How to convert string '6 dec 2019 9pm' to datetime object?

Question

I used strptime() with format '%d %a %Y %-Ip%' but it's throwing the error stating "-I" is bad directive.

`pd.to_datetime('6 dec 2019 9pm')` – luigigi Dec 16 '19 at 11:25 — luigigi, Dec 16 '19 at 11:25
Use `pd.to_datetime(date_col, format='%d %b %Y %I%p')` – Chris Adams Dec 16 '19 at 11:26 — Chris Adams, Dec 16 '19 at 11:26

score 0 · Answer 1 · answered Dec 16 '19 at 11:30

You have your time format wrong. %-Ip% is not a valid format (should be %I%p), and %a is also incorrect for parsing dec (should be %b%).

The following example shows the output with the format %d %b %Y %I%p:

import time

struct_time = time.strptime("6 dec 2019 9pm", "%d %b %Y %I%p")
print(struct_time)

Output:

time.struct_time(tm_year=2019, tm_mon=12, tm_mday=6, tm_hour=21, tm_min=0, tm_sec=0, tm_wday=4, tm_yday=340, tm_isdst=-1)

score 0 · Answer 2 · answered Dec 16 '19 at 11:32

0

You can use datetime

from datetime import datetime
datetime.strptime('6 dec 2019 9pm', '%d %b %Y %I%p')

answered Dec 16 '19 at 11:32

Bilal Zaib

2 Answers2