python - 在换行符处拆分python中的字符串

Question

我有一个包含一系列时间和活动的元组。因此，它将类似于：

('08:01: Woke Up\n08:05: Took Shower\n08:20: Ate Breakfast\n08:45: Left for work', )

我需要能够像这样进行每个活动并单独显示：

8:01 Woke Up
8:05 Took Shower
8:20 Ate Breakfast
8:45 Left for work

任何人都可以给我一个关于在 python 中执行此操作的最佳/最简单方法的建议吗？谢谢你。

Answer 1

您已经在元素中有换行符，所以只需打印：

>>> tup = ('08:01: Woke Up\n08:05: Took Shower\n08:20: Ate Breakfast\n08:45: Left for work',)
>>> print tup[0]
08:01: Woke Up
08:05: Took Shower
08:20: Ate Breakfast
08:45: Left for work

您可以使用以下方法将字符串拆分为单独的元素str.splitlines()：

>>> tup[0].splitlines()
['08:01: Woke Up', '08:05: Took Shower', '08:20: Ate Breakfast', '08:45: Left for work']
>>> for line in tup[0].splitlines():
...     print line
... 
08:01: Woke Up
08:05: Took Shower
08:20: Ate Breakfast
08:45: Left for work

如果您需要将时间和活动作为单独的字符串，请拆分字符串上的每一行': '：

>>> for line in tup[0].splitlines():
...     time, activity = line.split(': ', 1)
...     print time, activity
... 
08:01 Woke Up
08:05 Took Shower
08:20 Ate Breakfast
08:45 Left for work

Answer 2

>>> t = ('08:01: Woke Up\n08:05: Took Shower\n08:20: Ate Breakfast\n08:45: Left for work', )
>>> for elem in t:
...     print elem.split('\n')
...
['08:01: Woke Up', '08:05: Took Shower', '08:20: Ate Breakfast', '08:45: Left fo
r work']
>>> for elem in t:
...     for activity in elem.split('\n'):
...             print activity
...
08:01: Woke Up
08:05: Took Shower
08:20: Ate Breakfast
08:45: Left for work
>>> for elem in t:
...     print elem
...
08:01: Woke Up
08:05: Took Shower
08:20: Ate Breakfast
08:45: Left for work

Answer 3

你似乎没有元组，你有一个字符串，如果你打印，它会给你你想要的东西。

假设您有一个元组列表（我认为这就是您的意思），您应该执行以下操作：

input = [("8:01", "Woke Up"), ("8:05", "Took shower"), ("8:20", "Ate Breakfast")]
for time, description in input:
  print time, description

Answer 4

你有没有想过字典？您可以将时间信息存储为键，将活动存储为值，如下所示：

dictionary = {}
dictionary['8:01'] = 'Woke Up'
dictionary['8:05'] = 'Took Shower'
dictionary['8:20'] = 'Ate Breakfast'
dictionary['8:45'] = 'Left for work'

然后以您可以简单地执行的格式打印：

for k in dictionary:
    print k + ' ' + dictionary[k]