Python | Grouping similar substrings in list
Sometimes we have an application in which we require to group common prefix strings into one such that further processing can be done according to the grouping. This type of grouping is useful in the cases of Machine Learning and Web Development. Let’s discuss certain ways in which this can be done.
Method #1 : Using lambda + itertools.groupby() + split()
The combination of above three functions help us achieve the task. The split method is key as it defines the separator by which grouping has to be performed. The groupby function does the grouping of elements.
Python3
# Python3 code to demonstrate# group similar substrings# using lambda + itertools.groupby() + split()from itertools import groupby# initializing listtest_list = ['geek_1', 'coder_2', 'geek_4', 'coder_3', 'pro_3']# sort list# essential for groupingtest_list.sort()# printing the original listprint ("The original list is : " + str(test_list))# using lambda + itertools.groupby() + split()# group similar substringsres = [list(i) for j, i in groupby(test_list, lambda a: a.split('_')[0])]# printing resultprint ("The grouped list is : " + str(res)) |
The original list is : [‘coder_2’, ‘coder_3’, ‘geek_1’, ‘geek_4’, ‘pro_3’]
The grouped list is : [[‘coder_2’, ‘coder_3’], [‘geek_1’, ‘geek_4’], [‘pro_3’]]
Method #2 : Using lambda + itertools.groupby() + partition()
The similar task can also be performed replacing the split function with the partition function. This is more efficient way to perform this task as it uses the iterators and hence internally quicker.
Python3
# Python3 code to demonstrate# group similar substrings# using lambda + itertools.groupby() + partition()from itertools import groupby# initializing listtest_list = ['geek_1', 'coder_2', 'geek_4', 'coder_3', 'pro_3']# sort list# essential for groupingtest_list.sort()# printing the original listprint ("The original list is : " + str(test_list))# using lambda + itertools.groupby() + partition()# group similar substringsres = [list(i) for j, i in groupby(test_list, lambda a: a.partition('_')[0])]# printing resultprint ("The grouped list is : " + str(res)) |
The original list is : [‘coder_2’, ‘coder_3’, ‘geek_1’, ‘geek_4’, ‘pro_3’]
The grouped list is : [[‘coder_2’, ‘coder_3’], [‘geek_1’, ‘geek_4’], [‘pro_3’]]
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