2
$\begingroup$

In my current thesis I have two weight components. As I want to join those components, weighted by a percentage, I thought about normalizing(/scaling?) both components respectively by their max value.

Component $c1$ will therefore be:

$c1' = \frac{c1}{\max(c1)}$

respectively component $c2$

$c2' = \frac{c2}{\max(c2)}$

As I now have to write about it I can't find the correct term for this normalizing/scaling.

In short what is the name of normalizing/scaling process:

$$x' = \frac{x}{\max(x)}$$

Improve this question
$\endgroup$
3
  • $\begingroup$ Also known as ipsative scaling. methods.sagepub.com/reference/… $\endgroup$
    – user78229
    Commented Oct 27, 2020 at 13:35
  • 1
    $\begingroup$ Trivial but perhaps worth underlining: just dividing by the maximum won't yield a minimum of zero unless the minimum already is zero. The new minimum will be above or below zero if the old minimum is. Also, even there, the assumption is that the maximum is a positive value. $\endgroup$
    – Nick Cox
    Commented Oct 27, 2020 at 13:46
  • $\begingroup$ MaxAbsScaling according to scikit-learn api scikit-learn.org/stable/auto_examples/preprocessing/… $\endgroup$
    – Moreno
    Commented Oct 28, 2020 at 2:51

2 Answers 2

Reset to default
4
$\begingroup$

If the minimum value of both $c1$ and $c2$ is zero, then this is known as "min-max scaling": $$x' = \frac{x - x_{min}}{x_{max} - x_{min}}$$

This normalizes the variable range to $[0,1]$. Note that, depending on the variable range, a linear transformation to the range $[0,1]$ might not be appropriate (another occasionally used scaling function is the exponential function).

Another normalization method is "z score standardization", which normalizes to zero mean and variance one (and thus SD one too): $$x' = \frac{x-\mu}{\sigma}$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you, but I dont want to normalize to [0,1] as those components don't have 0 value because they equal the respective min-value. They still represent a factor and therefore I just want to normalize by max. $\endgroup$
    – Goddilein
    Commented Oct 27, 2020 at 14:32
  • $\begingroup$ Can you please explain, why you perform arithmetic operations (dividing by max) with factors, i.e. categorial variables? $\endgroup$
    – cdalitz
    Commented Oct 27, 2020 at 15:11
  • $\begingroup$ Maybe factor is the wrong wording. I have two values >= 0. Both represent a weight for their respective aspect. As I want to join those weights together with alpha*c1 + (1-alpha)*c2 I have to normalize c1,c2 to equal intervals. $\endgroup$
    – Goddilein
    Commented Oct 27, 2020 at 15:25
1
$\begingroup$

There is no specific name for this normalization, as far as I know....

I think, for a thesis, it is enough to mention that data were normalized because this is most probably marginal information and the reviewer should understand the process without providing more details.

If you want to be clearer, when your data are positive, you can say "Data were normalized to be between 0 and 1". When you have negative values, the normalized values will be between -1 and +1 and the max should be taken in absolute value.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ In a thesis, you can assume whatever people in your field are expected to know well, but on that you should be asking your supervisor/advisor/committee. But in general: just stating that a measure is normalized or standardized is very little information without stating the recipe used. $\endgroup$
    – Nick Cox
    Commented Oct 27, 2020 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.