# Python – Fill gaps in consecutive Records

Sometimes, while working with Python records, we can have a problem in which we have consecutive records, but a few missing and needs to be filled with any constant K. This kind of problem can have application in domains such as web development. Let’s discuss certain ways in which we need to perform this task.

Input :
test_list = [(1, 4), (9, 11)]
K = None
Output : [(1, 4), (2, None), (3, None), (4, None), (5, None), (6, None), (7, None), (8, None), (9, 11)]

Input :
test_list = [(1, 4), (2, 11)]
K = None
Output : [(1, 4), (2, 11)]

Method #1 : Using loop
This is one way in which this problem can be solved. In this, we check at each iteration if the element exists, if no, then it is filled with required value, if yes, the original value is retained.

 `# Python3 code to demonstrate working of  ` `# Fill gaps in consecutive Records ` `# Using loop ` ` `  `# initializing list ` `test_list ``=` `[(``1``, ``4``), (``3``, ``5``), (``4``, ``6``), (``7``, ``8``), (``9``, ``11``)] ` ` `  `# printing original list ` `print``(``"The original list is : "` `+` `str``(test_list)) ` ` `  `# initializing K value ` `K ``=` `"New"` ` `  `# Fill gaps in consecutive Records ` `# Using loop ` `res ``=` `[] ` `cnt ``=` `0` `for` `i, j ``in` `test_list: ` `    ``if` `i ``-` `cnt > ``1``: ` `        ``for` `k ``in` `range``(cnt ``+` `1``, i): ` `            ``res.append((k, K)) ` `    ``res.append((i, j)) ` `    ``cnt ``=` `i ` ` `  `# printing result  ` `print``(``"The list after filling gaps : "` `+` `str``(res))  `

Output :

The original list is : [(1, 4), (3, 5), (4, 6), (7, 8), (9, 11)]
The list after filling gaps : [(1, 4), (2, ‘New’), (3, 5), (4, 6), (5, ‘New’), (6, ‘New’), (7, 8), (8, ‘New’), (9, 11)]

Method #2 : Using `min() + max() + dict()` + list comprehension
This is yet another way in which this problem can be solved. In this, we check the range using min() and max() and use dictionary as a container of getting values that is required to fill in.

 `# Python3 code to demonstrate working of  ` `# Fill gaps in consecutive Records ` `# Using min() + max() + dict() + list comprehension ` ` `  `# initializing list ` `test_list ``=` `[(``1``, ``4``), (``3``, ``5``), (``4``, ``6``), (``7``, ``8``), (``9``, ``11``)] ` ` `  `# printing original list ` `print``(``"The original list is : "` `+` `str``(test_list)) ` ` `  `# initializing K value ` `K ``=` `"New"` ` `  `# Fill gaps in consecutive Records ` `# Using min() + max() + dict() + list comprehension ` `test_list ``=` `dict``(test_list) ` `mini, maxi ``=` `min``(test_list), ``max``(test_list) ` `res ``=` `[(idx, test_list.get(idx)) ``for` `idx ``in` `range``(mini, maxi ``+` `1``)] ` ` `  `# printing result  ` `print``(``"The list after filling gaps : "` `+` `str``(res))  `

Output :

The original list is : [(1, 4), (3, 5), (4, 6), (7, 8), (9, 11)]
The list after filling gaps : [(1, 4), (2, ‘New’), (3, 5), (4, 6), (5, ‘New’), (6, ‘New’), (7, 8), (8, ‘New’), (9, 11)]

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