python 消除序列的重復值，并保持原來順序

1、如果僅僅消除重復元素，可以簡單的構造一個集合

$ pythonPython 3.5.2 (default, Nov 23 2017, 16:37:01) [GCC 5.4.0 20160609] on linuxType "help", "copyright", "credits" or "license" for more information.>>> a = [1 , 3, 5, 1, 8, 1, 5]>>> set(a){8, 1, 3, 5}>>> 

2、利用集合或者生成器解決：值必須是hashable類型

$ pythonPython 3.5.2 (default, Nov 23 2017, 16:37:01) [GCC 5.4.0 20160609] on linuxType "help", "copyright", "credits" or "license" for more information.>>> def dupe(items):... seen = set()... for item in items:... if item not in seen:... yield item... seen.add(item)... >>> a = [1 , 3, 5, 1, 8, 1, 5]>>> list(dupe(a))[1, 3, 5, 8]>>>

3、消除元素不可哈希：如字典類型

Python 3.5.2 (default, Nov 23 2017, 16:37:01) [GCC 5.4.0 20160609] on linuxType "help", "copyright", "credits" or "license" for more information.>>> def rem(items, key=None):... seen = set()... for item in items:... va = item if key is None else key(item)... if va not in seen:... yield item... seen.add(va)... >>> a = [ {'x':1, 'y':2}, {'x':1, 'y':3}, {'x':1, 'y':2}, {'x':2, 'y':4}]>>> list(rem(a, key=lambda d: (d['x'],d['y'])))[{'y': 2, 'x': 1}, {'y': 3, 'x': 1}, {'y': 4, 'x': 2}]>>> list(rem(a, key=lambda d: d['x']))[{'y': 2, 'x': 1}, {'y': 4, 'x': 2}]>>>>>> #lambda is an anonymous function:... fuc = lambda : 'haha'>>> print (f())>>> print (fuc())haha>>> 

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