Python | Identical Strings Grouping
Last Updated :
09 Apr, 2023
Sometimes, we need to perform the conventional task of grouping some like Strings into a separate list and thus forming a list of list. This can also help in counting and also get the sorted order of elements. Let’s discuss certain ways in which this can be done.
Method #1: Using collections.Counter()
This particular function can prove to be quite useful to perform this particular task as it counts the frequency of Strings in the list and then we can pair them using the list comprehension.
Python3
import collections
test_list = ["Gfg", "best", "is", "Gfg", "is", "best", "Gfg", "best"]
print("The original list : " + str(test_list))
temp = collections.Counter(test_list)
res = [[i] * j for i, j in temp.items()]
print("The Strings after grouping are : " + str(res))
|
Output :
The original list : ['Gfg', 'best', 'is', 'Gfg', 'is', 'best', 'Gfg', 'best']
The Strings after grouping are : [['best', 'best', 'best'], ['Gfg', 'Gfg', 'Gfg'], ['is', 'is']]
The time complexity of the code is O(n), where n is the length of the input list.
The auxiliary space complexity of the code is also O(n), as the space required for the Counter object and the resulting list both depend on the number of unique strings in the input list, which can be at most n.
Method #2: Using itertools.groupby()
This problem can easily solved by the traditional groupby functionality that is offered by Python via groupby function, which groups the like elements as suggested by name.
Python3
import itertools
test_list = ["Gfg", "best", "is", "Gfg", "is", "best", "Gfg", "best"]
print("The original list : " + str(test_list))
res = [list(i) for j, i in itertools.groupby(sorted(test_list))]
print("The Strings after grouping are : " + str(res))
|
Output :
The original list : ['Gfg', 'best', 'is', 'Gfg', 'is', 'best', 'Gfg', 'best']
The Strings after grouping are : [['best', 'best', 'best'], ['Gfg', 'Gfg', 'Gfg'], ['is', 'is']]
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the list “test_list”.
Time complexity: The time complexity of this code is O(nlogn), where n is the length of the input list test_list.
Auxiliary space: The auxiliary space used by this code is O(n), where n is the length of the input list test_list.
Method #3 : Using count() method
Python3
test_list = ["Gfg", "best", "is", "Gfg", "is", "best", "Gfg", "best"]
print("The original list : " + str(test_list))
res=[]
x=list(set(test_list))
x.sort()
for i in x:
a=[i]*test_list.count(i)
res.append(a)
print("The Strings after grouping are : " + str(res))
|
Output
The original list : ['Gfg', 'best', 'is', 'Gfg', 'is', 'best', 'Gfg', 'best']
The Strings after grouping are : [['Gfg', 'Gfg', 'Gfg'], ['best', 'best', 'best'], ['is', 'is']]
Method #4 : Using operator.countOf() method
Python3
test_list = ["Gfg", "best", "is", "Gfg", "is", "best", "Gfg", "best"]
print("The original list : " + str(test_list))
res=[]
x=list(set(test_list))
x.sort()
import operator
for i in x:
a=[i]*operator.countOf(test_list,i)
res.append(a)
print("The Strings after grouping are : " + str(res))
|
Output
The original list : ['Gfg', 'best', 'is', 'Gfg', 'is', 'best', 'Gfg', 'best']
The Strings after grouping are : [['Gfg', 'Gfg', 'Gfg'], ['best', 'best', 'best'], ['is', 'is']]
Time Complexity : O(N)
Auxiliary Space : O(N)
METHOD 5: using a dictionary to group identical strings:
This method creates an empty dictionary res and iterates over the elements of the test_list. For each element s, it checks if it already exists in the dictionary. If it does, it appends s to the list corresponding to the key s. If it doesn’t, it creates a new list with s as its only element and assigns it to the key s in the dictionary. Finally, it converts the dictionary values to a list and prints the result.
Python3
test_list = ["Gfg", "best", "is", "Gfg", "is", "best", "Gfg", "best"]
print("The original list : " + str(test_list))
res = {}
for s in test_list:
if s in res:
res[s].append(s)
else:
res[s] = [s]
res = list(res.values())
print("The Strings after grouping are : " + str(res))
|
Output
The original list : ['Gfg', 'best', 'is', 'Gfg', 'is', 'best', 'Gfg', 'best']
The Strings after grouping are : [['Gfg', 'Gfg', 'Gfg'], ['best', 'best', 'best'], ['is', 'is']]
The time complexity of the above Python code is O(n), where n is the length of the input list test_list
The auxiliary space complexity of the code is O(k), where k is the number of unique elements in the input list.
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