1.list
宣告一個list很簡單，只需list1=[]（儘量不要將變數名起為關鍵字list）。
list有如下基本方法：
(1)append(x) 在list的末尾新增一個元素x，並且返回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的物件新增到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 <module>
r = list1.remove(5)
ValueError: list.remove(x): x not in list(5)pop([i]) 如果傳入引數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() 清空陣列所有元素，返回None。相當於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) 預設從小到大，如果reverse設為True則為從大到小。key為一個lambda表示式，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() 反轉陣列
list1 = [1, 2, 3, 4, 5]
list1.reverse()
print(list1)
#[5, 4, 3, 2, 1](10)copy() 淺拷貝陣列，相當於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]
（官網程式碼）也可以作為佇列使用，但是需要匯入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[:]為清空陣列,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)2.tuple
tuple1 = () 此時tuple1為tuple型別
tuple2 = (1) 此時tuple2非tuple型別，而是int型別
tuple3 = 1, 或者 tuple3 = (1,) 此時tuple3為tuple型別
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 <module>
tuple1[2] = 5
TypeError: `tuple` object does not support item assignment可以通過以下程式碼實現兩個變數值的互換
x = 5
y = 6
x, y = y, x
print(x)
print(y)
#6
53.set
執行set1={}這種語句，set1的型別為dict而非set
set1 = {}
print(type(set1))
#<class `dict`>當用set建構函式建立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，並且其中沒有重複元素，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，並集，差集，以及找出只存在於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`}
4.dict
初始化dict，需要將key與value都加上引號：
dict1 = {`name`: `Jero`, `age`: 20}獲取可採用更簡單的方法
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]