Python3 语法基础思维导图

深碍√TFBOYSˉ_ 2023-10-10 11:38 60阅读 0赞

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3pob3V6aGl3ZW5nYW5n_size_16_color_FFFFFF_t_70][]

示例代码 :

import io
    import sys
     
    sys.stdout = io.TextIOWrapper(sys.stdout.detach(),encoding='utf-8')
    print('元组定义')
    
    tup = ('Google','IE','sougou','360')
    for everyOne in tup:
        print(everyOne)
    
    print('元组访问')
    print ("tup[0]: ", tup[0])
    print ("tup[1:4]: ", tup[1:4])
    
    print('修改元组')
    tup1 = (12, 34.56)
    tup2 = ('abc', 'xyz')
    # 创建一个新的元组
    tup3 = tup1 + tup2
    print (tup3)
    
    print('删除元组')
    tup4 = ('Google', 'Runoob', 1997, 2000)
     
    print (tup4)
    del tup4
    
    
    print('集合创建方式一:\{\}')
    coll ={'one', 'two', 'three'}
    print(coll)
    print('集合创建方式一:set')
    colltwo = set('123456')
    print(colltwo)
    
    print('集合添加')
    collthree = set(("Google", "IE", "Sougou"))
    print('未添加之前:')
    print( collthree)
    collthree.add('360')
    print('添加之后:')
    print(collthree)
    
    print('集合删除')
    collthree.remove('IE')
    print('集合移除之后:')
    print(collthree)
    
    print('集合中个数总量计算')
    print(len(collthree))
    
    print('集合中的元素清空')
    collthree.clear()
    print(collthree)
    
    print('判断元素是否在集合中存在')
    collfour = set(("Google", "IE", "Sougou"))
    if 'IE' in collfour:
        print('包含')
    else:
         print('不包含')
    
    print('集合运算符')
    pf = set(['邓肯','加内特','马龙'])
    pd = set(['邓肯','奥尼尔','姚明'])
    
    
    print('集合运算符之交集运算:', pf & pd)
    print('集合运算符之并集运算:', pf | pd)
    print('集合运算符之差集运算:', pf - pd)
    
    print('字典创建')
    dict = {'编号': '0001', '姓名': '周晨曦', 'age': '2'}
    print(dict)
    
    print('字典访问')
    print('dict[\'编号\']',dict['编号'])
    
    print('字典遍历')
    for item in dict.items():
        print(item)
    
    print('字典修改')
    dict['编号'] = '0002'
    print(dict)
    
    print('字典删除')
    del dict['编号'] # 删除键 'Name'
    print(dict)
    dict.clear()     # 清空字典
    print(dict)
    del dict         # 删除字典
    
    # if...else 简单选择语句
    age = 3
    if age == 2:
        print('恭喜你符合年龄要求')
    else:
        print('提示你年龄不符合要求')
    
    # python3 if...elif...else
    score = 99
    if score ==100:
        print('满分')
    elif score >=90:
        print('优秀')
    elif score >=80:
        print('良好')
    elif score >=70:
        print('及格')
    else:
        print('不及格,请家长')
    
    # python3 if语句嵌套
    if score >= 90:
        if score == 100:
            print('A+')
        else:
            print('A')
    elif score >=80:
        if score >=85:
            print('B+')
        else:
            print('B')
    elif score >=70:
        print('C')
    else:
        print('D')
    
    # python3 and 选择语句
    age =45
    if(age >= 18 and age <=65):
        print('中国司机法定年龄为18~65岁之间，你的年龄是符合要求的')
    
    # python3 or 选择语句
    answer ='A'
    if(answer == 'A' or answer =='B'):
        print('这道选择题不管是选择A还是B 都是正确的')
    
    # python3 not 语句
    a ='1'
    b =['A','B','1']
    if a not in b:
        print('a 元素不存在于b序列中')
    else:
        print('a 元素存在于b序列中')
    
    # python3 for 遍历序列
    
    c =['A','B','C']
    for obj in c:
        print('value is', obj)
    
    
    # python3 for 遍历对象
    
    st = 'ABCDEFG'
    for obj in st:
        print('value is', obj)
    
    # python3 while 循环语句
    
    i = 1
    while i <=10:
        print('坚持就是胜利',i)
        i =i + 1
    
    # python3 break 跳转语句
    
    st = 'ABCDEFGH'
    for obj in st:
        if obj == 'D':
            print('中止循环')
            break
        print('value is', obj)
    
    # python3 continue 跳转语句
    
    st = 'ABCDEFGH'
    for obj in st:
        if obj == 'D':
            print('中止当前循环,进入下一个循环')
            continue
        print('value is', obj)

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