一、{}.fromkeys(list).keys()
list1 = [1,4,3,3,4,2,3,4,5,6,1]
list2 = {}.fromkeys(list1).keys() # [1, 2, 3, 4, 5, 6]
二、set
list1 = [1,4,3,3,4,2,3,4,5,6,1] list2 = list(set(list1)) # [1, 2, 3, 4, 5, 6]
三、itertools.grouby
import itertools
ids = [1,4,3,3,4,2,3,4,5,6,1]
ids.sort()
it = itertools.groupby(ids)
for k, g in it:
print k
四、笨方法
ids = [1,2,3,3,4,2,3,4,5,6,1]
news_ids = []
for id in ids:
if id not in news_ids:
news_ids.append(id)
print news_ids
这四种方法都有个特点，去重后元素排序变了，效率的话：据说第一种比第二种快一点。
五、索引再次排序-这种方法可以去重并且保持元素顺序
#要结果是[1, 4, 3, 2, 5, 6] 不要[1, 2, 3, 4, 5, 6]
ids = [1,4,3,3,4,2,3,4,5,6,1] news_ids = list(set(ids)) news_ids.sort(key=ids.index) print news_ids #[1, 4, 3, 2, 5, 6]
六、Reduce
ids = [1,4,3,3,4,2,3,4,5,6,1] func = lambda x,y:x if y in x else x + [y] print reduce(func, [[], ] + ids) #[1, 4, 3, 2, 5, 6]