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Python | Group consecutive list elements with tolerance
  • Last Updated : 27 Aug, 2019

Sometimes, we might need to group list according to the consecutive elements in list. But a useful variation of this can also be a case in which we need to consider a tolerance level, i.e allowing a skip value between number and not being exactly consecutive but a “gap” is allowed between numbers. Let’s discuss an approach in which this task can be performed.

Method : Using generator
The brute method to perform this task. Using loop and generator, one can perform this task. The slicing is taken care by the yield operator and hence this problem is solved by a small check of tolerance as well.




# Python3 code to demonstrate working of
# Group consecutive list elements with tolerance
# Using generator
  
# helper generator
def split_tol(test_list, tol):
    res = []
    last = test_list[0]
    for ele in test_list:
        if ele-last > tol:
            yield res
            res = []
        res.append(ele)
        last = ele
    yield res
  
# initializing list
test_list = [1, 2, 4, 5, 9, 11, 13, 24, 25, 26, 28]
  
# printing original list
print("The original list is : " + str(test_list))
  
# Group consecutive list elements with tolerance
# Using generator
res = list(split_tol(test_list, 2))
  
# printing result 
print("The splitted list is : " + str(res))
Output :
The original list is : [1, 2, 4, 5, 9, 11, 13, 24, 25, 26, 28]
The splitted list is : [[1, 2, 4, 5], [9, 11, 13], [24, 25, 26, 28]]

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