Python 列表去重怎么用暴力枚举法写出

set(x)

一楼正解，此贴终结

list(set(x))

赞同二楼说的一楼正解，此贴终结。

赞同三楼说的赞同二楼说的一楼正解，此贴终结。

如果要求保持顺序，list({i:None for i in x}.keys())

忘记说了，python3

@conn4575 python3.5 也是无序的，3.6 是有序的，但是不要依赖这一实现

@conn4575 3.6 以上 保证顺序

@0xABCD 3.6 以上的 cpython 保证顺序

@ericls 请问你们是指 3.6 的 set()有序还算 list(set())有序？

set 就够了

@so1n 指的是 Dict

@conn4575 这里不能保持有序吧，生成 dict 时失序了，再调用 keys 仍然无序

@xinhangliu 3.6 的 dict 变成有序的吗

楼主是否看过 python cookbook ？

If the values in the sequence are hashable, the problem can be easily solved using a set and a generator. For example:

def dedupe(items):
seen = set()
for item in items:
if item not in seen:
yield item seen.add(item)

Here is an example of how to use your function:
>>> a = [1, 5, 2, 1, 9, 1, 5, 10]
>>> list(dedupe(a))
[1, 5, 2, 9, 10]
>>>

This only works if the items in the sequence are hashable. If you are trying to eliminate duplicates in a sequence of unhashable types (such as dicts), you can make a slight change to this recipe, as follows:

def dedupe(items, key=None):
seen = set()
for item in items:
val = item if key is None else key(item)
if val not in seen:
yield item
seen.add(val)

Here, the purpose of the key argument is to specify a function that converts sequence items into a hashable type for the purposes of duplicate detection. Here ’ s how it works:
>>> a = [ {'x':1, 'y':2}, {'x':1, 'y':3}, {'x':1, 'y':2}, {'x':2, 'y':4}]
>>> list(dedupe(a, key=lambda d: (d['x'],d['y'])))
[{'x': 1, 'y': 2}, {'x': 1, 'y': 3}, {'x': 2, 'y': 4}]
>>> list(dedupe(a, key=lambda d: d['x']))
[{'x': 1, 'y': 2}, {'x': 2, 'y': 4}]
>>>
This latter solution also works nicely if you want to eliminate duplicates based on the value of a single field or attribute or a larger data structure.

res = []
lists = = [res.append(i) for i in target if i not in res]

我忘了在哪看过的一个以前没见过的操作,去重保留顺序
ccc = [1, 3, 5, 8, 9, 3, 5, 9, 10]
res = sorted(set(ccc), key=ccc.index)
print res
=>[1, 3, 5, 8, 9, 10]

python sorteddict 的实现已经默认了，原因是这个涉及到 dataclass 和 NamedTuple 的实现。
可以依赖 sorted dict 了