<每日一题>题目12：列表解析及zip、dict函数的简单应用

'''
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：