python中的list，tuple，set和dict（参考python文档）

1.list

声明一个list很简单，只需list1=[]（尽可能不要将变量名起为关键字list）。
list有以下基本方法：

(1)append(x) 在list的末尾添加一个元素x，而且返回None

list1 = [1, 2, 3, 4]
r = list1.append(5)
print(list1, r)
#[1, 2, 3, 4, 5] None

(2)extend(iterable) 将另外一个iterable的对象添加到list尾部，返回值为None。

list1 = [1, 2, 3, 4]
r = list1.extend([5, 6, 7, 8])
print(list1, r)
#[1, 2, 3, 4, 5, 6, 7, 8] None

(3)insert(i,x) 将元素x插入到索引i处，返回值为None。

list1 = [1, 2, 3, 4]
r = list1.insert(4, 5)
print(list1, r)
#[1, 2, 3, 4, 5] None

(4)remove(x) 删除值为x的元素（值而非索引），删除成功返回None，若没有x则报错

list1 = [1, 2, 3, 4]
r = list1.remove(3)
print(list1, r)
#[1, 2, 4] None


list1 = [1, 2, 3, 4]
r = list1.remove(5)
print(list1, r)


Traceback (most recent call last):
  File "E:/Programs/python/data-structure/list-demo.py", line 2, in <module>
    r = list1.remove(5)
ValueError: list.remove(x): x not in list

(5)pop([i]) 若是传入参数i，删除索引i处元素。若是不传参，删除尾部的元素。返回值为删掉的元素。

list1 = [1, 2, 3, 4]
r = list1.pop(3)
print(list1,r)

#[1, 2, 3] 4

(6)clear() 清空数组全部元素，返回None。至关于list1[:]=[]以及del list1[:]

list1 = [1, 2, 3, 4]
r = list1.clear()
print(list1, r)

#[] None

(7)count(x) 返回x在list中出现的次数

list1 = [1, 2, 3, 4, 1, 2, 3, 4, 2, 4]
r = list1.count(4)
print(r)

#3

(8)sort(key=None, reverse=False) 默认从小到大，若是reverse设为True则为从大到小。key为一个lambda表达式，传入当前元素，返回sort时依据德关键字。

list1 = [{'age': 20}, {'age': 15}, {'age': 18}, {'age': 25}]
list1.sort(key=lambda stu: stu['age'])
print(list1)

(9)reverse() 反转数组

list1 = [1, 2, 3, 4, 5]
list1.reverse()
print(list1)

#[5, 4, 3, 2, 1]

(10)copy() 浅拷贝数组，至关于list1[:]

list1 = [1, 2, 3, 4, {'name': 'John'}]
list2 = list1.copy()
list2[4]['name'] = 'Mike'
print(list1 is list2)
print(list1)
print(list2)

#
False
[1, 2, 3, 4, {'name': 'Mike'}]
[1, 2, 3, 4, {'name': 'Mike'}]

以上为list经常使用方法。
list能够做为栈使用。

>>> stack = [3, 4, 5]
>>> stack.append(6)
>>> stack.append(7)
>>> stack
[3, 4, 5, 6, 7]
>>> stack.pop()
7
>>> stack
[3, 4, 5, 6]
>>> stack.pop()
6
>>> stack.pop()
5
>>> stack
[3, 4]
（官网代码）

也能够做为队列使用，可是须要导入collections包的deque

>>> from collections import deque
>>> queue = deque(["Eric", "John", "Michael"])
>>> queue.append("Terry")           # Terry arrives
>>> queue.append("Graham")          # Graham arrives
>>> queue.popleft()                 # The first to arrive now leaves
'Eric'
>>> queue.popleft()                 # The second to arrive now leaves
'John'
>>> queue                           # Remaining queue in order of arrival
deque(['Michael', 'Terry', 'Graham'])

初始化list的小窍门：

list1 = [x ** 2 for x in range(10)]
print(list1)
#[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

list1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
list2 = [num for sub in list1 for num in sub]
print(list1)
print(list2)
#[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
[1, 2, 3, 4, 5, 6, 7, 8, 9]

list1 = [(x, y) for x in range(1, 4) for y in range(x, 4)]
print(list1)
#[(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]

matrix = [[1, 2, 4, 5], [5, 6, 7, 8], [9, 10, 11, 12]]
r = [[row[i] for row in matrix] for i in range(4)]
print(r)
#[[1, 5, 9], [2, 6, 10], [4, 7, 11], [5, 8, 12]]（官网案例）

del语句：del arr[i]删除索引为i的元素，del arr[:]为清空数组,del arr[i:j]为清空i到j以前的元素，包括i不包括j。

遍历方式

list1 = [1, 2, 3, 4, 5, 6, 7, 8]

for i in range(len(list1)):
    print(i, list1[i])

for i, val in enumerate(list1):
    print(i, val)

2.tuple

tuple1 = () 此时tuple1为tuple类型
tuple2 = (1) 此时tuple2非tuple类型，而是int类型
tuple3 = 1, 或者 tuple3 = (1,) 此时tuple3为tuple类型

tuple1 = ()
tuple2 = (1)
tuple3 = 1,
tuple4 = 1, 2, 3, 4
tuple5 = tuple4, 5
print(type(tuple1))
print(type(tuple2))
print(type(tuple3))
print(type(tuple4))
print(tuple5)

#<class 'tuple'>
<class 'int'>
<class 'tuple'>
<class 'tuple'>
((1, 2, 3, 4), 5)

tuple可使用切片，可是tuple不可变，不能给其中元素从新赋值。

tuple1 = 1, 2, 3, 4, 5, 6
tuple1[2] = 5

#Traceback (most recent call last):
  File "E:/Programs/python/data-structure/list-demo.py", line 2, in <module>
    tuple1[2] = 5
TypeError: 'tuple' object does not support item assignment

能够经过如下代码实现两个变量值的互换

x = 5
y = 6

x, y = y, x
print(x)
print(y)

#6
5

3.set

执行set1={}这种语句，set1的类型为dict而非set

set1 = {}
print(type(set1))

#<class 'dict'>

当用set构造函数建立set时要注意如下的问题：

set1 = set('abcedfg')
print(set1)

#{'f', 'c', 'g', 'e', 'b', 'a', 'd'}

set是无序的，而且其中没有重复元素，所以能够经过set实现去重。

list1 = [1, 1, 2, 3, 4, 4, 5, 6, 6, 7]
print(list(set(list1)))

#[1, 2, 3, 4, 5, 6, 7]

set能够进行求交集，并集，差集，以及找出只存在于其中一个的元素集合

a = set('abracadabra')
b = set('alacazam')

print(a)
print(b)

print(a & b)
print(a | b)
print(a - b)
print(a ^ b)

#
{'r', 'b', 'a', 'c', 'd'}
{'z', 'a', 'm', 'c', 'l'}
{'c', 'a'}
{'z', 'r', 'b', 'a', 'm', 'c', 'd', 'l'}
{'r', 'b', 'd'}
{'z', 'm', 'r', 'b', 'd', 'l'}

4.dict

初始化dict，须要将key与value都加上引号：

dict1 = {'name': 'Jero', 'age': 20}

获取可采用更简单的方法

dict2 = dict(name='Jero', age=20)

遍历dict时，有以下选择

dict1 = {'name': 'Jero', 'age': 20, 'gender': 'male'}

for i in dict1:
    print(i, dict1[i])
#name Jero
age 20
gender male


for val in dict1.values():
    print(val)
#Jero
20
male


for key, val in dict1.items():
    print(key, val)

运算：

list1 = [1, 2, 3]
list2 = [4, 5, 6]

print(list1 * 2)

list1 += list2

print(list1)

#[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6]