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Python - Remove lines prior to current month and year

I have a dataframe that contain arrival dates for vessels and I'd want to make python recognize the current year and month that we are at the moment and remove all entries that are prior to the current month and year.
I have a column with the date itself in the format '%d/%b/%Y' and columns for month and year separatly if needed.
For instance, if today is 01/01/2022. I'd like to remove everything that is from dec/2021 and prior.

Using pandas periods and boolean indexing:
# set up example
df = pd.DataFrame({'date': ['01/01/2022', '08/02/2022', '09/03/2022'], 'other_col': list('ABC')})
# find dates equal or greater to this month
keep = (pd.to_datetime(df['date'], dayfirst=False)
.dt.to_period('M')
.ge(pd.Timestamp('today').to_period('M'))
)
# filter
out = df[keep]
Output:
date other_col
1 08/02/2022 B
2 09/03/2022 C

from datetime import datetime
import pandas as pd
df = ...
# assuming your date column is named 'date'
t = datetime.utcnow()
df = df[pd.to_datetime(df.date) >= datetime(t.year, t.month, t.day)]

Let us consider this example dataframe:
import pandas as pd
import datetime
df = pd.DataFrame()
data = [['nao victoria', '21/Feb/2012'], ['argo', '6/Jun/2022'], ['kon tiki', '23/Aug/2022']]
df = pd.DataFrame(data, columns=['Vessel', 'Date'])
You can convert your dates to datetimes, by using pandas' to_datetime method; for instance, you may save the output into a new Series (column):
df['Datetime']=pd.to_datetime(df['Date'], format='%d/%b/%Y')
You end up with the following dataframe:
Vessel Date Datetime
0 nao victoria 21/Feb/2012 2012-02-21
1 argo 6/Jun/2022 2022-06-06
2 kon tiki 23/Aug/2022 2022-08-23
You can then reject rows containing datetime values that are smaller than today's date, defined using datetime's now method:
df = df[df.Datetime > datetime.datetime.now()]
This returns:
Vessel Date Datetime
2 kon tiki 23/Aug/2022 2022-08-23

Not all dates are captured when filtering by dates. Python Pandas

I am filtering a dataframe by dates to produce two seperate versions:
Data from only today's date
Data from the last two years
However, when I try to filter on the date, it seems to miss dates that are within the last two years.
date_format = '%m-%d-%Y' # desired date format
today = dt.now().strftime(date_format) # today's date. Will always result in today's date
today = dt.strptime(today, date_format).date() # converting 'today' into a datetime object
today = today.strftime(date_format)
two_years = today - relativedelta(years=2) # date is today's date minus two years.
two_years = two_years.strftime(date_format)
# normalizing the format of the date column to the desired format
df_data['date'] = pd.to_datetime(df_data['date'], errors='coerce').dt.strftime(date_format)
df_today = df_data[df_data['date'] == today]
df_two_year = df_data[df_data['date'] >= two_years]
Which results in:
all dates ['07-17-2020' '07-15-2020' '08-01-2019' '03-25-2015']
today df ['07-17-2020']
two year df ['07-17-2020' '08-01-2019']
The 07-15-2020 date is missing from the two year, even though 08-01-2019 is captured.

you don't need to convert anything to string, simply work with datetime dtype. Ex:
import pandas as pd
df = pd.DataFrame({'date': pd.to_datetime(['07-17-2020','07-15-2020','08-01-2019','03-25-2015'])})
today = pd.Timestamp('now')
print(df[df['date'].dt.date == today.date()])
# date
# 0 2020-07-17
print(df[(df['date'].dt.year >= today.year-1) & (df['date'].dt.date != today.date())])
# date
# 1 2020-07-15
# 2 2019-08-01
What you get from the comparison operations (adjust them as needed...) are boolean masks - you can use them nicely to filter the df.

Your datatype conversions are the problem here. You could do this:
today = dt.now() # today's date. Will always result in today's date
two_years = today - relativedelta(years=2) # date is today's date minus two years.
This prints '2018-07-17 18:40:42.704395'. You can then convert it to the date only format.
two_years = two_years.strftime(date_format)
two_years = dt.strptime(two_years, date_format).date()

Python datetime delta format

I am attempting to find records in my dataframe that are 30 days old or older. I pretty much have everything working but I need to correct the format of the Age column. Most everything in the program is stuff I found on stack overflow, but I can't figure out how to change the format of the delta that is returned.
import pandas as pd
import datetime as dt
file_name = '/Aging_SRs.xls'
sheet = 'All'
df = pd.read_excel(io=file_name, sheet_name=sheet)
df.rename(columns={'SR Create Date': 'Create_Date', 'SR Number': 'SR'}, inplace=True)
tday = dt.date.today()
tdelta = dt.timedelta(days=30)
aged = tday - tdelta
df = df.loc[df.Create_Date <= aged, :]
# Sets the SR as the index.
df = df.set_index('SR', drop = True)
# Created the Age column.
df.insert(2, 'Age', 0)
# Calculates the days between the Create Date and Today.
df['Age'] = df['Create_Date'].subtract(tday)
The calculation in the last line above gives me the result, but it looks like -197 days +09:39:12 and I need it to just be a positive number 197. I have also tried to search using the python, pandas, and datetime keywords.
df.rename(columns={'Create_Date': 'SR Create Date'}, inplace=True)
writer = pd.ExcelWriter('output_test.xlsx')
df.to_excel(writer)
writer.save()

I can't see your example data, but IIUC and you're just trying to get the absolute value of the number of days of a timedelta, this should work:
df['Age'] = abs(df['Create_Date'].subtract(tday)).dt.days)
Explanation:
Given a dataframe with a timedelta column:
>>> df
delta
0 26523 days 01:57:59
1 -1601 days +01:57:59
You can extract just the number of days as an int using dt.days:
>>> df['delta']dt.days
0 26523
1 -1601
Name: delta, dtype: int64
Then, all you need to do is wrap that in a call to abs to get the absolute value of that int:
>>> abs(df.delta.dt.days)
0 26523
1 1601
Name: delta, dtype: int64

here is what i worked out for basically the same issue.
# create timestamp for today, normalize to 00:00:00
today = pd.to_datetime('today', ).normalize()
# match timezone with datetimes in df so subtraction works
today = today.tz_localize(df['posted'].dt.tz)
# create 'age' column for days old
df['age'] = (today - df['posted']).dt.days
pretty much the same as the answer above, but without the call to abs().

How can I update a Pandas column safely (after some manipulation) if one row's value is formatted incorrectly?

I am essentially trying to take data in the Date column in my dataframe, and subtract it from the date today in order to get the timedelta (which I will be storing in a new column). The issue I am running into is that i the Date value is formatted incorrectly or not a date at all, that will either cause my program to crash, or when I try to handle that error with simply mess up the other row's data. Here is my code:
def add_delta_to_dataframe():
df = create_messages_dataframe()
date = pd.to_datetime(df['Date'], format='%m/%d/%Y', errors="ignore")
now = datetime.datetime.today()
try:
delta = ((date - now).dt.days) + 1
df['Delta'] = delta
except TypeError:
pass
return df
I have also tried to iterate through:
df['Date'] = pd.to_datetime(df['Date'], format='%m/%d/%Y', errors="ignore")
now = datetime.datetime.today()
for index, row in df.iterrows():
try:
delta = ((row['Date'] - now).days) + 1
df['Delta'] = delta
except TypeError:
continue
But no luck here either. Any ideas on doing this would be greatly appreciated. I either get an error if I don't catch the error, or the output leaves all Delta values as NaN. My expected output would be the columns with the correct date format to have the Delta value there, and the others to be NaN

IIUC, you can leverage the errors='coerce' argument of pd.to_datetime, which will set unformattable strings to NaT. Take the following df for an example:
df = pd.DataFrame({'date':['1999-01-01', 'xyz', '2000-05-05']})
>>> df
date
0 1999-01-01
1 xyz
2 2000-05-05
You can create your timedelta-like column using:
df['my_timedelta'] = pd.to_datetime('today') - pd.to_datetime(df['date'], errors='coerce')
Which results in:
>>> df
date my_timedelta
0 1999-01-01 7066 days
1 xyz NaT
2 2000-05-05 6576 days

Pandas - Python, deleting rows based on Date column

I'm trying to delete rows of a dataframe based on one date column; [Delivery Date]
I need to delete rows which are older than 6 months old but not equal to the year '1970'.
I've created 2 variables:
from datetime import date, timedelta
sixmonthago = date.today() - timedelta(188)
import time
nineteen_seventy = time.strptime('01-01-70', '%d-%m-%y')
but I don't know how to delete rows based on these two variables, using the [Delivery Date] column.
Could anyone provide the correct solution?

You can just filter them out:
df[(df['Delivery Date'].dt.year == 1970) | (df['Delivery Date'] >= sixmonthago)]
This returns all rows where the year is 1970 or the date is less than 6 months.
You can use boolean indexing and pass multiple conditions to filter the df, for multiple conditions you need to use the array operators so | instead of or, and parentheses around the conditions due to operator precedence.
Check the docs for an explanation of boolean indexing

Be sure the calculation itself is accurate for "6 months" prior. You may not want to be hardcoding in 188 days. Not all months are made equally.
from datetime import date
from dateutil.relativedelta import relativedelta
#http://stackoverflow.com/questions/546321/how-do-i-calculate-the-date-six-months-from-the-current-date-using-the-datetime
six_months = date.today() - relativedelta( months = +6 )
Then you can apply the following logic.
import time
nineteen_seventy = time.strptime('01-01-70', '%d-%m-%y')
df = df[(df['Delivery Date'].dt.year == nineteen_seventy.tm_year) | (df['Delivery Date'] >= six_months)]
If you truly want to drop sections of the dataframe, you can do the following:
df = df[(df['Delivery Date'].dt.year != nineteen_seventy.tm_year) | (df['Delivery Date'] < six_months)].drop(df.columns)