一、字典的实现
dic 可以使用list来实现
i（索引） = hash(key) % solt(槽位数)
此时i重复了怎么办（hash冲突）？
1、拉链法
每个槽位上拉一个List，就是拉链法
In [6]: solts = [] # 初始化一个list
In [7]: solts_num = 32 # 假设32个槽位
In [8]: for _ in range(solts_num):
...: solts.append([])
...:
In [9]: solts # 每个槽位上都是一个list
Out[9]:
[[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
## 定义dict的put方法
In [10]: def put(solts, key, value):
...: i = hash(key) % solts_num
...: solts[i].append((key, value))
...:
## 定义dict的get方法
In [12]: def get(solts, key):
...: i = hash(key) % solts_num
...: for k, v in solts[i]:
...: if k == key:
...: return v
...: raise KeyError(key)
...:
## 测试
In [23]: put(solts, 'a', 1)
In [24]: solts
Out[24]:
[[('a', 1)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [25]: put(solts, 'b', 2)
In [26]: put(solts, 'f', 9)
In [27]: solts
Out[27]:
[[('a', 1)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[('f', 9)],
[],
[('b', 2)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [28]: get(solts, 'b')
Out[28]: 2
In [29]: get(solts, 'f')
Out[29]: 9
In [107]: get(solts, 'x')
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
in ()
----> 1 get(solts, 'x')
in get(solts, key)
4 if k == key:
5 return v
----> 6 raise KeyError(key)
7
KeyError: 'x'
此时put一个已存在的key时：
In [33]: put(solts, 'a', 11)
In [34]: solts
Out[34]:
[[('a', 1), ('a', 11)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[('f', 9)],
[],
[('b', 2)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [35]: get(solts, 'a')
Out[35]: 1
重新定put方法：
In [87]: def put(solts, key, value):
...: i = hash(key) % solts_num
...: for p, (k, v) in enumerate(solts[i]):
...: if k == key:
...: solts[i][p] = (key, value)
...: else:
...: solts[i].append((key, value))
In [89]: solts
Out[89]: []
In [90]: for _ in range(solts_num):
...: solts.append([])
...:
In [92]: put(solts, 'a', 1)
In [94]: put(solts, 'b', 2)
In [95]: put(solts, 'f', 8)
In [96]: solts
Out[96]:
[[('a', 1)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[('f', 8)],
[],
[('b', 2)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [101]: get(solts, 'b')
Out[101]: 2
In [102]: get(solts, 'f')
Out[102]: 8
In [103]: get(solts, 'a')
Out[103]: 11
In [104]: put(solts, 'b', 22)
In [105]: get(solts, 'b')
Out[105]: 22
In [106]: solts
Out[106]:
[[('a', 11), ('a', 11)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[('f', 8)],
[],
[('b', 22), ('b', 22)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [107]: get(solts, 'x')
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
in ()
----> 1 get(solts, 'x')
in get(solts, key)
4 if k == key:
5 return v
----> 6 raise KeyError(key)
7
KeyError: 'x'
set在底层的实现就是一个忽略了value的dict
2、开地址法
使用某个算法重新计算i，就交开地址法
常用，效率更高，
i = fn(key, i)
待后续学习