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打印字典中最大和第二大元素

python

Pouya Ataei · 技术社区 · 6 年前

我已经编写了一个函数,它将字符串作为输入,并将一个字典交给用户,该字典指示最常出现的元素/字符。

def mostoccur(arr):

   n = len(arr)
   dict={}

   # filling the dictionary with itesm and their frequency using dict.get()

   for i in range(n):
       dict[arr[i]] = dict.get(arr[i],0) + 1


   return dict


string = "aabbccdddddddd"

# turning the string into a list, so it's iterable

ls = list(string)
ans = mostoccur(ls)
print("The dictionary of all the characters and their frequency: \n",ans)

maximum = max(ans)
print("\nThe most occuring character is:",maximum)

但后来我变得更好奇了,我想把字典里出现次数最多和出现次数次之的元素打印出来。因此,我写了这样的东西:

第2部分-好了,现在让我们找出第二个最常出现的字符

# defining a dictionary to store the most and second most occurring element

biggest = {'most-occuring':0, 'second-most-occuring':0}

# nested loops to go through all values in the dictionary

for a in ans.items():                                  # O(N)
   for b in ans.items():                               # O(N)
       if a < b:                                       
           a,b = b,a
           biggest['most-occuring'] = a
           biggest['second-most-occuring']= b

                                                       # Total = O(N^2)

print(biggest)

大O

你愿意告诉我更好的写作方法吗?

请记住,我不是在寻找一种利用

2 回复 | 直到 6 年前

TallChuck Sheetal 6 年前

下面是一个简单的算法 O(n) :

string = "aabbbccdddddddd"

def get_frequency(string):
    chars = {}
    for char in string:
        chars[char] = chars.get(char, 0) + 1 

    biggest = [None, 0]
    second = [None, 0]

    for entry in chars.items():
        char, count = entry
        if count > biggest[1]:
            second = biggest
            biggest = entry
        elif count > second[1]:
            second = entry

    return {
        "most-occurring": biggest[0],
        "second-most-occurring": second[0]
    }

print(get_frequency(string))

{'second-most-occurring': 'b', 'most-occurring': 'd'}

注意,我在 string 为了让它成为第二个最频繁的字母

T Burgis 6 年前

使用heapq.nlargest,如下所示:

from collections import Counter
import heapq
string = "aabbccdddddddd"
counts = Counter(string)
heapq.nlargest(2, counts, key=lambda k: counts[k])

并且不使用库,假设函数返回与Counter相同的内容:

keys = list(counts.keys())
keys.sort(key=lambda x: counts[x],reverse=True)
top_two = keys[:2] # just the keys
{ k : counts[k] for k in keys[:2] } # dict of top two

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