How to convert days to hours in pandas - pandas

Question

Example:
coupon expiration
Restaurant 1d
College 2d
Coffee House 2h
o/p:
coupon expiration
Restaurant 24h
College 48h
Coffee House 2h
How to convert days to hours in pandas

Answer 1

You can use pd.to_timedelta, but the values in the expiration column must be valid timedelta strings:
import pandas as pd
df = pd.read_clipboard() # Your df here
tds = pd.to_timedelta(df["expiration"])
# 0 1 days 00:00:00
# 1 2 days 00:00:00
# 2 0 days 02:00:00
# Name: expiration, dtype: timedelta64[ns]
# I would recommend stopping here, but you can reformat this into a string of hours:
df["expiration"] = tds.dt.total_seconds().div(3600).apply("{:g}h".format)
# coupon expiration
# 0 Restaurant 24h
# 1 College 48h
# 2 CoffeeHouse 2h

Answer 2

You can use str.replace on the expiration column and use a regex pattern to select those entries that have a day (d) suffix. You can also call a function for the repl parameter - which is where I chose to do the conversion to hours.
Code:
import pandas as pd
df = pd.DataFrame({"coupon":['Restaurant','College','Coffee House'], "expiration":['1d','2d','2h']})
def replacement(m):
x = int(m.group(0).split('d')[0]) * 24
return f"{x}h"
df.expiration = df.expiration.str.replace(pat=r'^\d+d$', repl=replacement, regex=True)
print(df)
Output:
coupon expiration
0 Restaurant 24h
1 College 48h
2 Coffee House 2h
Regex Pattern:
r'^\d+d$'
^ : start of string
\d+ : one or more digits [0-9]
d : followed by the letter d
$ : end of string
Note:
If you would rather a one-liner using a lambda function instead:
df.expiration = df.expiration.str.replace(pat=r'^\d+d$', repl= lambda m:f"{int(m.group(0).split('d')[0]) * 24}h", regex=True)

Answer 3

A simply Apply can help here
def convert(x):
if 'd' in x:
return f"{int(x.replace('d',''))*24}h"
return x
df['expiration']= df['expiration'].apply(lambda x:convert(x))
df
Out[57]:
coupon expiration
0 Restaurant 24h
1 College 48h
2 Coffee House 2h

Answer 4

Another possible solution, based on eval:
df['expiration'] = [str(eval(x)) + 'h' for x in
df['expiration'].str.replace('d', '*24').str.replace('h', '')]
Output:
coupon expiration
0 Restaurant 24h
1 College 48h
2 Coffee House 2h