Python function Converting an integer to bins as per given interval

Question

I am trying to convert an integer (num) to bins as per given interval. the bin size (intervals) are [1, 200), [200, 400), [400, 800), [800, 1200), [1200, num]

I'am doing it in a crude way ...

def create_bins(num):
    """Create Bins as per given intervals."""
    s1, s2, s3, s4 = 199, 200, 400, 400
    if num > 1200:
        res = [s1, s2, s3, s4, num - (s1 + s2 + s3 + s4)]
    elif num < 1200 and num >= 800:
        res = [s1, s2, s3, num - (s1 + s2 + s3)]
    elif num < 800 and num >= 400:
        res = [s1, s2, num - (s1 + s2)]
    elif num < 400 and num >= 200:
        res = [s1, num - s1]
    else:
        res = [num]

    return res

this function for create_bins(1041) returns [199, 200, 400, 242] which is correct. However, I'am sure there are better ways to get it done...

Will appreciate if you can lead me towards better solutions for these kind of problem.

@Julien I'am sure this is not an ideal solution for the problem. So rather going for code review, looking for better solutions. There are lots of repetition in the code, however, somehow that code is becoming elusive for me... — Arun Kumar Khattri, Jul 18 '23 at 04:50
That's the point of code-review. Also "better" is subjective, there is nothing wrong with your code. [ask] — Julien, Jul 18 '23 at 04:57
You could replace the sums `s1 + ...` by their values given they are constants. — Julien, Jul 18 '23 at 05:01
It took me a while to understand "converting to bins". If I am right, you want to decompose a number in pieces of "size" 199, 200, 400, 400, in this order, with a possible remainder. — Yves Daoust, Jul 18 '23 at 06:19
@YvesDaoust The intervals into which the entire range of the data is split into is called as the bin. Whereas decomposing is to express (a number or function) as a combination of simpler components. Here "intervals" are in the play, that's why i thought 'bins' will appropriately represent the final result... — Arun Kumar Khattri, Jul 18 '23 at 15:26
Are you happy to use numpy, if so there's a [digitize function](https://numpy.org/doc/stable/reference/generated/numpy.digitize.html), see [this answer](https://stackoverflow.com/a/6163403/2280890). — import random, Jul 19 '23 at 06:30

score 1 · Answer 1 · answered Jul 18 '23 at 06:08

1

this may be good idea.

  def create_bins_new(num):
        val = [199,200,400,400]
        out = []
        for v in val:
            if v < num:
                out.append(v)
                num -= v
        out.append(num)
        return out

answered Jul 18 '23 at 06:08

Jenil saliya

11
2

or not... this is substantially slower than OP's solution. – Julien Jul 18 '23 at 06:51
@Julien: substantially ? Can you substantiate ? – Yves Daoust Jul 18 '23 at 06:53
use %timeit to find that it's 50% slower. – Julien Jul 18 '23 at 06:53
@Julien: the expected time depends on the distribution of the values of `num`. – Yves Daoust Jul 18 '23 at 06:54
In the best case it's the same, in the worst case it;s slower. – Julien Jul 18 '23 at 06:55
s1 to s4 are fixed or it can be vary. i mean it can be s1 to s5 or anything else – Jenil saliya Jul 19 '23 at 10:53

score 0 · Answer 2 · answered Jul 18 '23 at 06:47

In case of a very large number of bins, compute the prefix sum of the bin sizes, locate num in this list by binary search and assemble the result.

import bisect

# Data arrays
bins= [199, 200, 400, 400]
prefix= [sum(bins[:i]) for i in range(len(bins)+1)]

# Try various values
for num in [1, 150, 199, 200, 201, 798, 799, 800, 801, 1041, 2000]:
    # Search and concatenete
    index= bisect.bisect_left(prefix, num) - 1
    print(num, bins[:index] + [num - prefix[index]])

Output:

1 [1]
150 [150]
199 [199]
200 [199, 1]
201 [199, 2]
798 [199, 200, 399]
799 [199, 200, 400]
800 [199, 200, 400, 1]
801 [199, 200, 400, 2]
1041 [199, 200, 400, 242]
2000 [199, 200, 400, 400, 801]

@Julien: "In case of a very large number of bins", can't you read ? And of course the prefix must be computed once for all. — Yves Daoust, Jul 18 '23 at 07:12

Python function Converting an integer to bins as per given interval

2 Answers2