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Python – Filter odd elements from value lists in dictionary
  • Last Updated : 22 Apr, 2020

Sometimes, while working with Python dictionaries we can have problem in which we need to perform the removal of odd elements from values list of dictionaries. This can have application in many domains including web development. Lets discuss certain ways in which this task can be performed.

Method #1 : Using list comprehension + dictionary comprehension
This is brute force one liner in which this task can be performed. In this, we remake new dictionary using dictionary comprehension in which filtering data and iteration in value list is performed using list comprehension.

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# Python3 code to demonstrate working of 
# Remove odd elements from value lists in dictionary
# Using list comprehension + dictionary comprehension
  
# initializing Dictionary
test_dict = {'gfg' : [1, 3, 4, 10], 'is' : [1, 2, 8], 'best' : [4, 3, 7]}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# Remove odd elements from value lists in dictionary
# Using list comprehension + dictionary comprehension
res = {key: [idx for idx in val if idx % 2
          for key, val in test_dict.items()}
  
# printing result 
print("The filtered values are : " + str(res)) 

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Output :

The original dictionary is : {‘gfg’: [1, 3, 4, 10], ‘best’: [4, 3, 7], ‘is’: [1, 2, 8]}
The filtered values are : {‘gfg’: [1, 3], ‘is’: [1], ‘best’: [3, 7]}

 



Method #2 : Using dictionary comprehension + filter() + lambda
This is yet another way to solve this problem. In this, the task performed by list comprehension is done using filter() + lambda.

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# Python3 code to demonstrate working of 
# Remove odd elements from value lists in dictionary
# Using filter() + lambda + dictionary comprehension
  
# initializing Dictionary
test_dict = {'gfg' : [1, 3, 4, 10], 'is' : [1, 2, 8], 'best' : [4, 3, 7]}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# Remove odd elements from value lists in dictionary
# Using filter() + lambda + dictionary comprehension
res = {key: list(filter(lambda ele: (ele % 2), val)) 
                  for key, val in test_dict.items()}
  
# printing result 
print("The filtered values are : " + str(res)) 

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Output :

The original dictionary is : {‘gfg’: [1, 3, 4, 10], ‘best’: [4, 3, 7], ‘is’: [1, 2, 8]}
The filtered values are : {‘gfg’: [1, 3], ‘is’: [1], ‘best’: [3, 7]}

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