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>>> 以上這篇python消除序列的重復值并保持順序不變的實例就是小編分享給大家的全部內容了，希望能給大家一個參考，也希望大家多多支持VEVB武林網。
新聞熱點
疑難解答
