CODEWITHSUNDEEP
pythonoptimizationlower-boundtext-normalization
Jaques duBalzac
Jaques duBalzac

Reputation: 3

Normalize vector such that sum equals 1, while satisfying a lower bound

Given a lower bound of 0.025, I want a vector consisting of weights that sum up to 1 and satisfy this lower bound. Starting from a vector with an arbitrary length and the values ranging from 0.025 (lower bound) to 1.

For example,

[0.025, 0.8, 0.7]

Then a normalization where you divide by the sum of the numbers gives you roughly the following:

[0.016, 0.524, 0.459]

Now this does not satisfy the lower bound, any ideas on how I can get this to work?

Upvotes: 0

Views: 3594

Answers (2)

Andre
Andre

Reputation: 788

If you want the weights (values in an array) to sum up to 1, you can divide each value by the sum of all values (i.e. normalize by the sum). This procedure keeps the relative sizes of each pair of values I means: before and after the step, the second item will be 5 times as large as the fourth item.

On the other hand you want all values to be larger than 0.025. Imagine if one item is 50 times larger than another before normalization, and the smallest value must be 0.025, the other item would need to be 1.25, which is already larger than the sum has to be.

You can figure out that you cannot (given any array) just scale all values equally so that they sum up to 1 AND the smallest value is 0.025.

Then the question is what relation between the values do you want to keep in the procedure?

As a side not, you cannot have more than 40 items, all bigger than 0.025, summ up to 1. So "arbitrary length" just cannot work either.

Upvotes: 2

Corralien
Corralien

Reputation: 120399

Add the lower bound to the dividend and divisor:

I used numpy for readability:

import numpy as np

v = np.array([0.025, 0.8, 0.7])

v2 = (v + min(v)) / sum(v + min(v))

Output:

>>> v2
array([0.03125 , 0.515625, 0.453125])

>>> sum(v2)
1.0

Upvotes: 0

Related Questions