python列表中的唯一項目

Question

我試圖在Python列表中創建一個獨特的日期集合。

僅在集合中尚未存在日期時才添加日期。

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)

我該如何排序？

Answer 1

你可以使用套裝。

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

Answer 2

這段代碼實際上什么都不做。 您正在引用相同的變量兩次（ timestamps ）。

所以你必須制作兩個單獨的列表：

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)

Answer 3

你的病情似乎是正確的。 如果您不關心日期的順序，則使用集合而不是列表可能更容易。 你不需要任何if在這種情況下：

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")