Python:在列表的 Pandas 列上进行字符串匹配

Question

对列表的一列进行字符串匹配的最佳方法是什么？
例如。我有一个数据集:

import numpy as np
import pandas as pd
list_items = ['apple', 'grapple', 'tackle', 'satchel', 'snapple']
df = pd.DataFrame({'id':xrange(3), 'L':[np.random.choice(list_items, 3).tolist() for _ in xrange(3)]})
df

    L                           id
0   [tackle, apple, grapple]    0
1   [tackle, snapple, satchel]  1
2   [satchel, satchel, tackle]  2

我想返回 L 中的任何项目与字符串匹配的行，例如“grap”应返回第 0 行，“sat”应返回第 1:2 行。

Answer 1

让我们用这个:

np.random.seed(123)
list_items = ['apple', 'grapple', 'tackle', 'satchel', 'snapple']
df = pd.DataFrame({'id':range(3), 'L':[np.random.choice(list_items, 3).tolist() for _ in range(3)]})
df
                             L  id
0    [tackle, snapple, tackle]   0
1   [grapple, satchel, tackle]   1
2  [satchel, grapple, grapple]   2

使用any和apply:

df[df.L.apply(lambda x: any('grap' in s for s in x))]

输出:

                             L  id
1   [grapple, satchel, tackle]   1
2  [satchel, grapple, grapple]   2

---

时间:

%timeit df.L.apply(lambda x: any('grap' in s for s in x))

10000 loops, best of 3: 194 µs per loop

%timeit df.L.apply(lambda i: ','.join(i)).str.contains('grap')

1000 loops, best of 3: 481 µs per loop

%timeit df.L.str.join(', ').str.contains('grap')

1000 loops, best of 3: 529 µs per loop