Python列表中的唯一项

Question

我正在尝试在Python列表中创建一个独特的日期集合。

只有在集合中不存在该日期时，才将其添加到集合中。

timestamps = []

timestamps = [
    '2011-02-22', '2011-02-05', '2011-02-04', '2010-12-14', '2010-12-13', 
    '2010-12-12', '2010-12-11', '2010-12-07', '2010-12-02', '2010-11-30', 
    '2010-11-26', '2010-11-23', '2010-11-22', '2010-11-16']

date = "2010-11-22"
if date not in timestamps:
    timestamps.append(date)

我该如何对列表进行排序？

- Rajeev

Answer 1

你可以使用集合来实现这个功能。

date = "2010-11-22"
timestamps = set(['2011-02-22', '2011-02-05', '2011-02-04', '2010-12-14', '2010-12-13', '2010-12-12', '2010-12-11', '2010-12-07', '2010-12-02', '2010-11-30', '2010-11-26', '2010-11-23', '2010-11-22', '2010-11-16'])
#then you can just update it like so
timestamps.update(['2010-11-16']) #if its in there it does nothing
timestamps.update(['2010-12-30']) # it does add it

- Jakob Bowyer

Answer 2

这段代码实际上并没有起到任何作用。你在两次引用同一个变量(timestamps)。

因此，你需要创建两个单独的列表：

unique_timestamps= []

timestamps = ['2011-02-22', '2011-02-05', '2011-02-04', '2010-12-14', '2010-12-13', '2010-12-12', '2010-12-11', '2010-12-07', '2010-12-02', '2010-11-30', '2010-11-26', '2010-11-23', '2010-11-22', '2010-11-16']

date="2010-11-22"
if(date not in timestamps):
   unique_timestamps.append(date)

- Ikke

Answer 3

你的条件似乎是正确的。但如果你不关心日期的顺序，使用集合(set)可能比使用列表(list)更容易。在这种情况下，您不需要任何if语句:

timestamps = set(['2011-02-22', '2011-02-05', '2011-02-04', '2010-12-14', 
                  '2010-12-13', '2010-12-12', '2010-12-11', '2010-12-07',
                  '2010-12-02', '2010-11-30', '2010-11-26', '2010-11-23',
                  '2010-11-22', '2010-11-16'])
timesteps.add("2010-11-22")

- Sven Marnach