Question

我可以使用re.finditer轻松提取文本中字符串的开始/结束位置，从而得出匹配项开始/结束位置的元组。

s1 = 'the quick quick brown fox jumps over the lazy dog'
s2 = 'Pack my box with five dozen liquor jugs'
s3 = 'How razorback jumping frogs can level six piqued gymnasts'

r1 = '(quick|fox|dog)'
r2 = '(box|five|jugs)'
r3 = '(frogs|six)'

t = [(s1,r1), (s2,r2), (s3,r3)]


for e in t:
    print([(f.start(), f.end()) for f in re.finditer(e[1],e[0])])

[(4, 9), (10, 15), (22, 25), (46, 49)]
[(8, 11), (17, 21), (35, 39)]
[(22, 27), (38, 41)]

我有一个DataFrame，其中一列为文本，而正则表达式为另一列

s = pd.DataFrame(data={'re':[r1,r2,r3], 'text':[s1,s2,s3]})

    re              text
0   (quick|fox|dog) the quick quick brown fox jumps over the lazy dog
1   (box|five|jugs) Pack my box with five dozen liquor jugs
2   (frogs|six)     How razorback jumping frogs can level six piqu...

我想使用Series方法将相同的位置信息提取到pandas.str对象中，但是没有finditer方法（在熊猫0.23中）。

是否有一种方法可以不借助for-each循环？

Answer 1

扩展@ user3483203的注释，您可以使用list comprehension执行以下操作：

import re
import pandas as pd

s1 = 'the quick quick brown fox jumps over the lazy dog'
s2 = 'Pack my box with five dozen liquor jugs'
s3 = 'How razorback jumping frogs can level six piqued gymnasts'

r1 = '(quick|fox|dog)'
r2 = '(box|five|jugs)'
r3 = '(frogs|six)'

t = [(s1,r1), (s2,r2), (s3,r3)]

s = pd.DataFrame(data={'re':[r1,r2,r3], 'text':[s1,s2,s3]})

result = pd.Series([[(f.start(), f.end()) for f in re.finditer(p, s)] for p, s in zip(s.re, s.text)])
print(result)

输出

0    [(4, 9), (10, 15), (22, 25), (46, 49)]
1             [(8, 11), (17, 21), (35, 39)]
2                      [(22, 27), (38, 41)]
dtype: object

另一种替代方法是使用apply，但我认为这可能会更慢：

def finditer(p, s):
    return [(f.start(), f.end()) for f in re.finditer(p, s)]


result = s[['re', 'text']].apply(lambda x: finditer(x[0], x[1]), axis=1)
print(result)

提取字符串中的匹配位置

1 个答案: