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Python – Dictionary with maximum count of pairs

Last Updated : 02 Sep, 2020
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Given dictionary list, extract dictionary with maximum keys.

Input : test_list = [{“gfg”: 2, “best” : 4}, {“gfg”: 2, “is” : 3, “best” : 4, “CS” : 9}, {“gfg”: 2}]
Output : 4
Explanation : 2nd dictionary has maximum keys, 4.

Input : test_list = [{“gfg”: 2, “best” : 4}, {“gfg”: 2}]
Output : 2
Explanation : 1st dictionary has maximum keys, 2.

Method #1 : Using len() + loop

In this, we iterate for each of dictionary and compare lengths of each, record and return one with maximum length.

Python3




# Python3 code to demonstrate working of 
# Dictionary with maximum keys
# Using loop + len()
  
# initializing list
test_list = [{"gfg": 2, "best" : 4}, 
             {"gfg": 2, "is" : 3, "best" : 4}, 
             {"gfg": 2}]
  
# printing original list
print("The original list is : " + str(test_list))
  
res = {} 
max_len = 0
for ele in test_list:
      
    # checking for lengths
    if len(ele) > max_len: 
        res = ele
        max_len = len(ele)
          
# printing results
print("Maximum keys Dictionary : " + str(res))


Output

The original list is : [{'gfg': 2, 'best': 4}, {'gfg': 2, 'is': 3, 'best': 4}, {'gfg': 2}]
Maximum keys Dictionary : {'gfg': 2, 'is': 3, 'best': 4}

Method #2 : Using max() + key=len

In this, we compute maximum length key using max() by passing additional key “len” for comparison based on lengths.

Python3




# Python3 code to demonstrate working of 
# Dictionary with maximum keys
# Using max() + key = len
  
# initializing list
test_list = [{"gfg": 2, "best" : 4}, 
             {"gfg": 2, "is" : 3, "best" : 4}, 
             {"gfg": 2}]
  
# printing original list
print("The original list is : " + str(test_list))
  
# maximum length dict using len param
res = max(test_list, key = len)
          
# printing results
print("Maximum keys Dictionary : " + str(res))


Output

The original list is : [{'gfg': 2, 'best': 4}, {'gfg': 2, 'is': 3, 'best': 4}, {'gfg': 2}]
Maximum keys Dictionary : {'gfg': 2, 'is': 3, 'best': 4}


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