优化Python代码

Question

I have written the following code to preprocess a dataset like this:

StartLocation   StartTime   EndTime
school          Mon Jul 25 19:04:30 GMT+01:00 2016  Mon Jul 25 19:04:33 GMT+01:00 2016
...             ...         ...

It contains a list of locations attended by a user with the start and end time. Each location may occur several times and there is no comprehensive list of locations. From this, I want to aggregate data for each location (frequency, total time, mean time). To do this I have written the following code:

def toEpoch(x):
    try:
        x = datetime.strptime(re.sub(r":(?=[^:]+$)", "", x), '%a %b %d %H:%M:%S %Z%z %Y').strftime('%s')
    except:
        x = datetime.strptime(x, '%a %b %d %H:%M:%S %Z %Y').strftime('%s')
    x = (int(x)/60)
    return x

#Preprocess data
df = pd.read_csv('...')
for index, row in df.iterrows():
    df['StartTime'][index] = toEpoch(df['StartTime'][index])
    df['EndTime'][index] = toEpoch(df['EndTime'][index])
    df['TimeTaken'][index] = int(df['EndTime'][index]) - int(df['StartTime'][index])
total = df.groupby(df['StartLocation'].str.lower()).sum()
av = df.groupby(df['StartLocation'].str.lower()).mean()
count = df.groupby(df['StartLocation'].str.lower()).count()
output = pd.DataFrame({"location": total.index, 'total': total['TimeTaken'], 'mean': av['TimeTaken'], 'count': count['TimeTaken']})
print(output)

This code functions correctly, however is quite inefficient. How can I optimise the code?

EDIT: Based on @Batman's helpful comments I no longer iterate. However, I still hope to further optimise this if possible. The updated code is:

df = pd.read_csv('...')
df['StartTime'] = df['StartTime'].apply(toEpoch)
df['EndTime'] = df['EndTime'].apply(toEpoch)
df['TimeTaken'] = df['EndTime'] - df['StartTime']
total = df.groupby(df['StartLocation'].str.lower()).sum()
av = df.groupby(df['StartLocation'].str.lower()).mean()
count = df.groupby(df['StartLocation'].str.lower()).count()
output = pd.DataFrame({"location": total.index, 'total': total['TimeTaken'], 'mean': av['TimeTaken'], 'count': count['TimeTaken']})
print(output)

Answer 1

First thing I'd do is stop iterating over the rows.

df['StartTime'] = df['StartTime'].apply(toEpoch)
df['EndTime'] = df['EndTime'].apply(toEpoch)
df['TimeTaken'] = df['EndTime'] - df['StartTime']

Then, do a single groupby operation.

gb = df.groupby('StartLocation')
total = gb.sum()
av = gb.mean()
count = gb.count()

Answer 2

• vectorize the date conversion
• take the difference of two series of timestamps gives a series of timedeltas
• use total_seconds to get the seconds from the the timedeltas
• groupby with agg

---

# convert dates
cols = ['StartTime', 'EndTime']
df[cols] = pd.to_datetime(df[cols].stack()).unstack()

# generate timedelta then total_seconds via the `dt` accessor
df['TimeTaken'] = (df.EndTime - df.StartTime).dt.total_seconds()

# define the lower case version for cleanliness
loc_lower = df.StartLocation.str.lower()

# define `agg` functions for cleanliness
# this tells `groupby` to use 3 functions, sum, mean, and count
# it also tells what column names to use
funcs = dict(Total='sum', Mean='mean', Count='count')
df.groupby(loc_lower).TimeTaken.agg(funcs).reset_index()

---

explanation of date conversion

• I define cols for convenience
• df[cols] = is an assignment to those two columns
• pd.to_datetime() is a vectorized date converter but only takes pd.Series not pd.DataFrame
• df[cols].stack() makes the 2-column dataframe into a series, now ready for pd.to_datetime()
• use pd.to_datetime(df[cols].stack()) as described and unstack() to get back my 2-columns and now ready to be assigned.