• python开发基础04-列表、元组、字典操作练习


    练习1:

    # l1    =  [11,22,33]
    # l2 = [22,33,44]
    # a. 获取内容相同的元素列表
    # b. 获取 l1 中有, l2 中没有的元素列表
    # c. 获取 l2 中有, l1 中没有的元素列表
    # d. 获取 l1 和 l2 中内容都不同的元素
    # l1    =    [11,22,33]
    # l2    =    [22,33,44]
    # for i in l2:
    #     if i not in l1:
    #         print(i)
    # for i in l1:
    #     if i not in l2:
    #         print(i)
    View Code

    练习2:

    12345678互相组合为一个2位数,两位数值不可重复

    # count =0
    # for i in range(1, 9):
    #     for v in range(1, 9):
    #         if i !=v:
    #             count += 1
    # # count = count -8
    # print(count)
    View Code

    练习3:

    99乘法表

    # for i in range(1,10):
    #     string = ""
    #     for j in range(1,i+1):
    #         string +=str(j) + " * "+str(i) + " = " +str(i*j)+"	"
    #     print(string)
    方法1
    # for i in range(1,10):
    #     for j in range(1,i+1):
    #         print(str(j) + " * "+str(i) + " = " +str(i*j)+"	",end="")
    #     print("
    ",end="")
    
    # 默认end=
    
    # print('alex', end='11')
    # print('alex','haifeng','gangniang',sep='11')
    方法2

    练习4:

    [1,2,3,4]这里面的数字两个组合 多少种不同样的,数字不重复的

    # li = [1,2,3,4]
    # l =len(li)
    # for i in range(0,l-1):
    #     for v in range(i+1,l):
    #         print(li[i],li[v])
    View Code

    练习5:

    自动计算方案

    百钱买小鸡

    公鸡5文钱1只,母鸡三文钱一只,小鸡一文钱三只。现在用100文钱共买了100只鸡,

    问这100只鸡中,公鸡,母鸡,小鸡各是多少只? 

    for x in range(1,100//5):
        for y in range(1,100//3):
            for z in range(1,100):
                if x + y + z == 100 and 5*x + 3*y + z/3 == 100:
                    print(x,y,z)
    View Code

    练习6:

     查找列表中元素,移除每个元素的空格,并查找以 a 或 A 开头 并且以 c 结尾的所有元素。

      li = ["alec", " aric", "Alex", "Tony", "rain"]

      tu = ("alec", " aric", "Alex", "Tony", "rain")

      dic = {'k1': "alex", 'k2': ' aric',"k3": "Alex", "k4": "Tony"}

    #!/usr/bin/env python
    # -*- coding:utf-8 -*-
    
    li = ["alec", " aric", "Alex", "Tony", "rain"]
    
    for S in li:
    S = S.strip()
    prefix_a = S.startswith('a')
    prefix_A = S.startswith('A')
    suffix_c = S.endswith('c')
    if prefix_a or prefix_A and suffix_c:
    print('要查找的元素:%s' % S)
    
    tu = ("alec", " aric", "Alex", "Tony", "rain")
    
    for S in tu:
    S = S.strip()
    prefix_a = S.startswith('a')
    prefix_A = S.startswith('A')
    suffix_c = S.endswith('c')
    if prefix_a or prefix_A and suffix_c:
    print('要查找的元素:%s' % S)
    
    
    dic = {'k1': "alex", 'k2': ' aric',"k3": "Alex", "k4": "Tony"}
    
    for S in dic.values():
    S = S.strip()
    prefix_a = S.startswith('a')
    prefix_A = S.startswith('A')
    suffix_c = S.endswith('c')
    if prefix_a or prefix_A and suffix_c:
    print('要查找的元素:%s' % S)
    View Code

    练习7:

    有如下变量,请实现要求的功能

    tu = ("alex", [11, 22, {"k1": 'v1', "k2": ["age", "name"], "k3": (11,22,33)}, 44])

    • 讲述元祖的特性

    • 请问 tu 变量中的"k2"对应的值是什么类型?是否可以被修改?如果可以,请在其中添加一个元素 “Seven”
    # 元组的一级元素不可被修改增加删除,有序,可迭代,可切片,可索引,可转换为列表。
    tu = ("alex", [11, 22, {"k1": 'v1', "k2": ["age", "name"], "k3": (11,22,33)}, 44])
    v = tu[1][2]["k2"]
    v.append("Seven")
    print(tu)
    View Code
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  • 原文地址:https://www.cnblogs.com/DaweiJ/p/8478554.html
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