regex demo

Question

我有一个我要解析的网址列表：

['https://www.richmondfed.org/-/media/richmondfedorg/press_room/speeches/president_jeff_lacker/2017/pdf/lacker_speech_20170303.pdf','http://www.federalreserve.gov/newsevents/speech/powell20160929a.htm','http://www.federalreserve.gov/newsevents/speech/fischer20161005a.htm']

我想使用Regex表达式创建一个新列表，其中包含字符串末尾的数字和标点符号前的任何字母（某些字符串包含两个位置的数字，如上面列表中的第一个字符串所示）。所以新列表看起来像：

['20170303', '20160929a', '20161005a']

这是我尝试过的，没有运气：

code = re.search(r'?[0-9a-z]*', urls)

更新

跑步 -

[re.search(r'(\d+)\D+$', url).group(1) for url in urls]

我收到以下错误 -

AttributeError: 'NoneType' object has no attribute 'group'

此外，如果有一封信，这似乎不会在数字后面收到一封信..！

Answer 1

假设：

>>> lios=['https://www.richmondfed.org/-/media/richmondfedorg/press_room/speeches/president_jeff_lacker/2017/pdf/lacker_speech_20170303.pdf','http://www.federalreserve.gov/newsevents/speech/powell20160929a.htm','http://www.federalreserve.gov/newsevents/speech/fischer20161005a.htm']

你可以这样做：

for s in lios:
    m=re.search(r'(\d+\w*)\D+$', s)
    if m:
        print m.group(1)

打印：

20170303
20160929a
20161005a

这是基于这个正则表达式：

(\d+\w*)\D+$
  ^              digits
     ^           any non digits
        ^        non digits
           ^     end of string

Answer 2

您可以使用此正则表达式(\d+[a-z]*)\.：

regex demo

<强>输出

20170303
20160929a
20161005a

Answer 3

# python3

from urllib.parse import urlparse
from os.path import basename

def extract_id(url):
    path = urlparse(url).path
    resource = basename(path)
    _id = re.search('\d[^.]*', resource)
    if _id:
        return _id.group(0)

urls =['https://www.richmondfed.org/-/media/richmondfedorg/press_room/speeches/president_jeff_lacker/2017/pdf/lacker_speech_20170303.pdf','http://www.federalreserve.gov/newsevents/speech/powell20160929a.htm','http://www.federalreserve.gov/newsevents/speech/fischer20161005a.htm']

# /!\ here you have None if pattern doesn't exist ;) in ids list
ids = [extract_id(url) for url in urls]

print(ids)

输出：

['20170303', '20160929a', '20161005a']

Answer 4

import re

patterns = {
    'url_refs': re.compile("(\d+[a-z]*)\."),  # YCF_L
}

def scan(iterable, pattern=None):
    """Scan for matches in an iterable."""
    for item in iterable:
        # if you want only one, add a comma:
        # reference, = pattern.findall(item)
        # but it's less reusable.
        matches = pattern.findall(item)
        yield matches

然后你可以这样做：

hits = scan(urls, pattern=patterns['url_refs'])
references = (item[0] for item in hits)

将references提供给您的其他功能。你可以通过这种方式浏览更多的东西，并且我想这样做更快。

使用python中的正则表达式从字符串中提取数字

4 个答案:

regex demo