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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

Creating year week based on date with different start date

I have a df
date
2021-03-12
2021-03-17
...
2022-05-21
2022-08-17
I am trying to add a column year_week, but my year week starts at 2021-06-28, which is the first day of July.
I tried:
df['date'] = pd.to_datetime(df['date'])
df['year_week'] = (df['date'] - timedelta(days=datetime(2021, 6, 24).timetuple()
.tm_yday)).dt.isocalendar().week
I played around with the timedelta days values so that the 2021-06-28 has a value of 1.
But then I got problems with previous & dates exceeding my start date + 1 year:
2021-03-12 has a value of 38
2022-08-17 has a value of 8
So it looks like the valid period is from 2021-06-28 + 1 year.
date year_week
2021-03-12 38 # LY38
2021-03-17 39 # LY39
2021-06-28 1 # correct
...
2022-05-21 47 # correct
2022-08-17 8 # NY8
Is there a way to get around this? As I am aggregating the data by year week I get incorrect results due to the past & upcoming dates. I would want to have negative dates for the days before 2021-06-28 or LY38 denoting that its the year week of the last year, accordingly year weeks of 52+ or NY8 denoting that this is the 8th week of the next year?

Here is a way, I added two dates more than a year away. You need the isocalendar from the difference between the date column and the dayofyear of your specific date. Then you can select the different scenario depending on the year of your specific date. use np.select for the different result format.
#dummy dataframe
df = pd.DataFrame(
{'date': ['2020-03-12', '2021-03-12', '2021-03-17', '2021-06-28',
'2022-05-21', '2022-08-17', '2023-08-17']
}
)
# define start date
d = pd.to_datetime('2021-6-24')
# remove the nomber of day of year from each date
s = (pd.to_datetime(df['date']) - pd.Timedelta(days=d.day_of_year)
).dt.isocalendar()
# get the difference in year
m = (s['year'].astype('int32') - d.year)
# all condition of result depending on year difference
conds = [m.eq(0), m.eq(-1), m.eq(1), m.lt(-1), m.gt(1)]
choices = ['', 'LY','NY',(m+1).astype(str)+'LY', '+'+(m-1).astype(str)+'NY']
# create the column
df['res'] = np.select(conds, choices) + s['week'].astype(str)
print(df)
date res
0 2020-03-12 -1LY38
1 2021-03-12 LY38
2 2021-03-17 LY39
3 2021-06-28 1
4 2022-05-21 47
5 2022-08-17 NY8
6 2023-08-17 +1NY8

I think
pandas period_range can be of some help
pd.Series(pd.period_range("6/28/2017", freq="W", periods=Number of weeks you want))

How to get a date from year, month, week of month and Day of week in Pandas?

I have a Pandas dataframe, which looks like below
I want to create a new column, which tells the exact date from the information from all the above columns. The code should look something like this:
df['Date'] = pd.to_datetime(df['Month']+df['WeekOfMonth']+df['DayOfWeek']+df['Year'])

I was able to find a workaround for your case. You will need to define the dictionaries for the months and the days of the week.
month = {"Jan":"01", "Feb":"02", "March":"03", "Apr": "04", "May":"05", "Jun":"06", "Jul":"07", "Aug":"08", "Sep":"09", "Oct":"10", "Nov":"11", "Dec":"12"}
week = {"Monday":1,"Tuesday":2,"Wednesday":3,"Thursday":4,"Friday":5,"Saturday":6,"Sunday":7}
With this dictionaries the transformation that I used with a custom dataframe was:
rows = [["Dec",5,"Wednesday", "1995"],
["Jan",3,"Wednesday","2013"]]
df = pd.DataFrame(rows, columns=["Month","Week","Weekday","Year"])
df['Date'] = (df["Year"] + "-" + df["Month"].map(month) + "-" + (df["Week"].apply(lambda x: (x - 1)*7) + df["Weekday"].map(week).apply(int) ).apply(str)).astype('datetime64[ns]')
However you have to be careful. With some data that you posted as example there were some dates that exceeds the date range. For example, for
row = ["Oct",5,"Friday","2018"]
The date displayed is 2018-10-33. I recommend using some logic to filter your data in order to avoid this kind of problems.

Let's approach it in 3 steps as follows:
Get the date of month start Month_Start from Year and Month
Calculate the date offsets DateOffset relative to Month_Start from WeekOfMonth and DayOfWeek
Get the actual date Date from Month_Start and DateOffset
Here's the codes:
df['Month_Start'] = pd.to_datetime(df['Year'].astype(str) + df['Month'] + '01', format="%Y%b%d")
import time
df['DateOffset'] = (df['WeekOfMonth'] - 1) * 7 + df['DayOfWeek'].map(lambda x: time.strptime(x, '%A').tm_wday) - df['Month_Start'].dt.dayofweek
df['Date'] = df['Month_Start'] + pd.to_timedelta(df['DateOffset'], unit='D')
Output:
Month WeekOfMonth DayOfWeek Year Month_Start DateOffset Date
0 Dec 5 Wednesday 1995 1995-12-01 26 1995-12-27
1 Jan 3 Wednesday 2013 2013-01-01 15 2013-01-16
2 Oct 5 Friday 2018 2018-10-01 32 2018-11-02
3 Jun 2 Saturday 1980 1980-06-01 6 1980-06-07
4 Jan 5 Monday 1976 1976-01-01 25 1976-01-26
The Date column now contains the dates derived from the information from other columns.
You can remove the working interim columns, if you like, as follows:
df = df.drop(['Month_Start', 'DateOffset'], axis=1)

Python: Pandas dataframe get the year to which the week number belongs and not the year of the date

I have a csv-file: https://data.rivm.nl/covid-19/COVID-19_aantallen_gemeente_per_dag.csv
I want to use it to provide insight into the corona deaths per week.
df = pd.read_csv("covid.csv", error_bad_lines=False, sep=";")
df = df.loc[df['Deceased'] > 0]
df["Date_of_publication"] = pd.to_datetime(df["Date_of_publication"])
df["Week"] = df["Date_of_publication"].dt.isocalendar().week
df["Year"] = df["Date_of_publication"].dt.year
df = df[["Week", "Year", "Municipality_name", "Deceased"]]
df = df.groupby(by=["Week", "Year", "Municipality_name"]).agg({"Deceased" : "sum"})
df = df.sort_values(by=["Year", "Week"])
print(df)
Everything seems to be working fine except for the first 3 days of 2021. The first 3 days of 2021 are part of the last week (53) of 2020: http://week-number.net/calendar-with-week-numbers-2021.html.
When I print the dataframe this is the result:
53 2021 Winterswijk 1
Woudenberg 1
Zaanstad 1
Zeist 2
Zutphen 1
So basically what I'm looking for is a way where this line returns the year of the week number and not the year of the date:
df["Year"] = df["Date_of_publication"].dt.year

You can use dt.isocalendar().year to setup df["Year"]:
df["Year"] = df["Date_of_publication"].dt.isocalendar().year
You will get year 2020 for date of 2021-01-01 but will get back to year 2021 for date of 2021-01-04 by this.
This is just similar to how you used dt.isocalendar().week for setting up df["Week"]. Since they are both basing on the same tuple (year, week, day) returned by dt.isocalendar(), they would always be in sync.
Demo
date_s = pd.Series(pd.date_range(start='2021-01-01', periods=5, freq='1D'))
date_s
0
0 2021-01-01
1 2021-01-02
2 2021-01-03
3 2021-01-04
4 2021-01-05
date_s.dt.isocalendar()
year week day
0 2020 53 5
1 2020 53 6
2 2020 53 7
3 2021 1 1
4 2021 1 2

You can simply subtract the two dates and then divide the days attribute of the timedelta object by 7.
For example, this is the current week we are on now.
time_delta = (dt.datetime.today() - dt.datetime(2021, 1, 1))
The output is a datetime timedelta object
datetime.timedelta(days=75, seconds=84904, microseconds=144959)
For your problem, you'd do something like this
time_delta = int((df["Date_of_publication"] - df["Year"].days / 7)
The output would be a number that is the current week since date_of_publication

How to find the median month between two dates?

I need to find the median month value between two dates in a date frame. I am simplifying the case by showing four examples.
import pandas as pd
import numpy as np
import datetime
df=pd.DataFrame([["1/31/2016","3/1/2016"],
["6/15/2016","7/14/2016"],
["7/14/2016","8/15/2016"],
["8/7/2016","9/6/2016"]], columns=['FromDate','ToDate'])
df['Month'] = df.ToDate.dt.month-df.FromDate.dt.month
I am trying to append a column but I am not getting the desired result.
I need to see these values: [2,6,7,8].

You can calculate the average date explicitly by adding half the timedelta between 2 dates to the earlier date. Then just extract the month:
# convert to datetime if necessary
df[df.columns] = df[df.columns].apply(pd.to_datetime)
# calculate mean date, then extract month
df['Month'] = (df['FromDate'] + (df['ToDate'] - df['FromDate']) / 2).dt.month
print(df)
FromDate ToDate Month
0 2016-01-31 2016-03-01 2
1 2016-06-15 2016-07-14 6
2 2016-07-14 2016-08-15 7
3 2016-08-07 2016-09-06 8

You need to convert the string to datetime before using dt.month.
This line calculates the average month number :
df['Month'] = (pd.to_datetime(df['ToDate']).dt.month +
pd.to_datetime(df['FromDate']).dt.month)//2
print(df)
FromDate ToDate Month
0 1/31/2016 3/1/2016 2
1 6/15/2016 7/14/2016 6
2 7/14/2016 8/15/2016 7
3 8/7/2016 9/6/2016 8
This only works with both dates in the same year.
jpp's solution is fine but will in some cases give the wrong answer:
["1/1/2016","3/1/2016"] one would expect 2 because February is between January and March, but jpp's will give 1 corresponding to January.