使用“else”np.nan公司'在列表中'

Question

我有一个网站列表，我只想提取社交媒体的个人资料（比如facebook、linkedin和pinterest）

import numpy as np

mylist = ['linkedin.com/profilexyz','facebook.com/profile374','bbcnews.com/USA_news','stackoverflow.com']

我使用列表理解来获取URL，如果找不到，则返回nan：

facebook = [x for x in mylist if 'facebook' in x else np.nan for x in mylist]
linkedin = [x for x in mylist if 'linkedin' in x else np.nan for x in mylist]
pinterest = [x for x in mylist if 'pinterest' in x else np.nan for x in mylist]

但是我得到了一个错误：

File "<ipython-input-329-578130619ae7>", line 1
facebook = [x for x in mylist if 'facebook' in x else np.nan for x in mylist]
                                                        ^
SyntaxError: invalid syntax

我已经检查了建议的副本，比如这个：if/else in a list comprehension?，但无法让我的理解发挥作用。你知道吗

Answer 1

你的订单搞砸了，还要给我加点钱

fb = [x if 'facebook' in x else np.nan for x in mylist]

Answer 2

只用括号：

facebook = [(x for x in mylist) if 'facebook' in x else np.nan for x in mylist]
linkedin = [(x for x in mylist) if 'linkedin' in x else np.nan for x in mylist]
pinterest = [(x for x in mylist) if 'pinterest' in x else np.nan for x in mylist]

Answer 3

这是一种使用collections.defaultdict和str.split的方法

例如：

from collections import defaultdict

result = defaultdict(list)
mylist = ['linkedin.com/profilexyz','facebook.com/profile374','bbcnews.com/USA_news','stackoverflow.com']

for url in mylist:
    result[url.split('/')[0]].append(url)
print(result)

输出：

defaultdict(<class 'list'>, {'linkedin.com': ['linkedin.com/profilexyz'], 'facebook.com': ['facebook.com/profile374'], 'bbcnews.com': ['bbcnews.com/USA_news'], 'stackoverflow.com': ['stackoverflow.com']})

---

以供参考

facebook = [x for x in mylist if 'facebook' in x]
print(facebook)