如何在python中使用一个print()来逐行打印列表项 - 而不是在一个条纹中？

Question

mister是一个列表列表.print(mister)在Python shell中给出了这个:

[['ququ.kz', 1], ['gp.kz', 1], ['gmail.ru', 1], ['mail.ru', 1], ['tlc.com', 1], ['mail.ko', 1], ['microsoft.jp', 1], ['hotmail.eu', 1], ['soman.com', 1], ['swedenborgen.sn', 1], ['customergoogle.com', 1], ['mail.jp', 2], ['gmail.com', 3], ['mail.ru', 3], ['hotmail.com', 3], ['mail.jp', 3], ['mail.com', 4], ['hotmail.com', 4], ['gmail.com', 4], ['mail.kz', 5], ['mail.cn', 7], ['hotmail.com', 9], ['customers.kz', 9], ['microsoft.com', 10], ['conestogamall.com', 13]]

我可以使用print()一次并获得不错的输出 - 逐行,而不是一个大条纹？或者是唯一的解决方案:

for email_date_entry in mister:
    print(email_date_entry)

有另一种优雅的方式只print使用一个电话吗？

Answer 1

您可以将列表作为单独的参数传递给print()使用*变量参数语法:

print(*mister, sep='\n')

现在,每个元素mister都被视为一个单独的参数,并使用\n分隔符打印:

>>> mister = [['ququ.kz', 1], ['gp.kz', 1], ['gmail.ru', 1], ['mail.ru', 1], ['tlc.com', 1], ['mail.ko', 1], ['microsoft.jp', 1], ['hotmail.eu', 1], ['soman.com', 1], ['swedenborgen.sn', 1], ['customergoogle.com', 1], ['mail.jp', 2], ['gmail.com', 3], ['mail.ru', 3], ['hotmail.com', 3], ['mail.jp', 3], ['mail.com', 4], ['hotmail.com', 4], ['gmail.com', 4], ['mail.kz', 5], ['mail.cn', 7], ['hotmail.com', 9], ['customers.kz', 9], ['microsoft.com', 10], ['conestogamall.com', 13]]
>>> print(*mister, sep='\n')
['ququ.kz', 1]
['gp.kz', 1]
['gmail.ru', 1]
['mail.ru', 1]
['tlc.com', 1]
['mail.ko', 1]
['microsoft.jp', 1]
['hotmail.eu', 1]
['soman.com', 1]
['swedenborgen.sn', 1]
['customergoogle.com', 1]
['mail.jp', 2]
['gmail.com', 3]
['mail.ru', 3]
['hotmail.com', 3]
['mail.jp', 3]
['mail.com', 4]
['hotmail.com', 4]
['gmail.com', 4]
['mail.kz', 5]
['mail.cn', 7]
['hotmail.com', 9]
['customers.kz', 9]
['microsoft.com', 10]
['conestogamall.com', 13]