Question

我有一个这样的数据框：

                                                    CreationDate
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]

我是CreationDate列中列表的计算长度，并按照以下方式创建新的Length列：

df['Length'] = df.CreationDate.apply(lambda x: len(x))

这给了我这个：

                                                    CreationDate  Length
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]       3
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]       4
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]       4

有更多的pythonic方法吗？

Answer 1

您也可以使用str访问器进行一些列表操作。在这个例子中，

df['CreationDate'].str.len()

返回每个列表的长度。请参阅resource management的文档。

df['Length'] = df['CreationDate'].str.len()
df
Out: 
                                                    CreationDate  Length
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]       3
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]       4
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]       4

对于这些操作，vanilla Python通常更快。熊猫虽然处理NaNs。这是时间：

ser = pd.Series([random.sample(string.ascii_letters, 
                               random.randint(1, 20)) for _ in range(10**6)])

%timeit ser.apply(lambda x: len(x))
1 loop, best of 3: 425 ms per loop

%timeit ser.str.len()
1 loop, best of 3: 248 ms per loop

%timeit [len(x) for x in ser]
10 loops, best of 3: 84 ms per loop

%timeit pd.Series([len(x) for x in ser], index=ser.index)
1 loop, best of 3: 236 ms per loop

Answer 2

pandas.Series.map(len)和pandas.Series.apply(len)的执行时间相同，并且比pandas.Series.str.len()快。
Difference between map, applymap and apply methods in Pandas

import pandas as pd

data = {'os': [['ubuntu', 'mac-osx', 'syslinux'], ['ubuntu', 'mod-rewrite', 'laconica', 'apache-2.2'], ['ubuntu', 'nat', 'squid', 'mikrotik']]}
index = ['2013-12-22 15:25:02', '2009-12-14 14:29:32', '2013-12-22 15:42:00']

df = pd.DataFrame(data, index)

# create Length column
df['Length'] = df.os.map(len)

# display(df)
                                                              os  Length
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]       3
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]       4
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]       4

`%timeit`

import pandas as pd
import random
import string

random.seed(365)

ser = pd.Series([random.sample(string.ascii_letters, random.randint(1, 20)) for _ in range(10**6)])

%timeit ser.str.len()
252 ms ± 12.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit ser.map(len)
220 ms ± 7.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit ser.apply(len)
222 ms ± 8.31 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

用于计算pandas dataframe列中列表长度的Pythonic方法

2 个答案:

`%timeit`