Python – Remove digits from Dictionary String Values List

Given list of dictionaries with string list values, remove all the numerics from all strings.

Input : test_dict = {'Gfg' : ["G4G is Best 4", "4 ALL geeks"], 
                    'best' : ["Gfg Heaven", "for 7 CS"]} 
                    
Output : {'Gfg': ['GG is Best ', ' ALL geeks'], 
                'best': ['Gfg Heaven', 'for CS']} 

Explanation : All the numeric strings are removed. 

Input : test_dict = {'Gfg' : ["G4G is Best 4", "4 ALL geeks"]} 

Output : {'Gfg': ['GG is Best ', ' ALL geeks']} 

Explanation : All the numeric strings are removed.

Method : Using regex + dictionary comprehension

This problem can be solved using combination of both functionalities. In this, we apply regex to replace each digit with an empty string and compile the result using dictionary comprehension.

Python3

# Python3 code to demonstrate working of
# Remove digits from Dictionary String Values List
# Using loop + regex() + dictionary comprehension
import re

# initializing dictionary
test_dict = {'Gfg' : ["G4G is Best 4", "4 ALL geeks"],
             'is' : ["5 6 Good"],
             'best' : ["Gfg Heaven", "for 7 CS"]}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# using dictionary comprehension to go through all keys
res = {key: [re.sub('\d', '', ele) for ele in val]
       for key, val in test_dict.items()}

# printing result
print("The filtered dictionary : " + str(res))

Output

The original dictionary is : {'Gfg': ['G4G is Best 4', '4 ALL geeks'], 'is': ['5 6 Good'], 'best': ['Gfg Heaven', 'for 7 CS']}
The filtered dictionary : {'Gfg': ['GG is Best ', ' ALL geeks'], 'is': ['  Good'], 'best': ['Gfg Heaven', 'for  CS']}

Method : Without using regex

Python3

# Python3 code to demonstrate working of
# Remove digits from Dictionary String Values List

def fun(s):
    x=""
    for i in s:
        if not i.isdigit():
            x+=i
    return x
# initializing dictionary
test_dict = {'Gfg' : ["G4G is Best 4", "4 ALL geeks"],
            'is' : ["5 6 Good"],
            'best' : ["Gfg Heaven", "for 7 CS"]}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# using dictionary comprehension to go through all keys
res=dict()
for i in test_dict:
    v=[]
    for j in test_dict[i]:
        v.append(fun(j))
    res[i]=v

# printing result
print("The filtered dictionary : " + str(res))

Output

The original dictionary is : {'Gfg': ['G4G is Best 4', '4 ALL geeks'], 'is': ['5 6 Good'], 'best': ['Gfg Heaven', 'for 7 CS']}
The filtered dictionary : {'Gfg': ['GG is Best ', ' ALL geeks'], 'is': ['  Good'], 'best': ['Gfg Heaven', 'for  CS']}

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n²)

Auxiliary Space: O(n)