Member Avatar for rasizzle

Hi all,

I want to take this code:

list_in_list = [[1,5,13,30,-4], [1,5,10,30,-4], [1,5,15,30,-4], [1,5,9,30,-4], [1,5,16,30,-8]]

and sort it by the 3rd item in each sublist (list_in_list[x][2])

I'd like to do it by the highest number so my output would be:

sorted_list_in_list = [[1,5,16,30,-8],[1,5,15,30,-4], [1,5,13,30,-4],[1,5,10,30,-4], [1,5,9,30,-4]]

Any help or suggestions?

Edited by rasizzle because: n/a
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  • Latest Post Latest Post by vegaseat

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Good call by — using an efficient key function is the right idea. Another short, idiomatic alternative is to use sorted() with a small key; that returns a new list rather than changing the original:

sorted_rows = sorted(rows, key=lambda r: r[2], reverse=True)

Note the practical differences: list.sort() sorts in place and returns None, while sorted() produces a new list. Python’s sort is stable, so you can either return a tuple from the key to sort by multiple fields at once, or do a stable two-pass sort when you need different directions for each key:

# two-pass: secondary ascending, then primary descending
rows.sort(key=lambda r: r[3])
rows.sort(key=lambda r: r[2], reverse=True)

Watch out for common pitfalls: IndexError if some sublists are shorter, and TypeError if the third items are mixed types (strings vs numbers). A safe key wrapper avoids exceptions and defines how to treat missing values:

def safe_third(row):
    return row[2] if len(row) > 2 else float('-inf')

rows_sorted = sorted(rows, key=safe_third, reverse=True)

As noted, there are many ways to do this; operator.itemgetter (what used) is slightly faster than a lambda for fixed-index access because it’s implemented in C. For practical tips, stability, and trade-offs, see the Python Sorting HOWTO: Python Sorting HOWTO.

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Member Avatar for rasizzle

It's difficult for me to translate the tuple and dictionary code to a simple list. Can anyone help explain it to me with my code?

Member Avatar for rasizzle

Figured it out:

import operator

myreturn = operator.itemgetter(2)

list_in_list = [[1,5,13,30,-4], [1,5,10,30,-4], [1,5,15,30,-4], [1,5,9,30,-4], [1,5,16,30,-8]]

list_in_list.sort(key=myreturn, reverse=True)

print list_in_list
Member Avatar for vegaseat

As you most likely found out, there are a number of ways to do this. You picked a good one.

There was also a snippet at:
http://www.daniweb.com/code/snippet237089.html

Edited by vegaseat because: n/a
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