'''分析：    1.列表解析：迭代机制的一种应用      语法：    　   [expression for iter_val in iterable]　　     [expression for iter_val in iterable if cond_expr]    2.zip函数：以可迭代的对象作为参数，将对应元素打包成一个元组,形如：zip([a,b],[c,d])→[(a,c),(b,d)]      dict函数：可以以可迭代方式创建字典，形如：dict([(a,b),(c,d)])→{a:b,c:d}'''A0 = dict(zip(('a','b','c','d','e'),(1,2,3,4,5)))A1 = range(10)A2 = [i for i in A1 if i in A0]A3 = [A0[s] for s in A0]A4 = [i for i in A1 if i in A3]A5 = {i:i*i for i in A1}A6 = [[i,i*i] for i in A1]'''解答：    A0→A0=dict([("a",1),("b",2),("c",3),("d",4),("e",5])→A0={"a":1,"b":2,"c":3,"d":4,"e":5}    A1→A1=[0,1,2,3,4,5,6,7,8,9]    A2→        A2 = []        for i in A1:            if i in A0:     #字典进行in判断的时候判断的是key               A2.append(i)     →A2=[]    A3→        A3 = []        for s in A0:            A3.append(A0[s]) →A3=[1, 2, 3, 4, 5]    A4→        A4=[]        for i in A1:            if i in A3:                A4.append(i)→A4=[1, 2, 3, 4, 5]    A5→        A5 = {}        for i in A1:            A5.update({i:i*i})→A5={0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64, 9: 81}    A6→        A6 = []        for i in A1:            A6.append([i,i*i])→A6=[[0, 0], [1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]'''

　　运行结果：