Python 按第二个值对元组列表进行排序，reverse=True，然后按键，reverse=False
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14466068/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Sort a list of tuples by second value, reverse=True and then by key, reverse=False
提问by Nicolas Hung
I need to sort a dictionary by first, values with reverse=True, and for repeating values, sort by keys, reverse=False
我需要首先对字典进行排序，值为reverse=True，对于重复值，按键排序，reverse=False
So far, I have this
到目前为止，我有这个
dict = [('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
sorted(dict.items(), key=lambda x: (x[1],x[1]), reverse=True)
which returns...
返回...
[('B', 3), ('A', 2), ('J', 1), ('I', 1), ('A', 1)]
but I need it to be:
但我需要它是：
[('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
as you can see, when values are equal, I can only sort the key in a decreasing fashion as specified... But how can I get them to sort in an increasing fashion?
如您所见，当值相等时，我只能按照指定的递减方式对键进行排序......但是我怎样才能让它们以递增的方式排序？
采纳答案by mgilson
The following works with your input:
以下适用于您的输入：
d = [('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
sorted(d,key=lambda x:(-x[1],x[0]))
Since your "values" are numeric, you can easily reverse the sort order by changing the sign.
由于您的“值”是数字，因此您可以通过更改符号轻松颠倒排序顺序。
In other words, this sort puts things in order by value (-x[1]) (the negative sign puts big numbers first) and then for numbers which are the same, it orders according to key (x[0]).
换句话说，这种排序按值 ( -x[1])排序（负号先放大数），然后对于相同的数字，它根据键 ( x[0])排序。
If your values can't so easily be "negated" to put big items first, an easy work-around is to sort twice:
如果您的值不能那么容易地被“否定”以将大项目放在首位，一个简单的解决方法是排序两次：
from operator import itemgetter
d.sort(key=itemgetter(0))
d.sort(key=itemgetter(1),reverse=True)
which works because python's sorting is stable.
之所以有效，是因为 python 的排序是稳定的。
回答by NPE
In [4]: l = [('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
In [5]: sorted(l, key=lambda (x,y):(-y,x))
Out[5]: [('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
回答by Ashwini Chaudhary
you can use collections.defaultdict:
你可以使用collections.defaultdict：
In [48]: from collections import defaultdict
In [49]: dic=[('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
In [50]: d=defaultdict(list)
In [51]: for x,y in dic:
d[y].append(x)
d[y].sort() #sort the list
now dis something like:
现在d是这样的：
defaultdict(<type 'list'>, {1: ['A', 'I', 'J'], 2: ['A'], 3: ['B']}
i.e. A new dictwith 1,2,3...as keys and corresponding alphabets stored in lists as values.
即一个新的dict与1,2,3...作为键和对应的存储在列表作为值字母表。
Now you can iterate over the sorted(d.items)and get the desired result using itertools.chain()and itertools.product().
现在，您可以sorted(d.items)使用itertools.chain()和迭代并获得所需的结果itertools.product()。
In [65]: l=[?product(y,[x]) for x,y in sorted(d.items(),reverse=True)]
In [66]: list(chain(*l))
Out[66]: [('B', 3), ('A', 2), ('A', 1), ('I', 1), ('J', 1)]
