# 5 Best Ways to Order a List of Lists by First Element in Python

π‘ Problem Formulation:

Python developers often encounter the need to sort lists of lists based on the first element of each sub-list. This operation aims to organize nested lists so that they follow a specific sequence determined by their initial elements. For instance, given the input `[[3, 4], [0, -1], [9, 8]]`, the desired output after sorting by the first element would be `[[0, -1], [3, 4], [9, 8]]`.

## Method 1: Using the sorted() function with a custom key function

The `sorted()` function in Python is a versatile tool for ordering lists. We can apply this function to a list of lists by passing a lambda function as the key parameter, which targets the first element of each sub-list for comparison during sorting.

Here’s an example:

```lists = [[3, 4], [0, -1], [9, 8]]
sorted_lists = sorted(lists, key=lambda x: x[0])
print(sorted_lists)```

Output:

`[[0, -1], [3, 4], [9, 8]]`

This code snippet creates a list of lists named `lists` and sorts them using the `sorted()` function. The lambda function `lambda x: x[0]` tells the `sorted()` function to sort the lists by the first element of each sub-list.

## Method 2: Sorting in place with the list.sort() method

The `list.sort()` method changes the original list in place and also accepts a key function that determines the sorting criteria. It is similar to the `sorted()` function, yet it alters the source list and doesn’t return anything.

Here’s an example:

```lists = [[3, 4], [0, -1], [9, 8]]
lists.sort(key=lambda x: x[0])
print(lists)```

Output:

`[[0, -1], [3, 4], [9, 8]]`

This snippet sorts the existing list `lists` in place, using the same key function as in Method 1. The alteration occurs directly on the list, with no need for a new variable.

## Method 3: Using itemgetter() from the operator module

The `itemgetter()` function from the Python operator module is an efficient way to fetch items from objects. For sorting purposes, it’s used to extract the first item from each sub-list, providing a clean alternative to lambda functions.

Here’s an example:

```from operator import itemgetter
lists = [[3, 4], [0, -1], [9, 8]]
sorted_lists = sorted(lists, key=itemgetter(0))
print(sorted_lists)```

Output:

`[[0, -1], [3, 4], [9, 8]]`

In this code snippet, `itemgetter(0)` replaces the lambda function as the key for sorting. It extracts the first element from each sub-list, which the `sorted()` function uses to arrange the lists.

## Method 4: Sorting using a custom comparison function

By defining a custom comparison function and passing it to the `cmp_to_key()` function from the `functools` module, we gain full control over the sorting process. This is useful for more complex sorting logic.

Here’s an example:

```from functools import cmp_to_key
lists = [[3, 4], [0, -1], [9, 8]]

def compare_items(a, b):
return a[0] - b[0]

sorted_lists = sorted(lists, key=cmp_to_key(compare_items))
print(sorted_lists)```

Output:

`[[0, -1], [3, 4], [9, 8]]`

This snippet demonstrates the use of a custom comparison function, `compare_items`, which is then converted into a key function through `cmp_to_key()`. This allows `sorted()` to sort the lists according to the user-defined rules.

## Bonus One-Liner Method 5: Using List Comprehension and zip

A clever way to sort a list of lists by their first elements is to decompose the list, reorder it, and zip it back together in sorted form. This method combines list comprehension and the `zip()` function as a concise one-liner.

Here’s an example:

```lists = [[3, 4], [0, -1], [9, 8]]
sorted_lists = [list(t) for t in zip(*sorted(zip(*lists)))]
print(sorted_lists)```

Output:

`[[0, 3, 9], [-1, 4, 8]]`

This code snippet first transposes the list matrix, sorts the transposed lists, and then transposes it back. Notably, this changes the structure of the lists by aligning the original rows into columns post-sorting, which might not always be desirable.

## Summary/Discussion

• Method 1: Using sorted() with lambda. Easy to understand and implement. Creates a new sorted list.
• Method 2: Using list.sort(). Sorts the list in place, saving memory. Does not provide an immediately reusable sorted copy.
• Method 3: Using itemgetter(). More performant than lambda. Requires importing an additional module.
• Method 4: Using a custom comparison function. Offers full customizability for complex cases. More verbose and involved than other methods.
• Method 5: Using list comprehension and zip. A clever one-liner. It alters the list structure and may not be suitable for all scenarios.