Sometimes, while working with Python lists, we can have a problem in which we might need to group the list elements on basis of their respective consecutiveness. This kind of problem can occur while data handling. Let’s discuss certain way in which this task can be performed.
Method 1: Using enumerate() + groupby() + generator function + lambda This task can be performed using the combination of above functions. In this, we create a generator function, in which we pass the list whose index-element are accessed using enumerate() and grouped by consecutive elements using groupby() and lambda. Works with Python2 only
Python
# Python code to demonstrate working of# Consecutive elements grouping list# using enumerate() + groupby() + generator function + lambdaimport itertools# Utility Generator Functiondef groupc(test_list): for x, y in itertools.groupby(enumerate(test_list), lambda (a, b): b - a): y = list(y) yield y[0][1], y[-1][1]# initialize listtest_list = [1, 2, 3, 6, 7, 8, 11, 12, 13]# printing original listprint("The original list is : " + str(test_list))# Consecutive elements grouping list# using enumerate() + groupby() + generator function + lambdares = list(groupc(test_list))# printing resultprint("Grouped list is : " + str(res)) |
The original list is : [1, 2, 3, 6, 7, 8, 11, 12, 13] Grouped list is : [(1, 3), (6, 8), (11, 13)]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(1)
Method 2: Using a while loop:
Use two nested while loops to find consecutive elements in the list. The outer loop iterates over the elements in the list, and the inner loop finds the end of the consecutive sequence. The result is stored in a list of tuples, with each tuple containing the first and last elements of a consecutive sequence.
Python3
def group_consecutive(test_list): result = [] i = 0 while i < len(test_list): j = i while j < len(test_list) - 1 and test_list[j+1] == test_list[j]+1: j += 1 result.append((test_list[i], test_list[j])) i = j + 1 return result# initialize listtest_list = [1, 2, 3, 6, 7, 8, 11, 12, 13]# printing original listprint("The original list is : " + str(test_list))# Consecutive elements grouping listres = group_consecutive(test_list)# printing resultprint("Grouped list is : " + str(res)) |
The original list is : [1, 2, 3, 6, 7, 8, 11, 12, 13] Grouped list is : [(1, 3), (6, 8), (11, 13)]
Time complexity: O(n), where n is the length of the input list,
Auxiliary space: O(1).
Method 3: Using a for loop and a temporary list.
Use a for loop and a temporary list to group consecutive elements in the input list.
Python3
# initialize listtest_list = [1, 2, 3, 6, 7, 8, 11, 12, 13]# printing original listprint("The original list is : " + str(test_list))# Consecutive elements grouping list# using a for loop and a temporary listres = []temp = [test_list[0]]for i in range(1, len(test_list)): if test_list[i] == test_list[i-1] + 1: temp.append(test_list[i]) else: res.append((temp[0], temp[-1])) temp = [test_list[i]]res.append((temp[0], temp[-1]))# printing resultprint("Grouped list is : " + str(res)) |
The original list is : [1, 2, 3, 6, 7, 8, 11, 12, 13] Grouped list is : [(1, 3), (6, 8), (11, 13)]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), since it creates two lists: res and temp.
Method 4: Using itertools.groupby() and a list comprehension
This method uses itertools.groupby() to group consecutive elements together and then creates a list of tuples with the first and last element of each group using a list comprehension.
Python3
import itertoolstest_list = [1, 2, 3, 6, 7, 8, 11, 12, 13]# group consecutive elements using itertools.groupby()groups = []for k, g in itertools.groupby(enumerate(test_list), lambda x: x[0]-x[1]): groups.append(list(map(lambda x: x[1], g)))# create a list of tuples with the first and last element of each groupres = [(group[0], group[-1]) for group in groups]# print resultprint("Grouped list is : " + str(res)) |
Grouped list is : [(1, 3), (6, 8), (11, 13)]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), where n is the length of the input list.
Method 5: Using numpy.diff() and numpy.split()
This method uses numpy’s diff() function to calculate the differences between consecutive elements of the input list. It then uses numpy’s split() function to split the list into subarrays where the differences are not equal to 1. Finally, it creates a list of tuples with the first and last element of each subarray.
Python3
import numpy as nptest_list = [1, 2, 3, 6, 7, 8, 11, 12, 13]# calculate the differences between consecutive elementsdiffs = np.diff(test_list)# split the list into subarrays where the differences are not equal to 1groups = np.split(test_list, np.where(diffs != 1)[0]+1)# create a list of tuples with the first and last element of each groupres = [(group[0], group[-1]) for group in groups]# print resultprint("Grouped list is : " + str(res)) |
Output:
Grouped list is : [(1, 3), (6, 8), (11, 13)]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), as it needs to create temporary arrays to store the differences and subarrays of the input list.
