrame({'col' : [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],
'datetime' : ['2019-01-01 00:00:00','2015-02-01 00:00:00','2015-03-01 00:00:00','2015-04-01 00:00:00',
'2018-05-01 00:00:00','2016-06-01 00:00:00','2017-07-01 00:00:00','2013-08-01 00:00:00',
'2015-09-01 00:00:00','2015-10-01 00:00:00','2015-11-01 00:00:00','2015-12-01 00:00:00',
'2014-01-01 00:00:00','2020-01-01 00:00:00','2014-01-01 00:00:00']})
df = df.sort_values('col')
df = df.iloc[0:10,:]
df
Output:
col datetime
0 1 2019-01-01 00:00:00
1 2 2015-02-01 00:00:00
2 3 2015-03-01 00:00:00
3 4 2015-04-01 00:00:00
4 5 2018-05-01 00:00:00
5 6 2016-06-01 00:00:00
6 7 2017-07-01 00:00:00
7 8 2013-08-01 00:00:00
8 9 2015-09-01 00:00:00
9 10 2015-10-01 00:00:00

I know there's a function called nlargest. I guess there should be an nsmallest counterpart. pandas.DataFrame.nsmallest
df.nsmallest(n=10, columns=['col'])
My bad, so your DateTimeIndex is a Hourly sampling. And you need the hour(s) with least events weekly.
...
Date n_events
2020-06-06 08:00:00 3
2020-06-06 09:00:00 3
2020-06-06 10:00:00 2
...
Well I'd start by converting each hour into columns.
1. Create an Hour column that holds the hour of the day.
df['hour'] = df['date'].hour
Pivot the hour values into columns having values as n_events.
So you'll then have 1 datetime index, 24 hour columns, with values denoting #events. pandas.DataFrame.pivot_table
...
Date hour0 ... hour8 hour9 hour10 ... hour24
2020-06-06 0 3 3 2 0
...
Then you can resample it to weekly level aggregate using sum.
df.resample('w').sum()
The last part is a bit tricky to do on the dataframe. But fairly simple if you just need the output.
for row in df.itertuples():
print(sorted(row[1:]))

Python: How to find the nth weekday of the year?

I have seen a lot of similar posts on "nth weekday of the month", but my question pertains to "nth weekday of the year".
Background:
I have a table that has daily sales data. There are 3 columns: date, day of week (Mon, Tue, Wed etc.) and sales. I would like to match nth weekday of Year 1 with Year 2 and compare sales that way.
Example1: 01/06/2020 matches with 01/04/2021, both are the 1st Monday of that year.
Example2: 11/02/2019 matches with 10/31/2020, both are the 44th Saturday of that year.
As you can see, I can't simply do a "nth weekday of the MONTH" because sometimes the matched nth weekday would fall in different months (as seen in 11/02/2019 & 10/31/2020).
I am manipulating the table in pandas. I am wondering if there's a quick way for me to create a column that helps me to calculate the "nth weekday of the year" for me, so that I could later match based on that value?
Thanks for your help.

The pandas package has some good time/date functions.
For example
import pandas as pd
s = pd.date_range('2020-01-01', '2020-12-31', freq='D').to_series()
print(s.dt.dayofweek)
gives you the weekdays as integers.
2020-01-01 2
2020-01-02 3
2020-01-03 4
2020-01-04 5
2020-01-05 6
2020-01-06 0
2020-01-07 1
2020-01-08 2
2020-01-09 3
2020-01-10 4
(Monday=0)
Then you can do
mondays = s.dt.dayofweek.eq(0)
If you want to find the first Monday of the year use.
print(mondays.idxmax())
Timestamp('2020-01-06 00:00:00', freq='D')
Or the 5th Monday:
n = 4
print(s[mondays].iloc[n])
Timestamp('2020-02-03 00:00:00')
If your sales dataframe is df then to compare sales on the first 5 Mondays of two different years you could do something like this:
mondays = df['Date'].dt.dayofweek.eq(0)
mondays_in_y1 = (df['Year'] == 2019) & mondays
mondays_in_y2 = (df['Year'] == 2020) & mondays
pd.DataFrame({
2019: df.loc[mondays_in_y1, 'Sales'].values[:5],
2020: df.loc[mondays_in_y2, 'Sales'].values[:5]
})

IIUC you can play from
import pandas as pd
import numpy as np
df = pd.DataFrame({"date":pd.date_range(start="2020-01-01",
end="2020-12-31")})
# weekday number Monday is 0
df["dow"] = df["date"].dt.weekday
# is weekday as int
df["is_weekday"] = (df["dow"]<5).astype(int)
df["n"] = df["is_weekday"].cumsum()
# remove weekends
df["n"] = np.where(df["n"]==df["n"].shift(), np.nan, df["n"])
df[df["n"]==100]["date"]
Edit
In two lines only
df["n"] = (df["date"].dt.weekday<5).astype(int).cumsum()
df["n"] = np.where(df["n"]==df["n"].shift(), np.nan, df["n"])

You can try using dt.week. It returns a series, but you can simply define a new column with these values.
For example:
import pandas as pd
rng = pd.date_range('2015-02-24', periods=5, freq='D')
df = pd.DataFrame({ 'Date': rng, 'Val' : np.random.randn(len(rng))})
Output:
Date Val
0 2015-02-24 -0.977278
1 2015-02-25 0.950088
2 2015-02-26 -0.151357
3 2015-02-27 -0.103219
4 2015-02-28 0.410599
The you should input df['Week_Number'] = df['Date'].dt.week, so you will make a new column with the week number:
Date Val Week_Number
0 2015-02-24 -0.977278 9
1 2015-02-25 0.950088 9
2 2015-02-26 -0.151357 9
3 2015-02-27 -0.103219 9
4 2015-02-28 0.410599 9
Hope it helps. It's my first contribution.

Count String Values in Column across 30 Minute Time Bins using Pandas

I am looking to determine the count of string variables in a column across a 3 month data sample. Samples were taken at random times throughout each day. I can group the data by hour, but I require the fidelity of 30 minute intervals (ex. 0500-0600, 0600-0630) on roughly 10k rows of data.
An example of the data:
datetime stringvalues
2018-06-06 17:00 A
2018-06-07 17:30 B
2018-06-07 17:33 A
2018-06-08 19:00 B
2018-06-09 05:27 A
I have tried setting the datetime column as the index, but I cannot figure how to group the data on anything other than 'hour' and I don't have fidelity on the string value count:
df['datetime'] = pd.to_datetime(df['datetime']
df.index = df['datetime']
df.groupby(df.index.hour).count()
Which returns an output similar to:
datetime stringvalues
datetime
5 0 0
6 2 2
7 5 5
8 1 1
...
I researched multi-indexing and resampling to some length the past two days but I have been unable to find a similar question. The desired result would look something like this:
datetime A B
0500 1 2
0530 3 5
0600 4 6
0630 2 0
....

There is no straightforward way to do a TimeGrouper on the time component, so we do this in two steps:
v = (df.groupby([pd.Grouper(key='datetime', freq='30min'), 'stringvalues'])
.size()
.unstack(fill_value=0))
v.groupby(v.index.time).sum()
stringvalues A B
05:00:00 1 0
17:00:00 1 0
17:30:00 1 1
19:00:00 0 1