python元组的列表或者列表的列表，如何简单快速计算每个item的指定单元？

Question

说的有点绕口，举个经常调用的例子：
对如下元组的list排序：students = [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]
使用lambda函数：sorted(students, key=lambda student : student[2])
或者itemgetter：sorted(students, key=operator.itemgetter(2))

黄舟 · Answer

The questioner doesn’t want to write the so-called loop, probably because he doesn’t want to write the displayed for while loop traversal.

To deal with some sequences, there is no other way except iteration and recursion.

If you don’t want to use an explicit for loop, you can write it in one sentence using list parsing, or using map, filter, reducers functional functions, or a combination of both

pythonIn [1]: students = [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]
In [2]: map(lambda t: t[0], students)
Out[2]: ['john', 'jane', 'dave']

In [3]: reduce(lambda x, y: x + y, map(lambda t: t[-1], students)) / float(len(students))
Out[3]: 12.333333333333334

伊谢尔伦 · Answer

Try list comprehension?

For example, extract the owner’s name

>>> students = [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]
>>> names = [ i[0] for i in students ]
>>> names
['john', 'jane', 'dave']

Find the average value

>>> students = [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]
>>> score = sum([ i[2] for i in students ])
>>> score
37
>>> score/len(students)
12.333333333333334