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练习

练习:

1.看代码写结果

v1 = [1,2,3,4,5]
v2 = [v1,v1,v1]
v1.append(6)
print(v1)
print(v2)

结果:
# v1 = [1,2,3,4,5,6]
# v2 = [[1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6]]

2.看代码写结果

v1 = [1,2,3,4,5]
v2 = [v1,v1,v1]
v2[1][0] = 111
v2[2][0] = 222
print(v1)
print(v2)

结果:
# v1 = [222,2,3,4,5]
# v2 = [[222,2,3,4,5], [222,2,3,4,5], [222,2,3,4,5] ]

3.看代码写结果,并解释每一步的流程。

v1 = [1,2,3,4,5,6,7,8,9]
v2 = {}
for item in v1:
    if item < 6:
        continue
    if 'k1' in v2:
        v2['k1'].append(item)
    else:
        v2['k1'] = [item ]
print(v2)
结果和解释:
# v2 = {'k1':[6,7,8,9]}
# 1-5执行第一个if 后边碰到continue后边的不执行
# 当item=6时执行第二个if的else 此时v2 = {'k1':[6]}
# 以后每一步都执行第二个if的append语句最终得到v2 = {'k1':[6,7,8,9]}

4.简述赋值和深浅拷贝?

赋值:只是两个变量名共用一个地址
浅拷贝:修改第一层的变量 两个变量互不影响,但是修改可变数据里面的元素两个都会改变
深拷贝:不可变数据类型共用一个地址,可变数据数据类型重新开辟一个新的地址,两个变量的修改都互不影响

5.看代码写结果

import copy
v1 = "alex"
v2 = copy.copy(v1)
v3 = copy.deepcopy(v1)
print(v1 is v2)
print(v1 is v3)

# True
# True

6.看代码写结果

import copy
v1 = [1,2,3,4,5]
v2 = copy.copy(v1)
v3 = copy.deepcopy(v1)
print(v1 is v2)
print(v1 is v3)

# False
# False

7.看代码写结果

import copy
v1 = [1,2,3,4,5]
v2 = copy.copy(v1)
v3 = copy.deepcopy(v1)

print(v1[0] is v2[0])
print(v1[0] is v3[0])
print(v2[0] is v3[0])


# True
# True
# True

8.看代码写结果

import copy

v1 = [1,2,3,4,[11,22]]
v2 = copy.copy(v1)
v3 = copy.deepcopy(v1)

print(v1[-1] is v2[-1])
print(v1[-1] is v3[-1])
print(v2[-1] is v3[-1])


# True
# False
# False

9.看代码写结果

import copy

v1 = [1,2,3,{"name":'太白',"numbers":[7,77,88]},4,5]
v2 = copy.copy(v1)

print(v1 is v2)

print(v1[0] is v2[0])
print(v1[3] is v2[3])

print(v1[3]['name'] is v2[3]['name'])
print(v1[3]['numbers'] is v2[3]['numbers'])
print(v1[3]['numbers'][1] is v2[3]['numbers'][1])


# False
# True
# True
# True
# True
# True

10.看代码写结果

import copy
v1 = [1,2,3,{"name":'太白',"numbers":[7,77,88]},4,5]
v2 = copy.deepcopy(v1)
print(v1 is v2)
print(v1[0] is v2[0])
print(v1[3] is v2[3])

print(v1[3]['name'] is v2[3]['name'])
print(v1[3]['numbers'] is v2[3]['numbers'])
print(v1[3]['numbers'][1] is v2[3]['numbers'][1])


# False
# True
# False
# True
# False
# True

11.请说出下面a,b,c三个变量的数据类型。
a = ('太白金星')
b = (1,)
c = ({'name': 'barry'})

# str
# tuple
# dict

12.按照需求为列表排序:

l1 = [1, 3, 6, 7, 9, 8, 5, 4, 2]
# 从大到小排序
# 从小到大排序
# 反转l1列表

# l1 = [1, 3, 6, 7, 9, 8, 5, 4, 2]
# l1.sort(reverse=True)
# print(l1)
#
# l1.sort()
# print(l1)

# l1.reverse()
# print(l1)

13.利用python代码构建一个这样的列表(升级题):

#[['_','_','_'],['_','_','_'],['_','_','_']]

ls = "_"
for i in range(2):
    ls = [ls,ls,ls]
print(ls)

14.看代码写结果:

l1 = [1,2,]
l1 += [3,4]
print(l1)

# [1,2,3,4]

15.看代码写结果:

dic = dict.fromkeys('abc',[])
dic['a'].append(666)
dic['b'].append(111)
print(dic)

# {'a':[666,111],'b':[666,111],'c':[666,111]}

16.l1 = [11, 22, 33, 44, 55],请把索引为奇数对应的元素删除(不能一个一个删除)

l1 = [11, 22, 33, 44, 55]
# l1 = l1[::2]
# print(l1)
for i in range(1, len(l1), 2):
    l1.pop(i)
print(l1)

17.dic = {'k1':'太白','k2':'barry','k3': '白白', 'age': 18} 请将字典中所有键带k元素的键值对删除.

dic = {'k1':'太白','k2':'barry','k3': '白白', 'age': 18}
dic1 = dic.copy()
for i in dic1:
    if "k" in i:
        dic.pop(i)
print(dic)

18.完成下列需求:
s1 = '太白金星'
将s1转换成utf-8的bytes类型。
将s1转化成gbk的bytes类型。
b = b'\xe5\xae\x9d\xe5\x85\x83\xe6\x9c\x80\xe5\xb8\x85'
b为utf-8的bytes类型,请转换成gbk的bytes类型。

s1 = '太白金星'
s1 = s1.encode("utf-8")
print(s1)
s1 = s1.decode("utf-8")
print(s1)
s1 = s1.encode("gbk")
print(s1)
b = b'\xe5\xae\x9d\xe5\x85\x83\xe6\x9c\x80\xe5\xb8\x85'
b = b.decode("utf-8")
print(b)
b = b.encode("gbk")
print(b)

19.用户输入一个数字,判断一个数是否是水仙花数。
水仙花数是一个三位数, 三位数的每一位的三次方的和还等于这个数. 那这个数就是一个水仙花数,
例如: 153 = 1**3 + 5**3 + 3**3

sx_num = input("请输入一个数字:")
if sx_num.isdecimal():
    sum = 0
    for i in sx_num:
        sum += int(i)**3
    if len(sx_num) == 3 and int(sx_num) == sum:
        print("这是一个水仙花数!")
    else:
        print("这不是一个水仙花数!")
else:
    print("请输入纯数字!")
    

20.把列表中所有姓周的⼈的信息删掉(此题有坑, 请慎重):
lst = ['周⽼⼆', '周星星', '麻花藤', '周扒⽪']
结果: lst = ['麻花藤']

lst = ['周⽼⼆', '周星星', '麻花藤', '周扒⽪']
for i in range(len(lst)-1,-1,-1):
    if "周" in lst[i]:
        lst.pop(i)
print(lst)

21.车牌区域划分, 现给出以下车牌. 根据车牌的信息, 分析出各省的车牌持有量. (选做题)
cars = ['鲁A32444','鲁B12333','京B8989M','⿊C49678','⿊C46555','沪 B25041']
locals = {'沪':'上海', '⿊':'⿊⻰江', '鲁':'⼭东', '鄂':'湖北', '湘':'湖南'}
结果: {'⿊⻰江':2, '⼭东': 2, '上海': 1}

cars = ['鲁A32444','鲁B12333','京B8989M','⿊C49678','⿊C46555','沪 B25041']
locals = {'沪':'上海', '⿊':'⿊⻰江', '鲁':'⼭东', '鄂':'湖北', '湘':'湖南'}

dic = {}
for i in cars:
    if i[0] in locals.keys() and locals[i[0]] not in dic.keys():
        dic[locals[i[0]]] = 1
    elif i[0] in locals.keys()and locals[i[0]] in dic.keys():
            dic[locals[i[0]]] += 1
print(dic)

posted @ 2019-07-12 18:10  学霸君主  阅读(174)  评论(0编辑  收藏  举报