Question

考虑以下代码

', '.join(['{} <{}> ({})'.format(user.nickname, user.email, user.employee_id)
                     for user in referrers])

现在，当employee_id不存在时，我想跳过()并清空空格。

referrer_strs = []
for user in referrers:
    if user.employee_id is None:
        referrer_strs.append('{} <{}>'.format(user.nickname, user.email))
    else:
        referrer_strs.append('{} <{}> ({})'.format(user.nickname, user.email, user.employee_id))

return ', '.join(referrer_strs)

我可以将它转换为理解，但有更多的pythonic方式吗？

Answer 1

现在代码更有趣= P

如果你想只用一行做所有事情，你可以使用“伪三元运算符”

class User:
    def __init__(self, nickname, email, employee_id=None):
        self.nickname = nickname
        self.email = email
        self.employee_id = employee_id

referrers = [
    User('macabeus', 'macabeus@gmail.com', 1),
    User('joão', 'joao@gmail.com'),
    User('josé', 'jose@gmail.com', 3)
]

print(
    ', '.join(
        ['{} <{}>{}'.format(user.nickname, user.email, ('', ' ({})'.format(user.employee_id))[user.employee_id is not None])
                     for user in referrers]
    )
)

Answer 2

使用此conditional list indices removal SO post，尝试添加if逻辑：

', '.join(['{} <{}> ({})'.format(user.nickname, user.email, user.employee_id)
                     for user in referrers if not (user.employee_id is None)])

有条件地格式化字符串

2 个答案: