Python | Filter String with substring at specific position
Sometimes, while working with Python string lists, we can have a problem in which we need to extract only those lists that have a specific substring at specific position. This kind of problem can come in data processing and web development domains. Lets discuss certain ways in which this task can be performed.
Method #1 : Using list comprehension + list slicing
The combination of above functionalities can be used to perform this particular task. In this, we check for substring range using list slicing and task of extraction list is compiled in list comprehension.
# Python3 code to demonstrate # Filter String with substring at specific position# using list comprehension + list slicing # Initializing listtest_list = ['geeksforgeeks', 'is', 'best', 'for', 'geeks'] # printing original listprint("The original list is : " + str(test_list)) # Initializing substringsub_str = 'geeks' # Initializing range i, j = 0, 5 # Filter String with substring at specific position# using list comprehension + list slicingres = [ele for ele in test_list if ele[i: j] == sub_str] # printing result print ("Filtered list : " + str(res)) |
The original list is : ['geeksforgeeks', 'is', 'best', 'for', 'geeks'] Filtered list : ['geeksforgeeks', 'geeks']
Method #2 : Using filter() + lambda
The combination of above methods can be used to perform this task. In this, we filter the elements using logic compiled using lambda using filter().
# Python3 code to demonstrate # Filter String with substring at specific position# using filter() + lambda # Initializing listtest_list = ['geeksforgeeks', 'is', 'best', 'for', 'geeks'] # printing original listprint("The original list is : " + str(test_list)) # Initializing substringsub_str = 'geeks' # Initializing range i, j = 0, 5 # Filter String with substring at specific position# using filter() + lambdares = list(filter(lambda ele: ele[i: j] == sub_str, test_list)) # printing result print ("Filtered list : " + str(res)) |
The original list is : ['geeksforgeeks', 'is', 'best', 'for', 'geeks'] Filtered list : ['geeksforgeeks', 'geeks']
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