# 删除重复元素，保持顺序不变

1数组去重

代码：

```python
def dedupe(items):
    seen=set()
    for item in items:
        if item not in seen:
            yield item
            seen.add(item)
a=[5,5,2,1,9,1,5,10]
print(a)
print(list(dedupe(a)))
```

结果：

```
[5, 5, 2, 1, 9, 1, 5, 10]
[5, 2, 1, 9, 10]
```

## 以上只能对散列元素才能之么做，在对象序列中去除重复项，保持顺序不变？

```python
def buha(items,key=None):
    seen=set()
    for item in items:
#         print(seen)
        val=item if key is None else key(item)
        print(val)
        if val not in seen:
            yield item
            seen.add(val)
a=[{'x':2,'y':3},{'x':1,'y':4},
      {'x':2,'y':3},{'x':10,'y':15}]
print(a)
print(list(buha(a,key=lambda x:(x['x'],x['y']))))
```

```
[{'x': 2, 'y': 3}, {'x': 1, 'y': 4}, {'x': 2, 'y': 3}, {'x': 10, 'y': 15}]
(2, 3)
(1, 4)
(2, 3)
(10, 15)
[{'x': 2, 'y': 3}, {'x': 1, 'y': 4}, {'x': 10, 'y': 15}]
```

## 自定义

```python
class Foo:
    def __init__(self, name, count):
        self.name = name
        self.count = count

    def __hash__(self):
        # 当两个变量的哈希值不相同时，就认为这两个变量是不同的。
        print("%s调用了哈希方法" % self.name)
        return hash(self.count)

    def __eq__(self, other):
        # 当两个变量哈希值一样时，调用__eq__方法，比较是否具有相同的各种属性，可以简单地理解为值是否相等
        # self是存在的一方对象，other是新来的和存在的冲突的对象
        print("%s调用了eq方法" % self.name)
        if self.count == other.count:
#         if self.__dict__ == other.__dict__:
            return True
        else:
            return False

    def __repr__(self, ):
        return 'print:%s' % (self.name)


f1 = Foo('f1', 1)
f2 = Foo('f2', 2)
f3 = Foo('f3', 3)
f4 = Foo('f4', 3)
f5 = Foo('f5', 3)
f6 = Foo('f6', 3)
ms = [f1, f2, f3, f4, f5,f6]
print(set(ms))
```

```
f1调用了哈希方法
f2调用了哈希方法
f3调用了哈希方法
f4调用了哈希方法
f3调用了eq方法
f5调用了哈希方法
f3调用了eq方法
f6调用了哈希方法
f3调用了eq方法
{print:f1, print:f2, print:f3}
```


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