Given a Tuple list, filter tuples that are made from consecutive elements, i.e diff is 1.
Input : test_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 4), (6, 4, 6, 3)]
Output : [(3, 4, 5, 6)]
Explanation : Only 1 tuple adheres to condition.Input : test_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)]
Output : [(3, 4, 5, 6), (1, 2, 3)]
Explanation : Only 2 tuples adhere to condition.
Method #1: Using loop
In this, for each tuple, we call consecutive elements utility which returns True if tuple is consecutive.
Python3
# Python3 code to demonstrate working of # Filter consecutive elements Tuples# Using loop# hlpr_func def consec_check(tup): for idx in range(len(tup) - 1): # returns false if any element is not consec. if tup[idx + 1] != tup[idx] + 1: return False return True# initializing listtest_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)]# printing original listprint("The original list is : " + str(test_list))res = []for sub in test_list: # calls fnc to check consec. if consec_check(sub): res.append(sub)# printing result print("The filtered tuples : " + str(res)) |
The original list is : [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)] The filtered tuples : [(3, 4, 5, 6), (1, 2, 3)]
Time Complexity: O(n*n), where n is the length of the input list.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the list “test_list”.
Method #2: Using list comprehension
In this, we perform a similar function as above, just in one-liner shorthand using list comprehension.
Python3
# Python3 code to demonstrate working of # Filter consecutive elements Tuples# Using list comprehension# hlpr_func def consec_check(tup): for idx in range(len(tup) - 1): # returns false if any element is not consec. if tup[idx + 1] != tup[idx] + 1: return False return True# initializing listtest_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)]# printing original listprint("The original list is : " + str(test_list))# one-liner to solve problem, using list comprehensionres = [sub for sub in test_list if consec_check(sub)]# printing result print("The filtered tuples : " + str(res)) |
The original list is : [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)] The filtered tuples : [(3, 4, 5, 6), (1, 2, 3)]
Method #3 : Using min(),max() and range() methods
Python3
# Python3 code to demonstrate working of# Filter consecutive elements Tuples# Using loop# initializing listtest_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)]# printing original listprint("The original list is : " + str(test_list))res = []for i in test_list: x=list(i) a=min(x) b=max(x) y=list(range(a,b+1)) if(x==y): res.append(i)# printing resultprint("The filtered tuples : " + str(res)) |
The original list is : [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)] The filtered tuples : [(3, 4, 5, 6), (1, 2, 3)]
Method #4: Using the set() function.
Here are the steps:
- Initialize an empty list res to store the filtered tuples.
- Loop through each tuple i in the test_list.
- Convert i into a set using the set() function and store it in a variable s.
- Create a set expected using the set() function with the values in the range between the minimum and maximum values of i inclusive.
- If s and expected are equal, append i to the res list.
- Return the res list.
Python3
test_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)]print("The original list is : " + str(test_list))res = []for i in test_list: s = set(i) expected = set(range(min(i), max(i)+1)) if s == expected: res.append(i)print("The filtered tuples : " + str(res)) |
The original list is : [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)] The filtered tuples : [(3, 4, 5, 6), (1, 2, 3)]
Time complexity: O(n*k), where n is the number of tuples in the input list and k is the length of the longest tuple.
Auxiliary space: O(k) to store the sets s and expected.
Method #5: Using lambda function and filter() function with slicing
Step by step algorithm:
- Define the input list of tuples.
- Use the filter() function along with a lambda function to filter out non-consecutive tuples.
- Inside the lambda function, for each tuple x in the input list, check if it is equal to a tuple of consecutive integers using the range() function.
- If the tuple is consecutive, it is added to a new list res.
- Convert the filtered list of tuples to a list using the list() function.
- Print the filtered list of tuples.
Python3
# Define a list of tuples to filtertest_list = [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)]print("The original list is : " + str(test_list))# Use the filter() function and a lambda function to filter out non-consecutive tuples# The lambda function checks if each tuple is equal to a tuple of consecutive integersres = list(filter(lambda x: x == tuple(range(x[0], x[-1]+1)), test_list))# Print the filtered tuplesprint("The filtered tuples : " + str(res)) |
The original list is : [(3, 4, 5, 6), (5, 6, 7, 2), (1, 2, 3), (6, 4, 6, 3)] The filtered tuples : [(3, 4, 5, 6), (1, 2, 3)]
Time Complexity: O(nm), where n is the length of the input list and m is the maximum length of a tuple in the list.
Space Complexity: O(n), where n is the length of the input list.
