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python-study-list

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inp = input("请输入你的名字")

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请输入你的名字李思涵

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print(inp)

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a
fa
asfasf

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李思涵

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inp = input("数字")

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数字3

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print(inp)

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3

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print(inp+"无语")

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3无语

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inp = int(input("再来"))
print(inp+"无语")

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再来6



---------------------------------------------------------------------------

TypeError                                 Traceback (most recent call last)

~\AppData\Local\Temp\ipykernel_13512\1933184884.py in <module>
      1 inp = int(input("再来"))
----> 2 print(inp+"无语")


TypeError: unsupported operand type(s) for +: 'int' and 'str'

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print(inp+3)

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9

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

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list[0]

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1

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list[:]

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

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list[0:6:2]

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[1, 3, 5]

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list[0:6:-1]

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[]

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list[0:6:3]

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[1, 4]

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list[-1]

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6

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list[0:6:-2] 

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[]

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list[::-2]

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[6, 4, 2]

增删改查

append在末尾添加一个新元素

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heros=['钢铁侠','绿巨人']

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heros.append('黑寡妇')

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heros

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['钢铁侠', '绿巨人', '黑寡妇']

extend在末尾添加几个元素

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heros.extend(["鹰眼","灭霸","雷神"])

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heros

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['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '灭霸', '雷神']

可以使用:操作(切片)来代替

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s=[1,2,3,4,5]

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s[len(s):]=[6]

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s

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

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s[len(s):]=[7,8,9]

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s

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[1, 2, 3, 4, 5, 6, 7, 8, 9]

insert任意位置插入元素

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s=[1,3,4,5]

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s.insert(1,2)

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s

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[1, 2, 3, 4, 5]

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s.insert(len(s),6)

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s

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

remove删除指定元素(内容)

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heros.remove("灭霸")

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heros

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['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']

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heros.remove("金莲")

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---------------------------------------------------------------------------

ValueError                                Traceback (most recent call last)

~\AppData\Local\Temp\ipykernel_4012\2787210069.py in <module>
----> 1 heros.remove("金莲")


ValueError: list.remove(x): x not in list

pop删除指定(位置)元素

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heros.pop(2)

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'黑寡妇'

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heros

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['钢铁侠', '绿巨人', '鹰眼', '雷神']

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heros.clear()

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heros

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[]

替换实际上是两步操作先删除，再插入

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heros=['蜘蛛侠', '绿巨人', '黑寡妇', '鹰眼', '灭霸', '雷神']

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heros[4]="钢铁侠"

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heros

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['蜘蛛侠', '绿巨人', '黑寡妇', '鹰眼', '钢铁侠', '雷神']

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heros[3:]=["武松","林冲","李逵"]

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heros

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['蜘蛛侠', '绿巨人', '黑寡妇', '武松', '林冲', '李逵']

sort排序

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nums=[3,1,9,6,8,3,5,3]

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nums.sort()

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nums

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[1, 3, 3, 3, 5, 6, 8, 9]

reverse排序

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nums.reverse()

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nums

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[9, 8, 6, 5, 3, 3, 3, 1]

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heros.reverse()

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heros

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['李逵', '林冲', '武松', '黑寡妇', '绿巨人', '蜘蛛侠']

参数设置同样效果

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nums=[3,1,9,6,8,3,5,3]

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nums.sort(reverse=True)

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nums

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[9, 8, 6, 5, 3, 3, 3, 1]

count查找某个元素出现的次数

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nums.count(3)

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3

index获取索引值(第一次出现)

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heros.index("绿巨人")

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4

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heros[heros.index("绿巨人")]="神奇女侠"

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heros

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['李逵', '林冲', '武松', '黑寡妇', '神奇女侠', '蜘蛛侠']

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nums=[3,1,9,6,8,3,5,3]

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nums.index(3)

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0

index(x,start,end)

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nums.index(3,1,7)

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5

copy拷贝

shallow copy

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nums_copy1=nums.copy()

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nums_copy1

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[3, 1, 9, 6, 8, 3, 5, 3]

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nums_copy2=nums

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nums_copy2

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[3, 1, 9, 6, 8, 3, 5, 3]

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nums_copy3=nums[:]

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nums_copy3

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[3, 1, 9, 6, 8, 3, 5, 3]

list的* +

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s=[1,2,3]

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t=[4,5,6]

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s+t

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

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s*3

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[1, 2, 3, 1, 2, 3, 1, 2, 3]

嵌套列表

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matrix=[[1,2,3],[4,5,6],[7,8,9]]

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matrix=[[1,2,3],
        [4,5,6],
        [7,8,9]]

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matrix

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[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

访问嵌套列表

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for i in matrix:
    for each in i:
        print(each)

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9

换行

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for i in matrix:
    for each in i:
        print(each,end=' ')
    print()

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1 2 3 
4 5 6 
7 8 9

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matrix[0]

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[1, 2, 3]

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matrix[0][0]

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1

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matrix[1][1]

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5

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matrix[2][2]

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9

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A = [0]*3

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A

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[0, 0, 0]

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for i in range(3):#从0到2
    A[i]=[0]*3

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A

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[[0, 0, 0], [0, 0, 0], [0, 0, 0]]

错误写法-虽然结果相同

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B=[[0]*3]*3

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B 

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[[0, 0, 0], [0, 0, 0], [0, 0, 0]]

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A[1][1]=1

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A

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[[0, 0, 0], [0, 1, 0], [0, 0, 0]]

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B[1][1]=1

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B

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[[0, 1, 0], [0, 1, 0], [0, 1, 0]]

每个嵌套中的元素都被改变-探究(is)

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x = "FishC"

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y = "FishC"

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x is y

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True

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 x=[1,2,3]

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y=[1,2,2]

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x is y

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False
可知同个位置存放，而list会开两个内存
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A[0] is A[1]

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False

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A[1] is A[2]

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False

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B[0] is B[1]

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True

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B[1] is B[2]

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True

A每个元素单独开辟一个空间

B每个元素始终指向同一个空间

变量不是盒子

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x=[1,2,3]

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y=x

变量的赋值与拷贝

赋值不会创建新空间，是引用，拷贝则是创建新空间。不互相影响

浅拷贝和深拷贝

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x = [1,2,3]

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y=x.copy()

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y

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[1, 2, 3]

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x[1]=1

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x

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[1, 1, 3]

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y

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[1, 2, 3]

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x=[1,2,3]

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y=x[:]

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x[1]=1

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x

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[1, 1, 3]

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y

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[1, 2, 3]

二维开始有区别

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x=[[1,2,3],[4,5,6],[7,8,9]]

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y=x.copy()

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x[1][1]=0

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x

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[[1, 2, 3], [4, 0, 6], [7, 8, 9]]

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y

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[[1, 2, 3], [4, 0, 6], [7, 8, 9]]

引入深拷贝 copy.deepcopy(x)

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import copy

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x=[[1,2,3],[4,5,6],[7,8,9]]

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y=copy.copy(x)

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x[1][1]=0

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x

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[[1, 2, 3], [4, 0, 6], [7, 8, 9]]

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y

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[[1, 2, 3], [4, 0, 6], [7, 8, 9]]

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x=[[1,2,3],[4,5,6],[7,8,9]]

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y=copy.deepcopy(x)

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x[1][1]=0

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x

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[[1, 2, 3], [4, 0, 6], [7, 8, 9]]

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y

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[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

列表推导式 [expression for target in iterable]

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list_1=[1,2,3]

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list_1*3

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[1, 2, 3, 1, 2, 3, 1, 2, 3]

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list

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list

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list_1=[1,2,3]

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for i in range(len(list_1)):
    list_1[i]=list_1[i]*3

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list_1

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[3, 6, 9]

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oho=[1,2,3,4,5]

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oho=[i * 2 for i in oho]

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oho

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[2, 4, 6, 8, 10]

不仅仅少写代码，speed will be faster

使用C，而python使用虚拟机pvm

循环是迭代替换

推导式是直接创建

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x = [i for i in range(10)]

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x

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[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

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x = [i + 1for i in range(10)]

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x

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[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

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x = []

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for i in range(10):
    x.append(i+1)

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x

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[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

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y = [c * 2 for c in "FishC"]

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y

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['FF', 'ii', 'ss', 'hh', 'CC']

ord将字符转化为Unicode编码

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code = [ord(c) for c in "FishC"]

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code

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[70, 105, 115, 104, 67]

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matrix=[[1,2,3],[4,5,6],[7,8,9]]

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col2=[row[1] for row in matrix]

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col2

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[2, 5, 8]

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diag = [matrix[i][i] for i in range(len(matrix))]

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diag

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[1, 5, 9]

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diag=[]

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A=[0]*3

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A

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[0, 0, 0]

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A=[[0]*3 for i in A]

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A=[0]*3

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A=[[0]*3 for i in range(3)]

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A

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[[0, 0, 0], [0, 0, 0], [0, 0, 0]]

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A=[[0]*3 for i in A]

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A

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[[0, 0, 0], [0, 0, 0], [0, 0, 0]]

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A[1][1]=1

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A

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[[0, 0, 0], [0, 1, 0], [0, 0, 0]]

列表推导式 [expression for target in iterable if condition]

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[i for i in range(10) if i % 2 == 0]

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[0, 2, 4, 6, 8]

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enve=[i for i in range(10) if i % 2 == 0]

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enve

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[0, 2, 4, 6, 8]

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enve=[i + 1 for i in range(10) if i % 2 == 0]

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enve

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[1, 3, 5, 7, 9]

先执行if语句再执行左侧for语句

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words=["Great","FishC","Brilliant","Excellent","Fantistic"]

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fwords=[w for w in words if w[0] == 'F']

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fwords

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['FishC', 'Fantistic']

列表推导式 [expression for target in iterable if condition

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                      for target in iterable2 if condition2
                      for target in iterable3 if condition3
                      --
                      for target in iterableN if conditionN]嵌套循环

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matrix=[[1,2,3],[4,5,6],[7,8,9]]#内存上横向存储

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flatten=[col for row in matrix for col in row]

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flatten

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[1, 2, 3, 4, 5, 6, 7, 8, 9]

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flatten = []

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for row in matrix:
    for col in row:
        flatten.append(col)

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flatten

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[1, 2, 3, 4, 5, 6, 7, 8, 9]

笛卡尔积

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[x + y for x in "fishc" for y in "FISHC" ]

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['fF',
 'fI',
 'fS',
 'fH',
 'fC',
 'iF',
 'iI',
 'iS',
 'iH',
 'iC',
 'sF',
 'sI',
 'sS',
 'sH',
 'sC',
 'hF',
 'hI',
 'hS',
 'hH',
 'hC',
 'cF',
 'cI',
 'cS',
 'cH',
 'cC']

临时变量 _

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_ = []

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for x in "fishc":
    for y in "FISHC":
        _.append(x+y)

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_

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['fF',
 'fI',
 'fS',
 'fH',
 'fC',
 'iF',
 'iI',
 'iS',
 'iH',
 'iC',
 'sF',
 'sI',
 'sS',
 'sH',
 'sC',
 'hF',
 'hI',
 'hS',
 'hH',
 'hC',
 'cF',
 'cI',
 'cS',
 'cH',
 'cC']

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[[x,y] for x in range(10) if x % 2 == 0 for y in range(10) if y % 3 == 0]

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[[0, 0],
 [0, 3],
 [0, 6],
 [0, 9],
 [2, 0],
 [2, 3],
 [2, 6],
 [2, 9],
 [4, 0],
 [4, 3],
 [4, 6],
 [4, 9],
 [6, 0],
 [6, 3],
 [6, 6],
 [6, 9],
 [8, 0],
 [8, 3],
 [8, 6],
 [8, 9]]

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_ = [];

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for x in range(10):
    if x % 2 == 0:
        for y in range(10):
            if y % 3 == 0:
                _.append([x,y])

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_

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[[0, 0],
 [0, 3],
 [0, 6],
 [0, 9],
 [2, 0],
 [2, 3],
 [2, 6],
 [2, 9],
 [4, 0],
 [4, 3],
 [4, 6],
 [4, 9],
 [6, 0],
 [6, 3],
 [6, 6],
 [6, 9],
 [8, 0],
 [8, 3],
 [8, 6],
 [8, 9]]

KISS原则:Keep It Simple & Stupid