使用"in"运算符,Python的结果不同

Question

这段代码的结果

items = ['j','ak',(4,5)]
tests = ['j','as',(4,5)]
for key in tests:
    for item in items:
        if item==key:
            print key,'was found'
            break
    else:
        print key,'not found'

是:
j被发现
未找到
(4,5)被发现

和这块代码的结果

items = ['j','ak',(4,5)]
tests = ['j','as',(4,5)]
for key in tests:
    if key in items:
        print key+' was found'
    else:
        print key+' not found'

是:
j被发现
没有找到

现在,问题是:第二个块中的为什么(4,5)在"测试"和"项目"中没有进行比较,而其中任何一个块的结果应该相同？它是"in"运算符的东西吗？

Answer 1

你的第二个代码将为TypeError最后一个元组引发一个元组,因为你用一个字符串连接键:

>>> items = ['j','ak',(4,5)]
>>> tests = ['j','as',(4,5)]
>>> for key in tests:
...     if key in items:
...         print key+' was found'
...     else:
...         print key+' not found'
... 
j was found
as not found
Traceback (most recent call last):
  File "<stdin>", line 3, in <module>
TypeError: can only concatenate tuple (not "str") to tuple

但是在第一个,因为你使用逗号分隔key字符串它不会引起任何错误.