Efficient way to sort by number thresholds in Python?

Question

Given a stream of dictionaries each with unique numerical ID, what would be the most efficient way to generate a list (or a format map-able into a sorted list) of IDs based on thresholds that do not increase linearly (<48, <103, <123...)? I've not looked extensively into itertools or other useful iteration libraries and I can't think of a much better way to group other than using elifs.

Example using if/elif/else:

dicts = [{'id':30},{'id':60},{'id':90},{'id':120},{'id':150}]
groups = [[] for _ in range(5)]

for a_dict in dicts:
    ID = a_dict['id']
    if ID < 50: groups[0].append(ID)
    elif ID < 100: groups[1].append(ID)
    elif ID < 150: groups[2].append(ID)
    elif ID < 200: groups[3].append(ID)
    else: groups[4].append(ID)

Output:

>>> print(groups)
[[30], [60, 90], [120], [150], []]

score 5 · Accepted Answer · answered Mar 23 '21 at 12:06

5

The bisect algorithm should be the most efficient way to decide in which group an item belongs (assumes your groups are sorted). In fact the bottom of the page has an example similar to what you want to achieve.

>>> import bisect
>>> bins = [50, 100, 150, 200]
>>> bisect.bisect(bins, 30)
0
>>> bisect.bisect(bins, 60)
1
>>> bisect.bisect(bins, 220)
4

All in all

for a_dict in dicts:
    ID = a_dict['id']   # don't use Python built-in names
    index = bisect.bisect(bins, ID)
    groups[index].append(ID)

answered Mar 23 '21 at 12:06

Reti43

9,656
3
28
44

Pleas see, https://stackoverflow.com/questions/9987483/elif-in-list-comprehension-conditionals/9987533 – Zeinab Mardi Mar 23 '21 at 12:18
2

@ZeinabMardi How is this relevant? – Reti43 Mar 23 '21 at 12:26
@Reti43 Thanks for the fast answer, this works great in my instance. – wish Mar 23 '21 at 12:31
2

@wish Please see [What should I do when someone answers my question?](https://stackoverflow.com/help/someone-answers) – Tomerikoo Mar 23 '21 at 12:32

Efficient way to sort by number thresholds in Python?

1 Answers1

Linked