# Guided Project: Exploring Hacker News Posts step 7 of 8

https://app.dataquest.io/m/356/guided-project%3A-exploring-hacker-news-posts/7/sorting-and-printing-values-from-a-list-of-lists

I got to the final step, and had no clue as to how to “convert the times to the time zone you live in.”

``````hour_format = "%H"
for row in sorted_swap[:5]:
hour = row[1]
hour = dt.datetime.strptime(hour, hour_format)
hour = hour.strftime("%H:%M")

print("{0} {1} average comments per post".format(hour, avg_comments))
``````

I’m sure I’m missing something obvious, but I’m drawing a blank.
FWIW, the times in the data set were Eastern Standard Time. I’m Central Standard Time, so I would need to subtract one hour.

Hey, Eroggenburg.

``````>>> import datetime as dt
>>> some_hour = "09"
``````

Here are two different (and incomplete) ways to do it.

### My prefered way

``````>>> dt.datetime.strptime(some_hour, "%H")
datetime.datetime(1900, 1, 1, 9, 0)
>>> dt.datetime.strptime(some_hour, "%H")+dt.timedelta(hours=1)
datetime.datetime(1900, 1, 1, 10, 0)
``````

### An alternative

``````>>> def add_one(s):
...     plus_one = int(s)+1
...     mod_24 = plus_one %24
...     return str(mod_24).zfill(2)
...
'10'
``````

I realize I added instead of subtracting. I’ll let you make the necessary changes.

1 Like

Thank you, Bruno! I fiddled around and ended up with a much less elegant solution:

``````print("Top 5 Hours for Ask Posts Comments:")
hour_format = "%H"
for row in sorted_swap[:5]:
hour = row[1]

# The original time zone was Eastern Standard.
# Here I am converting to Central Standard by subtracting an hour.
convert_to_cst = dt.datetime.strptime(hour, hour_format)
cst = convert_to_cst - dt.timedelta(hours=1)
cst = cst.strftime("%H:%M")