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Python | Find groups of strictly increasing numbers in a list

  • Last Updated : 08 Mar, 2019

Given a list of integers, write a Python program to find groups of strictly increasing numbers.

Examples:

Input : [1, 2, 3, 5, 6]
Output : [[1, 2, 3], [5, 6]]

Input : [8, 9, 10, 7, 8, 1, 2, 3]
Output : [[8, 9, 10], [7, 8], [1, 2, 3]]

Approach #1 : Pythonic naive

This is a naive approach which uses an extra input list space. It makes use of a for loop and in every iteration, it checks if the next element increments from previous by 1. If yes, append it to the current sublist, otherwise, create another sublist.




# Python3 program to Find groups 
# of strictly increasing numbers within 
  
def groupSequence(lst):
    res = [[lst[0]]]
  
    for i in range(1, len(lst)):
        if lst[i-1]+1 == lst[i]:
            res[-1].append(lst[i])
  
        else:
            res.append([lst[i]])
    return res
      
# Driver program 
l = [8, 9, 10, 7, 8, 1, 2, 3]
print(groupSequence(l))
Output:



[[8, 9, 10], [7, 8], [1, 2, 3]]

 
Approach #2 : Alternate naive

This is an alternative to the above mentioned naive approach. This method is quite simple and straight. It constructs a start_bound list and an end_bound list, which contains the position of starting and ending sequence of increasing integers. Thus simply return the bounds using for loops.




# Python3 program to Find groups 
# of strictly increasing numbers within 
  
def groupSequence(l):
    start_bound = [i for i in range(len(l)-1)
        if (l == 0 or l[i] != l[i-1]+1)
        and l[i + 1] == l[i]+1]
  
    end_bound = [i for i in range(1, len(l))
        if l[i] == l[i-1]+1 and
        (i == len(l)-1 or l[i + 1] != l[i]+1)]
  
    return [l[start_bound[i]:end_bound[i]+1]
    for i in range(len(start_bound))]
      
# Driver program 
l = [8, 9, 10, 7, 8, 1, 2, 3]
print(list(groupSequence(l)))
Output:
[[8, 9, 10], [7, 8], [1, 2, 3]]

 
Approach #3 : Using iterable and yield
This approach uses another list ‘res’ and an iterable ‘it’. A variable ‘prev’ is used for keeping the record of previous integer and start is used for getting the starting position of the increasing sequence. Using a loop, in every iteration, we check if start element is a successor of prev or not. If yes, we append it to res, otherwise, we simply yield the res + [prev] as list element.




# Python3 program to Find groups 
# of strictly increasing numbers within 
  
def groupSequence(x):
    it = iter(x)
    prev, res = next(it), []
  
    while prev is not None:
        start = next(it, None)
  
        if prev + 1 == start:
            res.append(prev)
        elif res:
            yield list(res + [prev])
            res = []
        prev = start
      
# Driver program 
l = [8, 9, 10, 7, 8, 1, 2, 3]
print(list(groupSequence(l)))
Output:
[[8, 9, 10], [7, 8], [1, 2, 3]]

 
Approach #4 : Using itertools
Python itertools provides operations like cycle and groupby which are used in this method. First we form another list ‘temp_list‘ using cycle. Cycle generates an infinitely repeating series of values. Then we group the temp_list accordingly using groupby operation and finally yield the desired output.




# Python3 program to Find groups 
# of strictly increasing numbers within 
from itertools import groupby, cycle
  
def groupSequence(l):
    temp_list = cycle(l)
  
    next(temp_list)
    groups = groupby(l, key = lambda j: j + 1 == next(temp_list))
    for k, v in groups:
        if k:
            yield tuple(v) + (next((next(groups)[1])), )
  
      
# Driver program 
l = [8, 9, 10, 7, 8, 1, 2, 3]
print(list(groupSequence(l)))
Output:
[(8, 9, 10), (7, 8), (1, 2, 3)]

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