(编辑：jimmy 日期: 2024/11/17 浏览：2)
python 消除序列的重复值，并保持原来顺序
1、如果仅仅消除重复元素，可以简单的构造一个集合
$ python
Python 3.5.2 (default, Nov 23 2017, 16:37:01)
[GCC 5.4.0 20160609] on linux
Type "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决：值必须是hashable类型
$ python Python 3.5.2 (default, Nov 23 2017, 16:37:01) [GCC 5.4.0 20160609] on linux Type "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：如字典类型
Python 3.5.2 (default, Nov 23 2017, 16:37:01)
[GCC 5.4.0 20160609] on linux
Type "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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