# Python | Group consecutive list elements with tolerance

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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