Because this bit me harder than I was ready for, I thought I’d make a note of it for the next victim.
In Python 2, suppose you have this:
Python 2.7.5>>> items = [(1, 'A number'), ('a', 'A letter'), (2, 'Another number')]
Sorting them, without specifying how, will automatically notice that it contains tuples:
Python 2.7.5>>> sorted(items)[(1, 'A number'), (2, 'Another number'), ('a', 'A letter')]
This doesn’t work in Python 3 because comparing integers and strings is not allowed. E.g.:
Python 3.3.3>>> 1 < '1'Traceback (most recent call last): File "", line 1, in TypeError: unorderable types: int() < str()
You have to convert them to stings first.
Python 3.3.3>>> sorted(items, key=lambda x: str(x[0]))[(1, 'A number'), (2, 'Another number'), ('a', 'A letter')]
If you really need to sort by 1 < '1' this won’t work. Then you need a more complex key function. E.g.:
Python 3.3.3>>> def keyfunction(x):... v = x[0]... if isinstance(v, int): v = '0%d' % v... return v...>>> sorted(items, key=keyfunction)[(1, 'A number'), (2, 'Another number'), ('1', 'Actually a string')]
That’s really messy but the best I can come up with at past 4PM on Friday.
原文地址：Sorting mixed type lists in Python 3, 感谢原作者分享。 人生就是一场旅行，不在乎目的地，
