# Python | Remove Front K elements

• Last Updated : 19 Feb, 2020

We often come to the situations in which we need to decrease the size of the list by truncating the first elements of the list. This particular problem occurs when we need to optimize memory. This has its application in the day-day programming when sometimes we require to get all the lists of similar size. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using `len()` + list slicing
List slicing can perform this particular task in which we just slice the last len(list) – K elements to be in the list and hence removing the first K elements.

 `# Python code to demonstrate ``# Remove Front K elements``# using len() + list slicing`` ` `# initializing list ``test_list ``=` `[``1``, ``4``, ``6``, ``3``, ``5``, ``8``]`` ` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# initializing K ``K ``=` `3`` ` `# using len() + list slicing``# Remove Front K elements``res ``=` `test_list[K:]`` ` `# printing result ``print` `(``"The list after removing first K elements : "` `+` `str``(res))`

Output :

```The original list is : [1, 4, 6, 3, 5, 8]
The list after removing first K elements : [3, 5, 8]
```

Method #2 : Using Negative list slicing
We can perform this particular task using the negative list slicing in which we start removing the elements from the first index of the list and hence remove all the K elements from the first. We remove None if 0 elements are asked to be removed.

 `# Python code to demonstrate ``# Remove Front K elements``# using negative list slicing`` ` `# initializing list ``test_list ``=` `[``1``, ``4``, ``6``, ``3``, ``5``, ``8``]`` ` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# initializing K ``K ``=` `3`` ` `# using negative list slicing``# Remove Front K elements``res ``=` `test_list[``-``(``len``(test_list) ``-` `K) ``or` `None``:]`` ` `# printing result ``print` `(``"The list after removing first K elements : "` `+` `str``(res))`

Output :

```The original list is : [1, 4, 6, 3, 5, 8]
The list after removing first K elements : [3, 5, 8]
```

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