'''
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：
转载于:https://www.cnblogs.com/shuimohei/p/10232668.html
