Python more efficient method than .apply()

I have a large dataframe with projected data 60 months into the future, and I need to drop the projections for months that haven't happened yet. I have a functioning way to do this but it's throwing memory errors for a 16 million row dataframe (I have removed all unnecessary columns):

from dateutil.relativedelta import relativedelta
from tqdm import tqdm
tqdm.pandas()

def add_months(start_date, delta_period):
  end_date = start_date + relativedelta(months=delta_period)
  return end_date

# Apply function on the dataframe using lambda operation.
snapshots["End_Date"] = snapshots.progress_apply(lambda row: add_months(row["startDate"], row["projectedMonth"]), axis = 1)

Then I would drop columns where end_date>today. I tried to import 'swifter' but my organisation's settings won't allow that. Is there a more efficient way to deal with this? I wondered about doing

snapshots['End_Date']=snapshots['startDate']+relativedelta(months=snapshots['projectedMonth'])

But get the error about relativedelta needing int not series. Thanks!

With a lambda function and instead using vectorized operations you could do that, for your case pd.DateOffset to add months directly to the date column would be good.

import pandas as pd

data = {
    "startDate": pd.date_range(start="2020-01-01", periods=10, freq="MS"),
    "projectedMonth": [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
}
snapshots = pd.DataFrame(data)

snapshots["End_Date"] = snapshots.apply(lambda row: row["startDate"] + pd.DateOffset(months=row["projectedMonth"]), axis=1)

today = pd.Timestamp.today()
filtered_snapshots = snapshots[snapshots["End_Date"] <= today]

print(filtered_snapshots)