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How to convert date format (dd/mm/yyyy) to days in python csv

I need a function to count the total number of days in the 'days' column between a start date of 1st Jan 1995 and an end date of 31st Dec 2019 in a dataframe taking leap years into account as well.
Example: 1st Jan 1995 - Day 1, 1st Feb 1995 - Day 32 .......and so on all the way to 31st.

If you want to filter a pandas dataframe using a range of 2 date you can do this by:
start_date = '1995/01/01'
end_date = '1995/02/01'
df = df[ (df['days']>=start_date) & (df['days']<=end_date) ]
and with len(df) you will see the number of rows of the filter dataframe.
Instead, if you want to calculate a range of days between 2 different date you can do without pandas with datetime:
from datetime import datetime
start_date = '1995/01/01'
end_date = '1995/02/01'
delta = datetime.strptime(end_date, '%Y/%m/%d') - datetime.strptime(start_date, '%Y/%m/%d')
print(delta.days)
Output:
31
The only thing is that this not taking into account leap years

Python - Convert 5 digit date to datetime from SAS date

I have a 5 digit date variable that was exported from SAS. I am having trouble converting it into a datetime format in Python. The variable is currently stored as an object.
Here is a background on SAS dates:
"The SAS System represents dates as the number of days since a reference date. The reference date, or date zero, used for SAS date values is 1 January 1960. Thus, for example, 3 February 1960 is represented by the SAS System as 33. The SAS date for 17 October 1991 is 11612."
Here are two examples:

Let's try pd.to_datetime with specified origin and unit
df['out'] = pd.to_datetime(df['Date'], unit='D', origin='1960-01-01')
print(df)
Date out
0 21032 2017-08-01
1 16387 2004-11-12
2 0 1960-01-01
3 33 1960-02-03
4 11612 1991-10-17

#Ynjxsjmh 's answer is better than mine, however you could retrieve the date you're looking for by adding the deltatime between your 'origin' date and the input(integer of amount of days to add upon origin) :
import datetime
date = datetime.datetime(1960, 1, 1) # Constant for SAS date
def SASdate(days:int):
"""
Convert SAS integer to date
"""
return date + datetime.timedelta(days = days)
print(SASdate(0))

How to convert a date into the nth day of the week in Alteryx or Python?

If I have a column of dates filed like the below;
Date
2021-08-01
2021-08-02
2021-08-03
2021-08-01
2021-08-02
What I wish to do is add a new column that will tell me the number of mondays for example that the date is in the year.
so I can see that for the first record the first of August was a Sunday and it was the 31st Sunday of the year, whereas the 12th was a Thursday and was the 32nd Thursday of the year.
Date Number Of WeekDay in Year
2021-08-01 31
2021-08-02 31
2021-08-03 31
2021-08-12 32
... ...
If it makes it easier is there a way to do it using the python tool within Alteryx?

The answer by johnjps111 explains it all, but here's an implementation using python only (no alteryx):
import math
from datetime import datetime
def get_weekday_occurrences(date_string):
year_day = datetime.strptime(date_string, '%Y-%m-%d').timetuple().tm_yday
return math.ceil(year_day / 7)
Which can be used as follows:
get_weekday_occurrences('2021-08-12')
Note we use datetime.timetuple to get the day of the year (tm_yday).

For Alteryx, try the formula Ceil(DateTimeFormat([date],'%j') / 7) ... explanation: regardless of day of week, if it's the first day of the year, it's also the first "of that weekday" of the year... at day number 8, it becomes the 2nd "of that weekday" of the year, and so on. Since Alteryx gives you "day of the year" for free using the given DateTimeFornat function, it's then a simple division and Ceil() function.

To get the week in the year from a date string:
from datetime import datetime
a = '2021-08-02'
b = datetime.fromisoformat(a)
print('week of the year:', b.strftime('%W'))
output:
week of the year: 31
For more information about datetime: link

how can I align different-day timeseries in pandas?

I have two time series, df1
day cnt
2020-03-01 135006282
2020-03-02 145184482
2020-03-03 146361872
2020-03-04 147702306
2020-03-05 148242336
and df2:
day cnt
2017-03-01 149104078
2017-03-02 149781629
2017-03-03 151963252
2017-03-04 147384922
2017-03-05 143466746
The problem is that the sensors I'm measuring are sensitive to the day of the week, so on Sunday, for instance, they will produce less cnt. Now I need to compare the time series over 2 different years, 2017 and 2020, but to do that I have to align (March, in this case) to the matching day of the week, and plot them accordingly. How do I "shift" the data to make the series comparable?

The ISO calendar is a representation of date in a tuple (year, weeknumber, weekday). In pandas they are the dt members year, weekofyear and weekday. So assuming that the day column actually contains Timestamps (convert if first with to_datetime if it does not), you could do:
df1['Y'] = df1.day.dt.year
df1['W'] = df1.day.dt.weekofyear
df1['D'] = df1.day.dt.weekday
Then you could align the dataframes on the W and D columns

March 2017 started on wednesday
March 2020 started on Sunday
So, delete the last 3 days of march 2017
So, delete the first sunday, monday and tuesday from 2020
this way you have comparable days
df1['ctn2020'] = df1['cnt']
df2['cnt2017'] = df2['cnt']
df1 = df1.iloc[2:, 2]
df2 = df2.iloc[:-3, 2]
Since you don't want to plot the date, but want the months to align, make a new dataframe with both columns and a index column. This way you will have 3 columns: index(0-27), 2017 and 2020. The index will represent.
new_df = pd.concat([df1,df2], axis=1)
If you also want to plot the days of the week on the x axis, check out this link, to know how to get the day of the week from a date, and them change the x ticks label.
Sorry for the "written step-to-stop", if it all sounds confusing, i can type the whole code later for you.

How can I find the fiscal Week of date in Pandas

I have a column with many dates: sample of the said list below
Dates
1 2019-02-01
2 2018-03-10
3 2019-08-01
4 2020-02-07
I would like to have it so that if input a date, of any year I can get the week number.
However, the fiscal year starts on Aug 1 of any given year.
I tried just shifting the date to Jan 1 but it's different for every year due to leap years.
data['Dates'] = pd.to_datetime(data['Dates'])
data['Week'] = (data['Dates'] - timedelta(days=215)).week
print(data)
how can I get a result similar to this one below
Dates Week
1 2019-02-01 27
2 2018-03-10 32
3 2019-08-01 1
4 2020-02-07 28
-Note: the weeks are probably incorrect.

The other answer ignores the fiscal year part of the OP. I am leaving the fiscal year start date calc to the reader but this will calculate the week number (where Monday is the start of the week) from an arbitrary start date.
from dateutil import relativedelta
from datetime import date, datetime, timedelta
NEXT_MONDAY = relativedelta.relativedelta(weekday=relativedelta.MO)
LAST_MONDAY = relativedelta.relativedelta(weekday=relativedelta.MO(-1))
ONE_WEEK = timedelta(weeks=1)
def week_in_fiscal_year(d: date, fiscal_year_start: date) -> int:
fy_week_2_monday = fiscal_year_start + NEXT_MONDAY
if d < fy_week_2_monday:
return 1
else:
cur_week_monday = d + LAST_MONDAY
return int((cur_week_monday - fy_week_2_monday) / ONE_WEEK) + 2
adapted from this post

Convert it to a datetime, then call datetime.date(2010, 6, 16).strftime("%V")4
You can also use isocalendar which will return a tuple, as opposed to a string above datetime.date(2010, 6, 16).isocalendar()[1]
How to get week number in Python?