在字典中打印最大和第二大元素

Question

我编写了一个函数，该函数将字符串作为输入并移交表示最常出现的元素/字符的字典。

第1部分-只是找到出现次数最多的字符

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)

但是后来我变得更加好奇，我想打印字典中出现次数最多和倒数第二的元素。因此，我写了这样的东西：

第2部分-好吧，现在让我们找到第二个最常出现的角色

# 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

我已经写了每个操作旁边的Big O，当我看着它时，我真的不喜欢我写的东西。我的意思是，O（N ^ 2）听起来太昂贵且效率低下。

您愿意以更好的方式阐明我的观点吗？

  

请记住，我不是在寻找一种利用   任何图书馆。

Answer 1

这是在UserPaymentAccountHistoryLog中执行的简单算法：

UserPaymentHistoryLog

这将打印O(n)

请注意，我在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)) 上添加了一个额外的'b'，以使其成为第二频繁出现的字母

Answer 2

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])

并且不使用任何库，假设您的函数返回的内容与Counter：

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