Python - 将字典转换为 K 大小的字典

给定一个字典，将字典分成 K 大小的不同字典列表。

Input : test_dict = {‘Gfg’ : 1, ‘is’ : 2, ‘best’ : 3, ‘for’ : 4, ‘geeks’ : 5, ‘CS’ : 6}, K = 3
Output : [{‘Gfg’: 1, ‘is’: 2, ‘best’: 3}, {‘for’: 4, ‘geeks’: 5, ‘CS’: 6}]
Explanation : Divided into size of 3 keys.

Input : test_dict = {‘Gfg’ : 1, ‘is’ : 2, ‘best’ : 3, ‘for’ : 4}, K = 2
Output : [{‘Gfg’: 1, ‘is’: 2}, {‘best’: 3, ‘for’: 4}]
Explanation : Divided into size of 2 keys.

方法：使用循环

在此，我们使用循环迭代字典中的所有键，并根据大小进行分叉并附加到新列表。

Python3

# Python3 code to demonstrate working of 
# Convert dictionary to K Keys dictionaries
# Using loop
  
# initializing dictionary
test_dict = {'Gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 4, 'geeks' : 5, 'CS' : 6}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# initializing K 
K = 2
  
res = []
count = 0
flag = 0
indict = dict()
for key in test_dict:
    indict[key] = test_dict[key]        
    count += 1
      
    # checking for K size and avoiding empty dict using flag
    if count % K == 0 and flag:
        res.append(indict)
          
        # reinitializing dictionary
        indict = dict()
        count = 0
    flag = 1
      
  
# printing result 
print("The converted list : " + str(res))

输出

The original dictionary is : {'Gfg': 1, 'is': 2, 'best': 3, 'for': 4, 'geeks': 5, 'CS': 6}
The converted list : [{'Gfg': 1, 'is': 2}, {'best': 3, 'for': 4}, {'geeks': 5, 'CS': 6}]