4_列表()

列表输出
ary = [1, 2, 3, 4, "是SV"]
print(ary)
for i in ary:
    print(i)
# 输出列表的最后一个元素
length = len(ary)
print(ary[length-1])
print(ary[-1])

# [1, 2, 3]
print(ary[:3])

# [1, 2]
print(ary[:-3])

# [4, '是SV']
print(ary[3:])

# [1, 2, 3, 4, '是SV']
print(ary[:])

# [1, 3, '是SV']
print(ary[0:5:2])
print(ary[::2])

# ['是SV', 3, 1]
print(ary[::-2])

# ['是SV', 4, 3, 2, 1]
print(ary[::-1])
列表增删改查
# append、extend增加
# append 每次只能增加一个元素
hero = ['钢铁侠', "绿巨人"]
hero.append("黑寡妇")
print(hero)
# extend 一次可以增加多个元素
hero.extend(["鹰眼", "灭霸", "雷神"])
print(hero)

s = [1, 2, 3, 4, 5]
#  [1, 2, 3, 4, 5, 6]
s[len(s):] = [6]
print(s)

# [1, 2, 3, 4, 5, 6, 7, 8, 9]
s[len(s):] = [7, 8, 9]
print(s)
----------------------------------
# insert增加元素
s = [1, 3, 4, 5]
# [1, 2, 3, 4, 5]
s.insert(1, 2)
print(s)

# [0, 1, 2, 3, 4, 5]
s.insert(0, 0)
print(s)

# [0, 1, 2, 3, 4, 5, 6]
s.insert(len(s), 6)
print(s)
-----------------------------------------
# remove、pop、clear删除列表中的数据
hero = ['钢铁侠', "绿巨人", "黑寡妇", "鹰眼", "灭霸", "雷神"]
# ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
# remove 如果有多个相同的元素,只会删除第一个元素,如果列表没有要删除的元素就会报错
# ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
hero.remove("灭霸")
print(hero)

# pop 删除的是对应元素的下标
# ['钢铁侠', '绿巨人', '鹰眼', '雷神']
hero.pop(2)
print(hero)

# 清除列表的所有元素
hero.clear()
print(hero)
-------------------------------------------
替换
# 替换指定元素
# hero = ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
hero = ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
hero[4] = "蜘蛛侠"
print(hero)

# 替换部分元素
# ['钢铁侠', '绿巨人', '黑寡妇', '武松', '林冲', '李逵']
hero[3:] = ["武松", "林冲", "李逵"]
print(hero)
-------------------------------------------
查找
# 查找元素出现的次数
# 4
num = [3, 4, 2, 3, 2, 45, 3, 65, 3]
print(num.count(3))
# 查找元素的下标
# 1
print(num.index(4))
# 从指定位置开始到指定位置结束查找
# 4
print(num.index(2, 3, 8))
# 将数组的某个元素替换
# [3, 40, 2, 3, 2, 45, 3, 65, 3]
num[num.index(4)] = 40
print(num)
# 浅拷贝
# [3, 40, 2, 3, 2, 45, 3, 65, 3]
num_copy1 = num.copy()
print(num_copy1)
# [3, 40, 2, 3, 2, 45, 3, 65, 3]
num_copy2 = num[:]
print(num_copy2)
---------------------------------------
数组排序
# 自动排序 sort(),从小到大排序
# [1, 2, 3, 4, 6, 9, 32, 45, 65]
num = [3, 4, 1, 32, 2, 45, 6, 65, 9]
num.sort()
print(num)
# sort(reverse=True) 的效果和先调用sort然后再调reverse的效果一模一样
# [65, 45, 32, 9, 6, 4, 3, 2, 1]
num.sort(reverse=True)
print(num)
# 自动排序 reverse(),只是将数组进行翻转,从大到小排序,前提是数组已经是从小到大排好了
# [9, 65, 6, 45, 2, 32, 1, 4, 3]
num1 = [3, 4, 1, 32, 2, 45, 6, 65, 9]
num1.reverse()
print(num1)
打印二维列表
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
for i in s:
    for x in i:
        print(x, end=" ")
    print()
# 二维数组赋值
# [[1, 1], [1, 1], [1, 1]]
A = [2]*3
print(A)
for i in range(3):
    A[i] = [1]*2
print(A)
浅拷贝和深拷贝
# 通过copy拷贝得到两个独立的列表
# [1, 2, 3]
x = [1, 2, 3]
y = x.copy()
x[1] = 1
print(y)
# 通过切片拷贝得到两个独立的列表
# [1, 2, 3]
x = [1, 2, 3]
y = x[:]
x[1] = 1
print(y)
# 浅拷贝只是拷贝了外层的对象,如果包涵嵌套对象,那么拷贝只是其引用
# [[1, 0, 3], [4, 1, 6], [7, 8, 9]]
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
y = s.copy()
s[0][1] = 0
s[1][1] = 1
print(y)
# 深拷贝
# deepcopy函数将原对象拷贝的同时,也将对象中所有引用的子对象一并进行了拷贝
# [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
import copy
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
y = copy.deepcopy(s)
s[0][1] = 0
s[1][1] = 1
print(y)
列表推导式
# 将数组元素都乘以2
# [2, 4, 6, 8, 10]
od = [1, 2, 3, 4, 5]
for i in range(len(od)):
    od[i] = od[i]*2
print(od)
# 用列表推导式效率更高
# [2, 4, 6, 8, 10]
od = [1, 2, 3, 4, 5]
od = [i * 2 for i in od]
print(od)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# for中的循环数据首先会存放在i中,列表中的i+1才是最终存放数据的表达式
x = [i+1 for i in range(10)]
print(x)
# 取出第二列的元素
# for循环将每一行的元素取出,row[1]每一行的第二个元素,每一行的第二个元素就是第二列元素
# [2, 5, 8]
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
clo2 = [row[1] for row in s]
print(clo2)
# 打印对角线的值
# [1, 5, 9]
diag = [s[i][i] for i in range(len(s))]
print(diag)
# [3, 5, 7]
diag = [s[i][-i+2] for i in range(len(s))]
print(diag)
---------------------------------------------
# 先执行for语句,然后再执行if语句,最后执行i+1
# [1, 3, 5, 7, 9]
even = [i+1 for i in range(10) if i % 2 == 0]
print(even)

# 输出f开头的单词
word = ["great", "fish", "brilliant", "excellent", "fantastic"]
fword = [w for w in word if w[0] == "f"]
print(fword)

# 降维
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
flatten = [col for row in s for col in row]
print(flatten)
# ['lL', 'lO', 'lV', 'lE', 'oL', 'oO', 'oV', 'oE', 'vL', 'vO', 'vV', 'vE', 'eL', 'eO', 'eV', 'eE']
a = [x+y for x in "love" for y in "LOVE"]
print(a)

# [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]
b = [[x, y] for x in range(10) if x % 2 == 0 for y in range(10) if y % 3 ==0]
print(b)
# 等价于
for x in range(10):
    if x % 2 == 0:
        for y in range(10):
            if y % 3 == 0:
                print([x, y], end="")
---------------------------------------------
清除列表中的数据,并非把整个列表清除
# []
s1 = [1, 2, 3, 4, 5]
s1.clear()
print(s1)
s = [1, 2, 3, 4, 5]
del s[:]
print(s)
————————
列表输出
ary = [1, 2, 3, 4, "是SV"]
print(ary)
for i in ary:
    print(i)
# 输出列表的最后一个元素
length = len(ary)
print(ary[length-1])
print(ary[-1])

# [1, 2, 3]
print(ary[:3])

# [1, 2]
print(ary[:-3])

# [4, '是SV']
print(ary[3:])

# [1, 2, 3, 4, '是SV']
print(ary[:])

# [1, 3, '是SV']
print(ary[0:5:2])
print(ary[::2])

# ['是SV', 3, 1]
print(ary[::-2])

# ['是SV', 4, 3, 2, 1]
print(ary[::-1])
列表增删改查
# append、extend增加
# append 每次只能增加一个元素
hero = ['钢铁侠', "绿巨人"]
hero.append("黑寡妇")
print(hero)
# extend 一次可以增加多个元素
hero.extend(["鹰眼", "灭霸", "雷神"])
print(hero)

s = [1, 2, 3, 4, 5]
#  [1, 2, 3, 4, 5, 6]
s[len(s):] = [6]
print(s)

# [1, 2, 3, 4, 5, 6, 7, 8, 9]
s[len(s):] = [7, 8, 9]
print(s)
----------------------------------
# insert增加元素
s = [1, 3, 4, 5]
# [1, 2, 3, 4, 5]
s.insert(1, 2)
print(s)

# [0, 1, 2, 3, 4, 5]
s.insert(0, 0)
print(s)

# [0, 1, 2, 3, 4, 5, 6]
s.insert(len(s), 6)
print(s)
-----------------------------------------
# remove、pop、clear删除列表中的数据
hero = ['钢铁侠', "绿巨人", "黑寡妇", "鹰眼", "灭霸", "雷神"]
# ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
# remove 如果有多个相同的元素,只会删除第一个元素,如果列表没有要删除的元素就会报错
# ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
hero.remove("灭霸")
print(hero)

# pop 删除的是对应元素的下标
# ['钢铁侠', '绿巨人', '鹰眼', '雷神']
hero.pop(2)
print(hero)

# 清除列表的所有元素
hero.clear()
print(hero)
-------------------------------------------
替换
# 替换指定元素
# hero = ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
hero = ['钢铁侠', '绿巨人', '黑寡妇', '鹰眼', '雷神']
hero[4] = "蜘蛛侠"
print(hero)

# 替换部分元素
# ['钢铁侠', '绿巨人', '黑寡妇', '武松', '林冲', '李逵']
hero[3:] = ["武松", "林冲", "李逵"]
print(hero)
-------------------------------------------
查找
# 查找元素出现的次数
# 4
num = [3, 4, 2, 3, 2, 45, 3, 65, 3]
print(num.count(3))
# 查找元素的下标
# 1
print(num.index(4))
# 从指定位置开始到指定位置结束查找
# 4
print(num.index(2, 3, 8))
# 将数组的某个元素替换
# [3, 40, 2, 3, 2, 45, 3, 65, 3]
num[num.index(4)] = 40
print(num)
# 浅拷贝
# [3, 40, 2, 3, 2, 45, 3, 65, 3]
num_copy1 = num.copy()
print(num_copy1)
# [3, 40, 2, 3, 2, 45, 3, 65, 3]
num_copy2 = num[:]
print(num_copy2)
---------------------------------------
数组排序
# 自动排序 sort(),从小到大排序
# [1, 2, 3, 4, 6, 9, 32, 45, 65]
num = [3, 4, 1, 32, 2, 45, 6, 65, 9]
num.sort()
print(num)
# sort(reverse=True) 的效果和先调用sort然后再调reverse的效果一模一样
# [65, 45, 32, 9, 6, 4, 3, 2, 1]
num.sort(reverse=True)
print(num)
# 自动排序 reverse(),只是将数组进行翻转,从大到小排序,前提是数组已经是从小到大排好了
# [9, 65, 6, 45, 2, 32, 1, 4, 3]
num1 = [3, 4, 1, 32, 2, 45, 6, 65, 9]
num1.reverse()
print(num1)
打印二维列表
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
for i in s:
    for x in i:
        print(x, end=" ")
    print()
# 二维数组赋值
# [[1, 1], [1, 1], [1, 1]]
A = [2]*3
print(A)
for i in range(3):
    A[i] = [1]*2
print(A)
浅拷贝和深拷贝
# 通过copy拷贝得到两个独立的列表
# [1, 2, 3]
x = [1, 2, 3]
y = x.copy()
x[1] = 1
print(y)
# 通过切片拷贝得到两个独立的列表
# [1, 2, 3]
x = [1, 2, 3]
y = x[:]
x[1] = 1
print(y)
# 浅拷贝只是拷贝了外层的对象,如果包涵嵌套对象,那么拷贝只是其引用
# [[1, 0, 3], [4, 1, 6], [7, 8, 9]]
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
y = s.copy()
s[0][1] = 0
s[1][1] = 1
print(y)
# 深拷贝
# deepcopy函数将原对象拷贝的同时,也将对象中所有引用的子对象一并进行了拷贝
# [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
import copy
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
y = copy.deepcopy(s)
s[0][1] = 0
s[1][1] = 1
print(y)
列表推导式
# 将数组元素都乘以2
# [2, 4, 6, 8, 10]
od = [1, 2, 3, 4, 5]
for i in range(len(od)):
    od[i] = od[i]*2
print(od)
# 用列表推导式效率更高
# [2, 4, 6, 8, 10]
od = [1, 2, 3, 4, 5]
od = [i * 2 for i in od]
print(od)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# for中的循环数据首先会存放在i中,列表中的i+1才是最终存放数据的表达式
x = [i+1 for i in range(10)]
print(x)
# 取出第二列的元素
# for循环将每一行的元素取出,row[1]每一行的第二个元素,每一行的第二个元素就是第二列元素
# [2, 5, 8]
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
clo2 = [row[1] for row in s]
print(clo2)
# 打印对角线的值
# [1, 5, 9]
diag = [s[i][i] for i in range(len(s))]
print(diag)
# [3, 5, 7]
diag = [s[i][-i+2] for i in range(len(s))]
print(diag)
---------------------------------------------
# 先执行for语句,然后再执行if语句,最后执行i+1
# [1, 3, 5, 7, 9]
even = [i+1 for i in range(10) if i % 2 == 0]
print(even)

# 输出f开头的单词
word = ["great", "fish", "brilliant", "excellent", "fantastic"]
fword = [w for w in word if w[0] == "f"]
print(fword)

# 降维
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
s = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]
flatten = [col for row in s for col in row]
print(flatten)
# ['lL', 'lO', 'lV', 'lE', 'oL', 'oO', 'oV', 'oE', 'vL', 'vO', 'vV', 'vE', 'eL', 'eO', 'eV', 'eE']
a = [x+y for x in "love" for y in "LOVE"]
print(a)

# [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]
b = [[x, y] for x in range(10) if x % 2 == 0 for y in range(10) if y % 3 ==0]
print(b)
# 等价于
for x in range(10):
    if x % 2 == 0:
        for y in range(10):
            if y % 3 == 0:
                print([x, y], end="")
---------------------------------------------
清除列表中的数据,并非把整个列表清除
# []
s1 = [1, 2, 3, 4, 5]
s1.clear()
print(s1)
s = [1, 2, 3, 4, 5]
del s[:]
print(s)