Question

如果存在长短划线（“ - ”），则返回完整字符串，如果为true，则返回第一个逗号（“，”）之前的所有内容。如何使用Python和Regex进行此操作？

from bs4 import BeautifulSoup
import requests
import json
import pandas as pd

request = requests.get('https://www.goodreads.com/quotes/tag/fun?page=1')
soup = BeautifulSoup(request.text, 'lxml')
# for loop
s = soup.find_all("div", class_="quoteText")[0].text
s = " ".join(s.split()) 
s[:s.index(",")]
s

原始输出：

“That does it," said Jace. "I\'m going to get you a dictionary for Christmas this year.""Why?" Isabelle said."So you can look up \'fun.\' I\'m not sure you know what it means.” ― Cassandra Clare, City of Ashes //<![CDATA[ function submitShelfLink(unique_id, book_id, shelf_id, shelf_name, submit_form, exclusive) { var checkbox_id = \'shelf_name_\' + unique_id + \'_\' + shelf_id; var element = document.getElementById(checkbox_id) var checked = element.checked if (checked && exclusive) { // can\'t uncheck a radio by clicking it! return } if(document.getElementById("savingMessage")){ Element.show(\'savingMessage\') } var element_id = \'shelfInDropdownName_\' + unique_id + \'_\' + shelf_id; Element.upda

期望的输出：

“That does it," said Jace. "I\'m going to get you a dictionary for Christmas this year.""Why?" Isabelle said."So you can look up \'fun.\' I\'m not sure you know what it means.” ― Cassandra Clare

Answer 1

我不确定我是否理解它，但我认为你的意思是：

example_string = "part to return,example__text"
if example_string.count('__') > 0:
    try:
        result = re.search('(.*?)\,', example_string).group(0)
    except:
        result = None
print(result)

这会打印＆＃39;部分返回＆＃39;

如果你的意思是，＆＃39; __＆＃39;之间的字符串部分。和＆＃39;，＆＃39;我会用：

example_string = "lala__part to return, lala"
try:
    result = re.search('__(.*?)\,', example_string).group(0)
except:
    result = None
print(result)

Answer 2

from bs4 import BeautifulSoup
from bs4.element import NavigableString
import requests


request = requests.get('https://www.goodreads.com/quotes/tag/fun?page=1')
soup = BeautifulSoup(request.text, 'html.parser')
# for loop
s = soup.find_all("div", class_="quoteText")[0]

text = ''

text += "".join([t.strip() for t in s.contents if type(t) == NavigableString])

for book_or_author_tag in s.find_all("a", class_ = "authorOrTitle"):
    text += "\n" + book_or_author_tag.text.strip()

print(text)

你想要的引用是在初始的quoteText div中分割的，但是在它上面调用text会返回你试图用正则表达式删除的所有CDATA垃圾。

通过循环遍历该div的每个子节点并检查它是否是可导航的字符串类型，我们只能提取您想要的实际文本数据。然后对作者和书进行处理，希望你的正则表达式变得简单得多。

Answer 3

这是一个解决方案：

import re

s = 'adflakjd, fkljlkjdf ― Cassandra Clare, City of Ash, adflak'

x = re.findall('.*―.*?(?=,)', s)


print x

['adflakjd, fkljlkjdf ― Cassandra Clare']

正则表达式/ Python：在一个符号之前查找所有内容，如果它在另一个符号之后

3 个答案: