如何在键的基础上首先对元组元素进行排序,然后在值的基础上对元组元素进行排序

Question

如何在python中对元素元组进行排序,首先是基于值,然后是基于键.考虑我将用户输入作为字符串的程序.我想找出每个字符的数量,并在字符串中打印3个最常见的字符.

#input string
strr=list(raw_input())
count=dict()

#store the count of each character in dictionary
for i in range(len(strr)):
count[strr[i]]=count.get(strr[i],0)+1

#hence we can't perform sorting on dict so convert it into tuple 
temp=list()
t=count.items()

for (k,v) in t:
    temp.append((v,k))

temp.sort(reverse=True)

 #print 3 most common element
for (v,k) in temp[:3]:
         print k,v

给予i/p -aabbbccde

上述代码的输出是:

3 b
2 c
2 a

但我希望输出为:

3 b
2 a
2 c

Answer 1

对元组列表进行排序,按降序排列第一个值(reverse=True),按升序排序第二个值(reverse=False默认情况下).这是一个MWE.

lists = [(2, 'c'), (2, 'a'), (3, 'b')]

result = sorted(lists, key=lambda x: (-x[0], x[1])) # -x[0] represents descending order

print(result)
# Output
[(3, 'b'), (2, 'a'), (2, 'c')]

可以直接用来collections.Counter 计算字符串中每个字母的频率.

import collections

s = 'bcabcab'

# If you don't care the order, just use `most_common`
#most_common = collections.Counter(s).most_common(3)

char_and_frequency = collections.Counter(s)
result = sorted(char_and_frequency.items(), key=lambda x:(-x[1], x[0]))[:3]    # sorted by x[1] in descending order, x[0] in ascending order

print(result)
# Output
[('b', 3), ('a', 2), ('c', 2)]