摘要：聲明一個很簡單，只需盡量不要將變量名起為關鍵字。有如下基本方法在的末尾添加一個元素，并且返回將另一個的對象添加到尾部，返回值為。返回值為刪掉的元素。為一個表達式，傳入當前元素，返回時依據德關鍵字。

聲明一個list很簡單，只需list1=[]（盡量不要將變量名起為關鍵字list）。
list有如下基本方法：

(1)append(x) 在list的末尾添加一個元素x，并且返回None

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

(2)extend(iterable) 將另一個iterable的對象添加到list尾部，返回值為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) 將元素x插入到索引i處，返回值為None。

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

(4)remove(x) 刪除值為x的元素（值而非索引），刪除成功返回None，若沒有x則報錯

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 
    r = list1.remove(5)
ValueError: list.remove(x): x not in list

(5)pop([i]) 如果傳入參數i，刪除索引i處元素。如果不傳參，刪除尾部的元素。返回值為刪掉的元素。

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

#[1, 2, 3] 4

(6)clear() 清空數組所有元素，返回None。相當于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中出現的次數

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

#3

(8)sort(key=None, reverse=False) 默認從小到大，如果reverse設為True則為從大到小。key為一個lambda表達式，傳入當前元素，返回sort時依據德關鍵字。

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

(9)reverse() 反轉數組

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

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

(10)copy() 淺拷貝數組，相當于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"}]

以上為list常用方法。
list可以作為棧使用。

>>> 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]
（官網代碼）

也可以作為隊列使用，但是需要導入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的小竅門：

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]]（官網案例）

del語句：del arr[i]刪除索引為i的元素，del arr[:]為清空數組,del arr[i:j]為清空i到j之前的元素，包括i不包括j。

遍歷方式

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)

tuple1 = () 此時tuple1為tuple類型
tuple2 = (1) 此時tuple2非tuple類型，而是int類型
tuple3 = 1, 或者 tuple3 = (1,) 此時tuple3為tuple類型

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)

#



((1, 2, 3, 4), 5)

tuple可以使用切片，但是tuple不可變，不能給其中元素重新賦值。

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 
    tuple1[2] = 5
TypeError: "tuple" object does not support item assignment

可以通過以下代碼實現兩個變量值的互換

x = 5
y = 6

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

#6
5

執行set1={}這種語句，set1的類型為dict而非set

set1 = {}
print(type(set1))

#

當用set構造函數創建set時要注意以下的問題：

set1 = set("abcedfg")
print(set1)

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

set是無序的，并且其中沒有重復元素，因此可以通過set實現去重。

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

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

set可以進行求交集，并集，差集，以及找出只存在于其中一個的元素集合

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"}

初始化dict，需要將key與value都加上引號：

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

獲取可采用更簡單的方法

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

遍歷dict時，有如下選擇

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)

運算：

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]