Question

我有一个名为times的字典，其中包含一个datetime时间戳作为键，而元组作为值。格式示例如下：

{datetime.datetime(2019, 11, 4, 20, 2): ('A', 'B'),
datetime.datetime(2019, 11, 4, 19, 59): ('C', 'D'), 
datetime.datetime(2019, 11, 4, 19, 55): ('E', 'F'), …, }

我正在尝试按升序对列表中的时间戳进行排序，循环遍历列表以计算连续时间戳之间的差异，如果差异大于10分钟阈值，则将开始时间存储在新字典中（并以相应的元组作为值）。

到目前为止，这是我要编写的代码。我相信我需要先在列表time_with_breaks中存储时间戳之间的差异，然后使用大于字典中阈值的if语句存储增量，但我不确定如何执行此操作。

deltas = {} # store timestamp and seconds until next entry
time_with_breaks = [] # store timing information of breaks
# sorted list of time stamps (ascending order)
timelist = sorted(playlist_dict.keys(), reverse=False)
for i in timelist:

我该怎么做？在此先感谢您的帮助。

Answer 1

以下应该可以工作

# This will return an ordered list of the timestamps
times_sorted = list(sorted(playlist_dict.keys()))
time_with_breaks = []
# "zipping" the list with itself offset by 1 will give you an iterable
# of each timestamp and the next timestamp in the list
for time_1, time_2 in zip(times_sorted, times_sorted[1:]):
    # datetime - datetime results in a timedelta
    # 600 seconds is 10 minutes
    if (time_2 - time_1).seconds > 600:
        # Append the datetime where the next datetime in the list is more than
        # 10 minutes away
        time_with_breaks.append(time_1)

计算时间戳之间的差异并将其存储在字典中

1 个答案: