Python3 基础语法从入门到精通二

痛定思痛。 2023-10-17 11:59 61阅读 0赞

#Python3  变量赋值
    i = 100; #整型变量
    d = 100.0; #浮点型
    s = "Python3"; #字符串
    
    print(i);
    print(d);
    print(s);
    
    #Python3 列表
    list = ['one','two','three','four','five'];
    
    print(list);
    print(list[0]);
    print(list[1:3]);
    print(list[2:]);
    print(list * 2);
    
    #Python3 元组
    arraylist = ('1','2','3','4','5');
    print(arraylist);
    print(arraylist[0]);
    print(arraylist[1:3]);
    print(arraylist[2:]);
    print(arraylist * 2);
    
    # Python3 字典
    dirt ={};
    dirt['one'] = 'one';
    dirt['two'] = 'two';
    dirt['thress'] = 'three';
    dirt['four'] = 'four';
    dirt['five'] = 'five';
    dirt[0] =0;
    dirt[1] =1;
    dirt[2] =2;
    dirt[3] =3;
    
    print(dirt['one']);
    print(dirt[0]);
    print(dirt);
    print(dirt.keys());
    print(dirt.values());
    
    #Python3 数据类型转换
    a = int('1');
    print("type is {0}".format(type(a))); #type 基础数据类型判断
    
    b = float(1);
    print("type is {0}".format(type(b))); #type 基础数据类型判断
    
    c = str(1);
    print("type is {0}".format(type(c))); #type 基础数据类型判断
    
    d = repr(1.0);
    print("type is {0}".format(type(d))); #type 基础数据类型判断
    
    #Python3 成员运算符
    e = 1;
    f = 6;
    lists = [1,2,3,4,5];
    
    def whether(a,b):
        if(a in b):
            return 'true';
        else:
            return 'false';
    
    print("e is it inside {0}".format(whether(e,lists)));
    print("f is it inside {0}".format(whether(f,lists)));
    
    #Python3 身份运算符（基于对象）
    class Dog:
        def __init__(self,weight):
            self.weight = weight;
        def getWeight(self):
            print("my weight is {0}".format(self.weight));
    
    toady = Dog(10);
    boogie = Dog(15);
    
    if(toady is boogie):
        print("true");
    else:
        print("false");
    
    #Python3 身份运算符（基于数据类型）
    a = 10;
    b = 10;
    if(a is b):
        print("true");
    else:
        print("false");