Python | Prefix key match in dictionary
Last Updated :
08 May, 2023
Sometimes, while working with dictionaries, we can have a problem in which we need to find the dictionary items that have some constraints on keys. One such constraint can be a prefix match on keys. Let’s discuss certain ways in which this task can be performed.
Method #1: Using dictionary comprehension + startswith()
The combination of above two methods can be used to perform this particular task. In this, dictionary comprehension does the basic task of dictionary construction and startswith() performs the utility task of checking keys starting with specific prefix.
Python3
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
res = {key: val for key, val in test_dict.items()
if key.startswith(test_pref)}
print("Filtered dictionary keys are : " + str(res))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
Time complexity: O(n), where n is the number of keys in the dictionary.
Auxiliary Space: O(m), where m is the number of keys in the filtered dictionary
Method #2 : Using map() + filter() + items() + startswith() This particular task can also be performed using the combination of above functions. The map function ties the filter logic of startswith() to each dictionary’s items extracted by items()
Python3
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
res = dict(filter(lambda item: item[0].startswith(test_pref),
test_dict.items()))
print("Filtered dictionary keys are : " + str(res))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
Time complexity: O(n), where n is the number of items in the dictionary. This is because we are iterating over the items in the dictionary and applying the startswith function to each key, which takes O(m), where m is the length of the key.
Auxiliary Space: O(k), where k is the number of items that match the prefix, because the filtered dictionary contains only the items that match the prefix.
Method #3: Using find() method
Python3
test_dict = {'tough' : 1, 'to' : 2, 'do' : 3, 'todays' : 4, 'work' : 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
res = {key:val for key, val in test_dict.items()
if key.find(test_pref)==0}
print("Filtered dictionary keys are : " + str(res))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
Time complexity: O(n) where n is the number of items in the dictionary.
Auxiliary space: O(m) where m is the number of items in the filtered dictionary.
Method #4 : use re.match()
Python3
import re
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
res = {key: val for key, val in test_dict.items()
if re.match(test_pref, key)}
print("Filtered dictionary keys are : " + str(res))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5 : Using slicing
Python3
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
res=dict()
for i in list(test_dict.keys()):
if(i[:len(test_pref)]==test_pref):
res[i]=test_dict[i]
print("Filtered dictionary keys are : " + str(res))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 6: Using a list comprehension
Step-by-step approach
- Initialize the original dictionary.
- Print the original dictionary.
- Initialize the prefix string.
- Use a list comprehension to filter out the keys that start with the given prefix.
- Create a new dictionary using the filtered keys and their corresponding values from the original dictionary.
- Print the filtered dictionary.
Python3
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
filtered_keys = [key for key in test_dict.keys() if key.startswith(test_pref)]
filtered_dict = {key: test_dict[key] for key in filtered_keys}
print("Filtered dictionary keys are : " + str(filtered_dict))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
The time complexity of this method is O(n), where n is the number of keys in the dictionary.
The space complexity is O(k), where k is the number of keys that start with the given prefix.
Method #7: Using a for loop
Step-by-step explanation:
- Initialize the dictionary test_dict.
- Print the original dictionary using print().
- Initialize the prefix test_pref.
- Initialize an empty dictionary filtered_dict.
- Use a for loop to iterate through each key in test_dict.
- Use the startswith() method to check if the key starts with test_pref.
- If the key starts with test_pref, add the key-value pair to filtered_dict.
- Print the filtered dictionary using print().
Python3
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
print("The original dictionary : " + str(test_dict))
test_pref = 'to'
filtered_dict = {}
for key in test_dict.keys():
if key.startswith(test_pref):
filtered_dict[key] = test_dict[key]
print("Filtered dictionary keys are : " + str(filtered_dict))
|
Output
The original dictionary : {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}
Filtered dictionary keys are : {'tough': 1, 'to': 2, 'todays': 4}
Time complexity: O(n), where n is the number of keys in the dictionary.
Auxiliary space: O(k), where k is the number of keys that start with
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