时间计算，平均值，中位数，众数

Question

(	Name	Gun_time	Net_time	Pace
John	28:48:00	28:47:00	4:38:00
George	29:11:00	29:10:00	4:42:00
Mike	29:38:00	29:37:00	4:46:00
Sarah	29:46:00	29:46:00	4:48:00
Roy	30:31:00	30:30:00	4:55:00

Q1. How can I add another column stating difference between Gun_time and Net_time? Q2. How will I calculate the mean for Gun_time and Net_time. Please help!

I have tried doing the following but it doesn't work

df['Difference'] = df['Gun_time'] - df['Net_time']

for mean value I tried df['Gun_time'].mean

but it doesn't work either, please help!

Q.3 What if we have times in 28:48 (minutes and seconds) format and not 28:48:00 the function gives out a value error.

ValueError: expected hh:mm:ss format

Answer 1

Convert your columns to dtype timedelta, eg like

for col in ("Gun_time", "Net_time", "Pace"):
    df[col] = pd.to_timedelta(df[col])

Now you can do calculations like

df['Gun_time'].mean()
# Timedelta('1 days 05:34:48')  

or

df['Difference'] = df['Gun_time'] - df['Net_time']

#df['Difference']
# 0   0 days 00:01:00
# 1   0 days 00:01:00
# 2   0 days 00:01:00
# 3   0 days 00:00:00
# 4   0 days 00:01:00
# Name: Difference, dtype: timedelta64[ns]

If you need nicer output to string, you can use

def timedeltaToString(td):
    hours, remainder = divmod(td.total_seconds(), 3600)
    minutes, seconds = divmod(remainder, 60)
    return f"{int(hours):02d}:{int(minutes):02d}:{int(seconds):02d}"


df['diffString'] = df['Difference'].apply(timedeltaToString)

# df['diffString']
# 0    00:01:00
# 1    00:01:00
# 2    00:01:00
# 3    00:00:00
# 4    00:01:00
#Name: diffString, dtype: object

See also Format timedelta to string .