如何首先根据键然后根据值对元组元素进行排序 [英] how to sort tuple element first on the basis of key and then on the basis of value

问题描述

如何在python中对元素的元组进行排序，首先根据值，然后根据键.考虑我将用户输入作为字符串输入的程序.我想找出每个字符的计数并打印字符串中最常见的 3 个字符.

How to sort a tuple of elements in python, first on the basis of value and then on the basis of key. Consider the program in which I am taking input from the user as a string. I want to find out the count of each character and print the 3 most common characters in a string.

#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

关于给 i/p -aabbbccde

on giving the i/p -aabbbccde

以上代码的输出为:

3 b
2 c
2 a

但我希望输出为:

3 b
2 a
2 c

推荐答案

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

Sort a list of tuples, the first value in descending order (reverse=True) and the second value in ascending order (reverse=False, by default). Here is a 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')]

<小时>

使用collections.Counter计算字符串中每个字母的频率.

---

It is straightforward to use collections.Counter to count each letter's frequency in a string.

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)]

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