我正在阅读https://docs.python.org/2/tutorial/controlflow.html#lambda-expressions处的lambda表达式，我不明白sort（）从何处获取其参数。它显示：
>>> pairs = [(1, 'one'), (2, 'two'), (3, 'three'), (4, 'four')]
>>> pairs.sort(key=lambda pair: pair[1])
>>> pairs
[(4, 'four'), (1, 'one'), (3, 'three'), (2, 'two')]
其中＆＃34;对：对[1]＆＃34;在用于lambda之前没有命名。我也做了：
In [23]: pairs = [(1, 'one'), (2, 'two'), (3, 'three'), (4, 'four')]
In [24]: pairs.sort(key=lambda x: x[1])
In [25]: pairs
Out[25]: [(4, 'four'), (1, 'one'), (3, 'three'), (2, 'two')]
In [26]: pairs = [(1, 'one'), (2, 'two'), (3, 'three'), (4, 'four')]
In [27]: pairs.sort(key=lambda randomname: randomname[1])
In [28]: pairs
Out[28]: [(4, 'four'), (1, 'one'), (3, 'three'), (2, 'two')]
In [29]: pairs = [(1, 'one'), (2, 'two'), (3, 'three'), (4, 'four')]
In [30]: pairs.sort(key=lambda randomname: mismatch[1])
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
...
NameError: global name 'mismatch' is not defined
这告诉我lambda可以接受任何随机参数名称，并且知道从对列表中获取[1]点。
我也不明白＆＃34; key＆＃34;甚至采取，正如文档说它需要一个参数的函数，然后使用key=str.lower作为一个例子。但是使用他们的确切示例失败了：
student_tuples
Out[55]: [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10), ('John', 'b', 5)]
In [60]: sorted(student_tuples, key=str.lower)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-60-86a57d272cfb> in <module>()
----> 1 sorted(student_tuples, key=str.lower)
TypeError: descriptor 'lower' requires a 'str' object but received a 'tuple'
In [61]: sorted(student_tuples, key=str.lower())
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-61-983b5ca19cdb> in <module>()
----> 1 sorted(student_tuples, key=str.lower())
TypeError: descriptor 'lower' of 'str' object needs an argument
即使将字符串作为可迭代项提供，str.lower方法也会失败：
In [68]: strings
Out[68]: ['str', 'str2', 'str3']
In [69]: sorted(strings, str.lower)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-69-271d28a2618f> in <module>()
----> 1 sorted(strings, str.lower)
TypeError: lower() takes no arguments (1 given)
我的问题是
如何知道lambda函数在当前iterable中分配位置（调用iterable本身得到TypeErro:'tuple' object is not callable）
究竟是什么关键，只做lambdas？
谢谢