按元素的频率排序python列表

Question

I have this code which sorts python list by frequency of elements. It works for all other cases except when frequency of two elements are same. If the frequency is same then I want to place smaller value first before higher value:

counts = collections.Counter(arr)
new_list = sorted(arr, key=lambda x: counts[x])

for item in new_list:
    print item

In case of [3,1,2,2,4] the output should be [1,3,4,2,2] but I get [3,1,4,2,2] . How do I resolve this error?

Answer 1

你可以将你的密钥lambda函数设置为一个元组，因此，它将首先按counts[x]排序，如果有一个平局，它将按x排序，值本身。

 new_list = sorted(arr, key=lambda x: (counts[x], x))

Answer 2

You are only sorting by the number of item. Under this logic, all the items which appear once have the same "weight", so python retains their original relevant position. Eg, 3 and 1 both appear once, so as far as their sorting is concerned, they are equivalent. Since 3 comes before 1 in the original list, it is also placed first in the result.

Your required output calls for a secondary sort criteria - the value of the element. To do so, you must specify this explicitly:

new_list = sorted(arr, key=lambda x: (counts[x], x))