The problem of flattening a list is quite common, but sometimes, we need to perform the flattening but in the grouped manner, i.e getting the sublists of size K and flatten them. This particular utility has use-cases in many domains including web-development and day-day programming as well. Let’s discuss certain ways in which this can be done.
Method #1 : Using list comprehension + list slicing This problem can be approached by using the list comprehension and also applying list slicing to limit the grouping and binding the list is done as a part of list comprehension logic.
Python3
# Python3 code to demonstrate # group flattening of list # using list comprehension + list slicing# initializing list of liststest_list = [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]]# printing original list print("The original list : " + str(test_list))# declaring K K = 3# using list comprehension + list slicing# group flattening of list res = [[i for sub in test_list[j : j + K] for i in sub] for j in range(0, len(test_list), K)]# printing result print("The grouped flattened list : " + str(res)) |
The original list : [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]] The grouped flattened list : [[1, 3, 3, 4, 6, 5], [4, 5, 7, 6, 7, 9]]
Time Complexity: O(N * K), where N is the number of sublists in test_list and K is the size of the sublist.
Space Complexity: O(N * K)..
Method #2 : Using list comprehension + zip() + map() This problem can also be solved using the combination of the above 3 functions. The zip function is used to get all the groups into one and map function is used to convert the iterator into list and list comprehension performs the task of grouping.
Python3
# Python3 code to demonstrate # group flattening of list # using list comprehension + zip() + map()# initializing list of liststest_list = [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]]# printing original list print("The original list : " + str(test_list))# declaring K K = 3# using list comprehension + zip() + map()# group flattening of list res = list(map(list, zip(*[iter([i for sub in test_list for i in sub])]*(K * len(test_list[0])))))# printing result print("The grouped flattened list : " + str(res)) |
The original list : [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]] The grouped flattened list : [[1, 3, 3, 4, 6, 5], [4, 5, 7, 6, 7, 9]]
Method #3: Using itertools.zip_longest()
Iterate over the sublists in the input list in groups of size K and Flatten the group and append it to the result list
Python3
import itertools# Python3 code to demonstrate# group flattening of list# using list comprehension + list slicing# initializing list of listsl = [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]]# printing original listprint("The original list : " + str(l))# declaring KK = 3# Initialize an empty result listresult = []# Iterate over the sublists in the input list in groups of size Kfor group in itertools.zip_longest(*[iter(l)] * K): # Flatten the group and append it to the result list result.append([item for sublist in group for item in sublist])# Return the result listprint(result)#This code is contributed by Edula Vinay Kumar Reddy |
The original list : [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]] [[1, 3, 3, 4, 6, 5], [4, 5, 7, 6, 7, 9]]
The time complexity of the above code is O(n), where n is the number of elements in the input list. This is because the code iterates over the input list once to create the groups and then iterates over each group to flatten it.
The space complexity is also O(n), as the result list has the same number of elements as the input list. The itertools.zip_longest() function creates an iterator that returns tuples containing the elements of the input list in groups of size K, but this iterator is not stored in memory, so it does not contribute to the space complexity.
Method #4: Using nested loops
Step-by-step approach:
- Initialize the list of lists named test_list.
- Print the original list using the print() function.
- Declare a variable K with value 3.
- Initialize an empty list res.
- Use a for loop to iterate through the elements of the list test_list. The loop starts at 0, ends at the length of test_list, and increments by K.
- Create an empty list named temp inside the outer loop.
- Use another for loop to iterate through the sublists of test_list. The loop iterates from j to j+K, where j is the current value of the outer loop.
- Use another for loop inside the previous for loop to iterate through the elements of each sublist.
- Append each element to the list temp.
- Append temp to the list res.
- Print the result using the print() function.
Python3
# Python3 code to demonstrate # group flattening of list # using for loop and append()# initializing list of liststest_list = [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]]# printing original list print("The original list : " + str(test_list))# declaring K K = 3# using for loop and append()# group flattening of list res = []for j in range(0, len(test_list), K): temp = [] for sub in test_list[j:j+K]: for i in sub: temp.append(i) res.append(temp)# printing result print("The grouped flattened list : " + str(res)) |
The original list : [[1, 3], [3, 4], [6, 5], [4, 5], [7, 6], [7, 9]] The grouped flattened list : [[1, 3, 3, 4, 6, 5], [4, 5, 7, 6, 7, 9]]
Time complexity: O(N^2), where N is the length of the input list.
Auxiliary space: O(NK), where N is the length of the input list and K is the group size.
