Given a list of elements, the task is to write a Python program to perform a join of consecutive strings according to the suffix of each string.
Input : test_list = [“Geeks-“, “for-“, “Geeks”, “is”, “best-“, “for”, “geeks”, suff = ‘-‘
Output : [‘Geeks-for-Geeks’, ‘is’, ‘best-for’, ‘geeks’]
Explanation : Strings are joined to next which have “-” as suffix.Input : test_list = [“Geeks*”, “for*”, “Geeks”, “is”, “best*”, “for”, “geeks”, suff = ‘*’
Output : [‘Geeks*for*Geeks’, ‘is’, ‘best*for’, ‘geeks’]
Explanation : Strings are joined to next which have “*” as suffix.
Approach : Using loop + endswith() + join()
In this we perform the task of joining using join() and endswith() performs the task of conditional checks for suffix as defined.
Python3
# Python3 code to demonstrate working of # Selective consecutive Suffix Join # Using loop + endswith() + join() # initializing list test_list = [ "Geeks-" , "for-" , "Geeks" , "is" , "best-" , "for" , "geeks" ] # printing original list print ( "The original list is : " + str (test_list)) # initializing suffix suff = '-' res = [] temp = [] for ele in test_list: temp.append(ele) # conditionally test values if not ele.endswith(suff): res.append(''.join(temp)) temp = [] if temp: res.append(''.join(temp)) # printing result print ( "The joined result : " + str (res)) |
Output:
The original list is : [‘Geeks-‘, ‘for-‘, ‘Geeks’, ‘is’, ‘best-‘, ‘for’, ‘geeks’]
The joined result : [‘Geeks-for-Geeks’, ‘is’, ‘best-for’, ‘geeks’]
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