阅读下面的代码，写出A0,A1,至An的最终值
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
zip接收两个可迭代对象，返回一个zip对象，这个zip对象可以用next()取值，取出的值是一个个元组，python中关于zip的定义如下：
class zip(object):
"""
zip(iter1 [,iter2 [...]]) --> zip object
Return a zip object whose .__next__() method returns a tuple where
the i-th element comes from the i-th iterable argument. The .__next__()
method continues until the shortest iterable in the argument sequence
is exhausted and then it raises StopIteration.
"""
zip的使用方法 ```python obj = zip(('a', 'b', 'c', 'd', 'e'), (1, 2, 3, 4, 5)) print(obj, type(obj)) #
try:
for i in range(10):
print(next(obj))
except StopIteration:
print("结束了~")
由上面的演示可以很容易得出
```python
A0 = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5 }
A1就是生成器的简单测试，结果为
A1 = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
A2为一个列表推导式，元素为满足以下两个条件的值： - i在A1中 - i在A0中 注意A0是一个字典，i在A0中等价于i在A0的键组成的列表(A0.keys())里，而A0的键都是字母，所以没有满足条件的i ```python A2 = []
<br>
s为A0的键,A0[s]为A0的值，因此结果是A0值组成的列表
```python
A3 = [1, 2, 3, 4, 5]
```python A4 = [1, 2, 3, 4, 5]
<br>
```python
A5 = {0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64, 9: 81}
```python A6 = [[0, 0]， [1, 1]， [2, 4]， [3, 9]， [4, 16]， [5, 25]， [6, 36]，[7, 49]，[8, 64] [9, 81]]