I am grouping integers into a nested list. But I end up with with two sublists each time instead of multiple

Question

Given an integer list and a key, I am required to group the list by greater than or equal to the key, or less than the key.

For example, When

key = 6 and lst = [1, 1, 1, 0, 0, 6, 10, 5, 10]

Output should be:

[[1, 1, 1, 0, 0], [6, 10], [5], [10]]

A new sublist must be created if while looping a new condition is met. When the process of grouping is over, the sublists are placed into a list. If the list is empty, it returns an empty list.

def group_ints(lst, key):
    greater_or_equal = []
    lessthan = []
    result = []
    
    if lst == None:
        return []
    else: 
        for i in lst:
            if i >= key:
                greater_or_equal.append(i)
            elif i < key:
                lessthan.append(i)
        if greater_or_equal:
            result.append(greater_or_equal)
        if lessthan:
            result.append(lessthan)
    return result

result.append(greater_or_equal) and result.append(lessthan) are the only two places in the code that add something to result. So yes, of course result is only going to have, at most, two things in it... — John Gordon
– John Gordon, Commented Jun 24, 2024 at 16:34

trincot · Accepted Answer · 2024-06-24 16:29:10Z

4

You're only building two lists: greater_or_equal and lessthan, yet it is clear that the expected result could have any number of sublists, not just two. So every time you switch category, you'll have to create a new list (to serve as sub list).

Here is how you could do it:

def group_ints(lst, key):
    if not lst:
        return []
    
    result = [lst[:1]]  # start with the first value already in its own sublist

    for i in lst:
        # Is the current value of a different category than the last one?
        if (i >= key) != (result[-1][0] >= key):  # We need a new list
            result.append([])  # create and add it to the result
        result[-1].append(i)
    return result

answered Jun 24, 2024 at 16:29

trincot

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S.B · Accepted Answer · 2024-06-24 16:32:34Z

1

You can use itertools.groupby() and pass lambda x: x >= key as the key= parameter to achieve this:

from itertools import groupby


def split_list(lst: list[int], key: int):
    return [list(g) for _k, g in groupby(lst, key=lambda x: x >= key)]


assert split_list([1, 1, 1, 0, 0, 6, 10, 5, 10], key=6) == [[1, 1, 1, 0, 0], [6, 10], [5], [10]]
assert split_list([], key=6) == []

answered Jun 24, 2024 at 16:32

S.B

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