Open In App

Python – Filter index similar values

Improve
Improve
Like Article
Like
Save
Share
Report

Sometimes, while working with Python dictionaries, we can have a problem in which we need to extract all the values in values lists that match the filtered indices from matching values from list with some key in dictionary. This kind of application can occur in web development.

Input : test_dict = {“Gfg” : [4, 5, 7], “is” : [5, 6, 8], “best” : [10, 7, 4]} Output : {“Gfg” : [4, 5, 7], “is” : [5, 6, 8], “best” : [10, 7, 4]} Input : test_dict = {“Gfg” : [4, 20, 5, 7], “is” : [5, 17, 6, 8], “best” : [10, 11, 7, 4]} Output : {‘Gfg’: [4, 5, 7], ‘is’: [5, 6, 8], ‘best’: [10, 7, 4]}

Method #1 : Using loop + zip() + defaultdict() The combination of above methods can be used to solve this problem. In this, we initialize the defaultdict with list, bind all the keys with zip() and use loop to append the required elements. 

Python3




# Python3 code to demonstrate working of
# Filter index similar values
# Using loop + zip() + defaultdict()
from collections import defaultdict
 
# initializing dictionary
test_dict = {"Gfg" : [1, 4, 5, 6, 7], "is" : [5, 6, 8, 9, 10],
                                 "best" : [10, 7, 4, 11, 23]}
 
# printing original dictionary
print("The original dictionary : " + str(test_dict))
 
# initializing value list
filt_list = [4, 5, 7]
 
# Filter index similar values
# Using loop + zip() + defaultdict()
res = defaultdict(list)
 
for x, y, z in zip(test_dict['Gfg'], test_dict['is'], test_dict['best']):
    if x in filt_list:
        res['Gfg'].append(x)
        res['is'].append(y)
        res['best'].append(z)
     
# printing result
print("The filtered dictionary : " + str(dict(res)))


Output : 

The original dictionary : {‘Gfg’: [1, 4, 5, 6, 7], ‘is’: [5, 6, 8, 9, 10], ‘best’: [10, 7, 4, 11, 23]} The filtered dictionary : {‘Gfg’: [4, 5, 7], ‘is’: [6, 8, 10], ‘best’: [7, 4, 23]}

  Method #2 : Using list comprehension + dictionary comprehension The combination of above functionalities can be used to solve this problem. In this, we extract indices using list comprehension and filtering from other keys is done using dictionary comprehension. 

Python3




# Python3 code to demonstrate working of
# Filter index similar values
# Using list comprehension + dictionary comprehension
 
# initializing dictionary
test_dict = {"Gfg" : [1, 4, 5, 6, 7], "is" : [5, 6, 8, 9, 10],
                                 "best" : [10, 7, 4, 11, 23]}
 
# printing original dictionary
print("The original dictionary : " + str(test_dict))
 
# initializing value list
filt_list = [4, 5, 7]
 
# Filter index similar values
# Using list comprehension + dictionary comprehension
temp = [test_dict['Gfg'].index(idx) for idx in filt_list]
res = {key : [test_dict[key][idx] for idx in temp] for key in test_dict.keys()}
     
# printing result
print("The filtered dictionary : " + str(res))


Output : 

The original dictionary : {‘Gfg’: [1, 4, 5, 6, 7], ‘is’: [5, 6, 8, 9, 10], ‘best’: [10, 7, 4, 11, 23]} The filtered dictionary : {‘Gfg’: [4, 5, 7], ‘is’: [6, 8, 10], ‘best’: [7, 4, 23]}

Using a set to store the unique values:

Approach:

Create an empty set to store unique values.
Iterate through the values in the dictionary.
Convert each value to a tuple to make it hashable.
Check if the tuple is already in the set of unique values.
If the tuple is already in the set, remove the current key-value pair from the dictionary.
If the tuple is not in the set, add it to the set of unique values.
Return the filtered dictionary.

Python3




def filter_dict(test_dict):
    unique_values = set()
    for value in test_dict.values():
        unique_value = tuple(value)
        if unique_value in unique_values:
            del test_dict[key]
        else:
            unique_values.add(unique_value)
    return test_dict
test_dict = {"Gfg" : [4, 5, 7], "is" : [5, 6, 8], "best" : [107, 4]}
print(filter_dict(test_dict)) # Output: {'Gfg': [4, 5, 7], 'is': [5, 6, 8], 'best': [4]}


Output

{'Gfg': [4, 5, 7], 'is': [5, 6, 8], 'best': [10, 7, 4]}

Time complexity: O(n)
Auxiliary Space: O(n)



Last Updated : 11 Apr, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads