【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 
解构与封装
一、解构
In [1]: x = 1
In [2]: y = 2
In [3]: x, y = y, x # Python 的解构
In [4]: print(x, y)
2 1
In [5]: lst = [1, 2]
In [6]: first = lst[0]
In [7]: second = lst[1]
In [8]: first, second = lst
In [9]: print(first, second)
1 2
In [10]:
解构：按元素顺序，把线性结构的元素，赋值给变量
二、封装
In [10]: t = 1, 2
In [11]: t
Out[11]: (1, 2)
In [12]: type(t)
Out[12]: tuple
In [13]:
封装：定义一个元组，可以忽略小括号，封装出来的一定是元组
In [21]: t = 1, 2
In [22]: t
Out[22]: (1, 2)
In [23]: t1 = 1, 2
In [24]: t2 = (1, 2)
In [25]: t1 == t2
Out[25]: True
In [26]: t1 is t2
Out[26]: False
三、Python3 中的解构变化
In [1]: lst = list(range(1000))
In [2]: head, *mid, tail = lst # 通过使用星号 * 加上变量，可以接受 所有变量
In [3]: head
Out[3]: 0
In [4]: mid
Out[4]:
[1,
2,
3,
4,
5,
6
...
996,
997,
998]
In [5]: tail
Out[5]: 999
In [6]: *mid
File "<ipython-input-6-bad835e70ef0>", line 1
*mid
^
SyntaxError: can't use starred expression here
In [7]: head, *tail = lst # *tail 代表的是 除了第一个元素之外的 所有元素
In [8]: head
Out[8]: 0
In [9]: tail
Out[9]:
[1,
2,
3,
4,
5,
6,
...
996,
997,
998,
999]
In [10]:
In [10]: lst = [0, 1, 2, 3, 4]
In [11]: *head tail = lst # *head 代表除了最后一个元素外的所有元素
File "<ipython-input-11-d740de60225c>", line 1
*head tail = lst
^
SyntaxError: invalid syntax
In [12]: lst = [0, 1, 2, 3, 4]
In [13]: *head, tail = lst
In [14]: head
Out[14]: [0, 1, 2, 3]
In [15]: *lst2 = lst # 不能单一的使用星号作为变量接收， 如果可以，相当于lst[0:0]
File "<ipython-input-15-419bc512b767>", line 1
*lst2 = lst
^
SyntaxError: starred assignment target must be in a list or tuple
In [16]: head *m1, *m2, tail = lst # 同一个解构中，只能使用一个星号 *
File "<ipython-input-16-6a6d2cc4b0f7>", line 1
head *m1, *m2, tail = lst
^
SyntaxError: can't assign to operator
In [17]:
In [2]: lst = [0, 1, 2, 3, 4]
In [3]: *head, tail = lst
In [4]: head
Out[4]: [0, 1, 2, 3]
In [5]: first, second, *others, last = lst
In [6]: print(first, second, others, last)
0 1 [2, 3] 4
In [7]: v1, v2, v3, v4, v5, v6, v7 = lst # 当左边变量超过右边元素个数的时候，是不允许的
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-7-af83bc79cde9> in <module>()
----> 1 v1, v2, v3, v4, v5, v6, v7 = lst
ValueError: not enough values to unpack (expected 7, got 5)
In [8]: v1, v2 = lst # 数量不匹配
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-8-881bb3bee854> in <module>()
----> 1 v1, v2 = lst- 元素按照顺序赋值给变量
- 变量个数 和 元素个数必须匹配
- 加星号变量， 可以接受任意个数的元素
- 加星号的变量不能单独出现
In [9]: f, *t = (1, 2, 3, 4)
In [10]: f
Out[10]: 1
In [11]: t # 无论右边元素是什么类型， 左边的星号变量都是列表
Out[11]: [2, 3, 4]
In [16]: lst
Out[16]: [0, 1, 2, 3, 4]
In [17]: head, *_, tail = lst # 在Python中， 使用下划线_的方式 来丢弃该变量
In [18]: _
Out[18]: [1, 2, 3]
In [19]:
In [20]: for i, _ in enumerate(l):
...: print(i)
...:
0
1
In [21]:
单个下划线 是python 的合法标识符，但是如果不是要丢弃一个变量，通常不要用单个下划线表示一个有意义的变量
# 如何取出索引1， 3， 倒数第二个数
In [1]: lst = [1, 2, 3, 4, 5, 6, 7, 8, 9]
In [2]: _, v1, _, v2, *_, last, _ = lst
In [3]: v1
Out[3]: 2
In [4]: v2
Out[4]: 4
In [5]: last
Out[5]: 8
In [6]:
In [6]: lst = [1, (2, 3), 5]
In [7]: _, v, *_ = lst
In [8]: v
Out[8]: (2, 3)
In [9]: _, val = v
In [10]: val
Out[10]: 3
In [11]: _, (*_, tail), *_ = lst # 解构是支持多层次的
In [12]: tail
Out[12]: 3
In [13]: _, [*_, tail], *_ = lst
In [14]: tail
Out[14]: 3
集 合
一、定义
数学意义上的集合：没有重复元素
In [1]: s = set() # set 是一个无序的集合
In [2]: s
Out[2]: set()
In [3]: s = {1, 2, 3}
In [4]: s
Out[4]: {1, 2, 3}
In [5]: type(s)
Out[5]: set
In [6]: s = set(range(3))
In [7]: type(s)
Out[7]: set
二、增加
In [8]: s
Out[8]: {0, 1, 2}
In [9]: s.add(3) # 通过 add 的方法对集合进行添加元素
In [10]: s
Out[10]: {0, 1, 2, 3}
In [11]: help(s.update)
Help on built-in function update:
update(...) method of builtins.set instance
Update a set with the union of itself and others.
~
(END)
In [12]: s
Out[12]: {0, 1, 2, 3}
In [13]: s.update(range(8,10)) # update 方法 原地修改，返回None， update 方法是对可迭代对象进行操作
In [14]: s
Out[14]: {0, 1, 2, 3, 8, 9}
In [15]: s.update(range(4, 10))
In [16]: s
Out[16]: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
In [20]: for e in range(10):
...: s.add(e)
...: print(e)
...:
0
1
2
3
4
5
6
7
8
9
In [21]: 三、删除
1、s.remove()
In [22]: s
Out[22]: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
In [23]: s.remove(6)
In [24]: s
Out[24]: {0, 1, 2, 3, 4, 5, 7, 8, 9}
In [25]: s.remove(6)
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-25-077f15baad77> in <module>()
----> 1 s.remove(6)
KeyError: 6
In [26]: s.remove(999) # remove方法 删除一个set 中的元素；若元素不存在，则抛出KeyError
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-26-5cd7a055ca70> in <module>()
----> 1 s.remove(999)
KeyError: 999
In [27]:
2、s.pop()
In [27]: help(s.pop)
pop(...) method of builtins.set instance
Remove and return an arbitrary set element.
Raises KeyError if the set is empty.
~
(END)
In [2]: s = set(range(10))
In [3]: s
Out[3]: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
In [4]: s.pop()
Out[4]: 0
In [5]: s
Out[5]: {1, 2, 3, 4, 5, 6, 7, 8, 9}
In [6]: lst = [1, (2, 3), 5]
In [7]: lst.pop()
Out[7]: 5
In [8]: lst
Out[8]: [1, (2, 3)]
In [9]: lst.pop()
Out[9]: (2, 3)
In [10]: s = {'a', 1, 'b',-1}
In [11]: s
Out[11]: {-1, 1, 'a', 'b'}
In [12]: s.pop()
Out[12]: 'a'
In [13]: s.pop()
Out[13]: 1
In [14]: s.pop()
Out[14]: 'b'
In [15]: s.pop()
Out[15]: -1
In [16]: s.pop() # s.pop() 随机删除s中一个元素，若集合s为空，则抛出 KeyError
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-16-c88c8c48122b> in <module>()
----> 1 s.pop()
KeyError: 'pop from an empty set'
In [17]:
3、s.clear()
In [17]: s
Out[17]: set()
In [18]: s = {'a', 1, 'b',-1}
In [19]: s.clear()
In [20]: s
Out[20]: set()
In [21]:
4、s.discard()
In [21]: help(s.discard)
Help on built-in function discard:
discard(...) method of builtins.set instance
Remove an element from a set if it is a member.
If the element is not a member, do nothing.
~
(END)
In [23]: s = {1, 2}
In [24]: s.discard(1)
In [25]: s
Out[25]: {2}
In [26]: s.discard(2)
In [27]: s
Out[27]: set()
In [28]: s.discard(1)
In [29]: s.discard(999) # discard 并不会报错
In [30]:
- remove 删除给定的元素， 若元素不存在，抛出KeyError
- discard 删除给定的元素，若元素不存在，什么也不做
- pop 随机删除一个元素并返回，集合为空时，抛出KeyError
- clear 清空集合
四、修改
集合不能修改单个元素
五、查找
集合不能通过索引访问，因为集合不是一个线性结构
集合没有访问单个元素的方法
集合不是线性结构， 集合元素没有顺序
成员运算符
- in
- not in
判断一个元素，是否在容器中
In [30]: 3 in [1, 2, 3, 4]
Out[30]: True
In [31]: 10 in [1, 2, 3, 4]
Out[31]: False
In [32]: 10 not in [1, 2, 3, 4]
Out[32]: True
In [33]: 'a' in 'i love miracle'
Out[33]: True
In [34]: 'i' not in 'apple'
Out[34]: True
In [35]: 1 in 1 # 这里第二个 1 不是一个容器
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-35-ca0393ac6151> in <module>()
----> 1 1 in 1
TypeError: argument of type 'int' is not iterable
In [36]: b'b' in bytearray(b'abc')
Out[36]: True
In [37]:
集合的成员运算 和 其他线性结构的时间复杂度是不同的
In [41]: lst = list(range(1000000))
In [42]: s = set(range(1000000))
In [43]: %%timeit
...: -1 in lst
...:
8.61 ms ± 37.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [44]: %%timeit
...: -1 in s
...:
32.4 ns ± 0.219 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
In [45]:
能够使用set 做成员运算的时候，尽量使用集合，而不要使用list，tuple等线性结构。
- 集合的效率 和 集合的规模无关，O(1)
- 列表的效率 和 列表的规模成正比，O(n)
个人经验，集合可以用于去重
来源：oschina
链接：https://my.oschina.net/u/3552459/blog/1559554