Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have some questions regarding the normalisation of a data set between 0-1. The members of the set that I try to normalise are:

X1 = 4.803
X2 = 24.30
X3 = 280.32
X4 = 44.78

If I add them all up I get 354.203. As I want to map each data point to the 0-1 range, I use the following equation:

$$ X_{i,0\;to\;1} = \frac{X_i - X_{min}}{X_{max} - X_{min}} $$

I get:

X1_norm = 0
X2_norm = 0.0707651433486863
X3_norm = 1
X4_norm = 0.14509812461663

I then have two questions: 1) Can I somehow get 354.203 using the normalised values? 2) What should I do in order to map the entire data set to 0-1? That is, if I add all the normalised terms (X1_norm to X4_norm) I want to get exactly 1.

Thanks in advance.

share|improve this question
    
You have a typo in the formula (should be $-Xmin$ both on top and bottom) and $X_4^{norm} \approx 0.145098125$. –  gt6989b Oct 23 '12 at 16:21
1  
Fixed now, thanks –  limp Oct 23 '12 at 16:25

1 Answer 1

up vote 1 down vote accepted

(1) Once you have normalized, your data is in $[0,1]$, so it's not possible to tell what was the original data range, unless you keep original min and max.

(2) Divide all normalized terms by their sum (roughly, $1.215863268$). You get $(0, \ 0.058201564, \ 0.822460902, \ 0.119337534)$

share|improve this answer
    
Thanks a lot for your answer. One last thing: Let's say that I keep the original min and max. What should I do to retrieve the rest data points (X2 and X4 in my case) by only having the normalised data? –  limp Oct 23 '12 at 16:39
    
@limp Let $x$ be the rescaled point. The original will be $X_\min + x*(X_\max - X_\min)$. If you have rescaled further as in (2), say by another factor $f$, you would need to compute the original value as $X_\min + x*f*(X_\max - X_\min)$ –  gt6989b Oct 23 '12 at 16:48
    
That was really helpful, thanks again. –  limp Oct 23 '12 at 16:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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