如何确定pandas数据框列中列表的长度

Question

如何在没有迭代的情况下确定列中列表的长度？

我有一个这样的数据帧：

                                                    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]

我正在计算CreationDate列中列表的长度，并创建一个新的Length列，如下所示：

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

这给了我这样的结论：

                                                    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

有没有更好的方法来做到这一点呢？

Answer 1

您还可以将str访问器用于某些列表操作。在此示例中，

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

返回每个列表的长度。请参阅str.len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，普通Python通常更快。不过，熊猫会处理NaNs。以下是时间安排：

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

与pandas.Series.str.len().相比，

• pandas.Series.map(len)和pandas.Series.apply(len)在执行时间上是相等的，并且略快一些

- [`pandas.Series.map`](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.map.html)
- [`pandas.Series.apply`](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.apply.html)
- [`pandas.Series.str.len`](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.len.html)

• 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)