将 function 传递到数据框中的列 - Python

Question

I am trying to pass a function that truncates a timestamp to a single column. It's performing the function but returning a list. I'm hoping to keep the data structure.

df = pd.DataFrame({
    'Time' : ['8:03:001','8:17:004','8:20:003','8:28:002','8:35:004','8:40:006','8:42:002','8:45:004','8:50:009'],                 
    'Place' : ['House 1','House 1','House 1','House 2','House 2','House 2','House 3','House 3','House 3'],                 
     })

def truncate_time(col):
    col = [x[:-2] for x in col]
    return col

df1 = (truncate_time(df['Time']))

Intended Output:

       Time    Place
0  8:03:0    House 1
1  8:17:0    House 1
2  8:20:0    House 1
3  8:28:0    House 2
4  8:35:0    House 2
5  8:40:0    House 2
6  8:42:0    House 3
7  8:45:0    House 3
8  8:50:0    House 3

Answer 1

You can assign back:

df['Time'] = truncate_time(df['Time'])
print (df)
     Time    Place
0  8:03:0  House 1
1  8:17:0  House 1
2  8:20:0  House 1
3  8:28:0  House 2
4  8:35:0  House 2
5  8:40:0  House 2
6  8:42:0  House 3
7  8:45:0  House 3
8  8:50:0  House 3

But here is also possible use str with indexing:

df['Time'] = df['Time'].str[:-2]

Or lambda function:

df['Time'] = df['Time'].apply(lambda col: col[:-2])

Or for function simplify solution with remove list comprehension with Series.apply :

def truncate_time(col):
    return col[:-2]

df['Time'] = df['Time'].apply(truncate_time)

And last solution with list comprehension:

df['Time'] = [x[:-2] for x in df['Time']]

EDIT: Performance with possible missing values - depends of number of values and also number of missing values:

#added one row with missing value
df = pd.DataFrame({
    'Time' : ['8:03:001','8:17:004','8:20:003','8:28:002','8:35:004','8:40:006','8:42:002','8:45:004','8:50:009',np.nan],                 
    'Place' : ['House 1','House 1','House 1','House 2','House 2','House 2','House 3','House 3','House 3','House 3'],                 
     })

def truncate_time(col):
    return col[:-2] if col == col else col

#[1000000 rows x 2 columns]
df = pd.concat([df] * 100000, ignore_index=True)

In [104]: %timeit df['Time1'] = df['Time'].str[:-2]
460 ms ± 20.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [105]: %timeit df['Time2'] = [x[:-2] if x == x else x for x in df['Time']]
445 ms ± 9.72 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [106]: %timeit df['Time3'] = df['Time'].apply(lambda col: col[:-2] if col == col else col)
428 ms ± 18.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [107]: %timeit df['Time4'] = df['Time'].apply(truncate_time)
416 ms ± 8.28 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)