计算字典频率的Python程序

给定一个字典列表，这里的任务是编写一个Python程序来提取每个字典的字典频率。

Input : test_list = [{‘gfg’ : 1, ‘is’ : 4, ‘best’ : 9},

             {‘gfg’ : 6, ‘is’ : 3, ‘best’ : 8},

             {‘gfg’ : 1, ‘is’ : 4, ‘best’ : 9},

             {‘gfg’ : 1, ‘is’ : 1, ‘best’ : 9},

             {‘gfg’ : 6, ‘is’ : 3, ‘best’ : 8}]

Output : [({‘gfg’: 1, ‘is’: 4, ‘best’: 9}, 2), ({‘gfg’: 6, ‘is’: 3, ‘best’: 8}, 2), ({‘gfg’: 1, ‘is’: 1, ‘best’: 9}, 1)]

Explanation : Dictionaries with their frequency appended as result in list.

Input : test_list = [{‘gfg’ : 1, ‘is’ : 4, ‘best’ : 9},

             {‘gfg’ : 6, ‘is’ : 3, ‘best’ : 8},

             {‘gfg’ : 1, ‘is’ : 4, ‘best’ : 9},

             {‘gfg’ : 1, ‘is’ : 1, ‘best’ : 9}]

Output : [({‘gfg’: 1, ‘is’: 4, ‘best’: 9}, 2), ({‘gfg’: 6, ‘is’: 3, ‘best’: 8}, 1), ({‘gfg’: 1, ‘is’: 1, ‘best’: 9}, 1)]

Explanation : Dictionaries with their frequency appended as result in list.

方法 1：使用index()和循环

在这种情况下，迭代每个字典，并使用 index() 获取字典的索引与其增加的频率映射，并在重复字典的情况下增加计数器。

例子：

Python3

# initializing list
test_list = [{'gfg': 1, 'is': 4, 'best': 9},
             {'gfg': 6, 'is': 3, 'best': 8},
             {'gfg': 1, 'is': 4, 'best': 9},
             {'gfg': 1, 'is': 1, 'best': 9},
             {'gfg': 6, 'is': 3, 'best': 8}]
  
# printing original list
print("The original list is : " + str(test_list))
  
res = []
for sub in test_list:
  
    flag = 0
    for ele in res:
  
        # checking for presence and incrementing frequency
        if sub == ele[0]:
            res[res.index(ele)] = (sub, ele[1] + 1)
            flag = 1
  
    if not flag:
        res.append((sub, 1))
  
# printing result
print("Dictionaries frequencies : " + str(res))

输出：

The original list is : [{‘gfg’: 1, ‘is’: 4, ‘best’: 9}, {‘gfg’: 6, ‘is’: 3, ‘best’: 8}, {‘gfg’: 1, ‘is’: 4, ‘best’: 9}, {‘gfg’: 1, ‘is’: 1, ‘best’: 9}, {‘gfg’: 6, ‘is’: 3, ‘best’: 8}]

Dictionaries frequencies : [({‘best’: 9, ‘gfg’: 1, ‘is’: 4}, 2), ({‘best’: 8, ‘gfg’: 6, ‘is’: 3}, 2), ({‘best’: 9, ‘gfg’: 1, ‘is’: 1}, 1)]

方法 2：使用Counter()和sorted()

在这种情况下，字典元素被转换为元组对，然后使用 Counter 来获取每个元素的频率。在最后一步，每本词典都被重新转换成原来的形式。

例子：

蟒蛇3

from collections import Counter
  
# initializing list
test_list = [{'gfg': 1, 'is': 4, 'best': 9},
             {'gfg': 6, 'is': 3, 'best': 8},
             {'gfg': 1, 'is': 4, 'best': 9},
             {'gfg': 1, 'is': 1, 'best': 9},
             {'gfg': 6, 'is': 3, 'best': 8}]
  
# printing original list
print("The original list is : " + str(test_list))
  
# getting frequencies
temp = Counter(tuple(sorted(sub.items())) for sub in test_list)
  
# converting back to Dictionaries
res = [(dict([tuple(ele) for ele in sub]), temp[sub]) for sub in temp]
  
# printing result
print("Dictionaries frequencies : " + str(res))

输出：

The original list is : [{‘gfg’: 1, ‘is’: 4, ‘best’: 9}, {‘gfg’: 6, ‘is’: 3, ‘best’: 8}, {‘gfg’: 1, ‘is’: 4, ‘best’: 9}, {‘gfg’: 1, ‘is’: 1, ‘best’: 9}, {‘gfg’: 6, ‘is’: 3, ‘best’: 8}]

Dictionaries frequencies : [({‘best’: 9, ‘gfg’: 1, ‘is’: 4}, 2), ({‘best’: 8, ‘gfg’: 6, ‘is’: 3}, 2), ({‘best’: 9, ‘gfg’: 1, ‘is’: 1}, 1)]