我想删除整个链接：
https://www.linkedin.com/in/ACoAAAJv1l4BATlBOVqhEEaqrVNojJPWnID9Nk0
当链接包含ACo正则表达式应该从我的模式中删除整个链接。
regex2 = re.compile(r"\bhttps?://www.linkedin.com/in/\b[^in]+")
由于某种原因，我没有让这个工作，想法是当链接的行为以“ACo”（大写 A 和大写 C）开头时删除/in/
我们有 4 个链接，我只想打印，https://www.linkedin.com/in/joao1 and https://www.linkedin.com/in/joao2.
unique_hrefs = ['https://www.linkedin.com/in/joao1','https://www.linkedin.com/in/joao2','https://www.linkedin.com/in/ACoAAAI3JyABlHv1LxXa27GHFneEbdrqAtMu9eY','https://www.linkedin.com/in/ACoAABWYG0kB8IXhFzDTCFGOwAZ18YbXprOLcmg']
regex = re.compile(r"\bhttps?://www.linkedin.com/in/\b[^in]+")
regex2 = re.compile(r"""\bhttps?://www\.linkedin\.com/in/ACo[^<>"'\s]*""")
filtered = [i for i in unique_hrefs if regex.search(i) and regex2.search(i)]
for i in filtered:
print(i)
Use
import re
unique_hrefs = ['https://www.linkedin.com/in/joao1','https://www.linkedin.com/in/joao2','https://www.linkedin.com/in/ACoAAAI3JyABlHv1LxXa27GHFneEbdrqAtMu9eY','https://www.linkedin.com/in/ACoAABWYG0kB8IXhFzDTCFGOwAZ18YbXprOLcmg']
pattern = re.compile(r'https?://www\.linkedin\.com/in/ACo')
results = list(filter(lambda x: not pattern.match(x), unique_hrefs))
print(results)
See Python证明.
Results: ['https://www.linkedin.com/in/joao1', 'https://www.linkedin.com/in/joao2'].
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