Question

有人请告诉我有更简洁的方式来写这个！？我想删除所有＆＃39;功能＆amp; amp;优点＆＃39 ;.必须有更简洁的方法为列表中的字符串执行此操作。

lov_headers = [x.find('displayname').text for x in soup.find_all('attributedefinition')]


if 'Part number' in lov_headers:
    lov_headers.remove('Part number')
if 'Product Features' in lov_headers:
    lov_headers.remove('Product Features')
if 'Packaging' in lov_headers:
    lov_headers.remove('Packaging')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')
if 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')

Answer 1

将要删除的内容放入set()，然后使用列表推导过滤掉您不想要的结果：

to_remove = set(['Part number', 'Features & Benefits', 'Product Features', 'Packaging'])

new_list = [i for i in lov_headers if i not in to_remove]

Answer 2

您可以使用while循环：

while 'Features & Benefits' in lov_headers:
    lov_headers.remove('Features & Benefits')

从列表中删除所有出现的字符串

2 个答案: