How to normalize data between 0 and 1?

I have seen the min-max normalization formula in several answers (e.g. , , ), where data is normalized into the interval $$\left[0,1 \right]$$.

However, is there a method to normalize data into the interval $$\left(0,1 \right)$$, i.e. excluding 0 and 1?

EDIT:

My data is a sample from a uniform distribution within the range $$\left[a,b \right]$$. I would like to normalize it into the interval $$\left(0,1 \right)$$ while remaining uniformly distributed.

• $$\frac{1}{1 + \exp(-x)} \in (0,1)$$ for any $x\in \mathbb{R}$. Do you have some other requirements that would exclude this? – Sycorax Dec 4 '18 at 15:47
• Thanks @Sycorax, to clarify, i just edited my question to point out that my data sample should be uniformly distributed. – skoestlmeier Dec 4 '18 at 16:01

Using the property that the CDF is uniformly distributed on $$[0,1]$$, you can compute the empirical CDF for $$x$$. This is essentially the same as ranking the data and then rescaling by the number of elements $$n$$. To enforce the requirement that the scaled data exclude 0 and 1, you can deviate from the standard ECDF procedure and construct the scale so that the outputs are $$\frac{1}{n+1}, \frac{2}{n+1},\cdots, \frac{n}{n+1}$$, which is likewise uniform.

• There's a whole class of symmetric versions of your scaling procedure: $u_\alpha(i) = \frac{i-\alpha}{n+1-2\alpha}$ (with $0\leq\alpha\leq 1$, of which the above has $\alpha=0$. (There's also asymmetric ones which have uses in some applications) – Glen_b Dec 6 '18 at 5:12
• Does this have any particular name? – Sycorax Dec 6 '18 at 13:42
• Several, I think but I can't recall any right now. It comes up in probability plotting. Blom 1958 "Statistical Estimates and Transformed Beta Variables" is the standard reference for this thing (and variations). – Glen_b Dec 7 '18 at 8:49

A uniform distribution on $$(a, b)$$ is the same as a uniform distribution on $$[a, b]$$, since for any $$X$$ distributed uniformly on $$[a, b]$$, $$P(X = a) = P(X = b) = 0$$. So, just use the formulae for translating to $$[0, 1]$$. On the other hand, if your sample has a value equal to $$a$$ or $$b$$, then you can safely conclude that you don't actually have a continuous uniform distribution.

• I don't agree with your latter statement. Following the same logic, you could exclude any data from ever being sampled from a uniform distribution. – dedObed Dec 4 '18 at 19:47
• @dedObed The argument works for any countable set of points, because any such set has Lebesgue measure zero, but not for uncountable sets. – Kodiologist Dec 4 '18 at 20:27
• I agree that a uniform distribution on (a, b) is the same as a uniform on [a, b]. The claim I challenge is "if your sample has a value equal to a or b [...] you don't actually have a continuous uniform distribution." – dedObed Dec 4 '18 at 20:34
• @dedObed I know. I'm saying that the argument works because $\{a, b\}$, the set of just the two values $a$ and $b$, is countable. It wouldn't if you used a non-null set, which is what would be required to "follow the same logic" to "exclude any data from ever being sampled from a uniform distribution". – Kodiologist Dec 4 '18 at 20:36
• @dedObed I guess the chief thing to keep in mind is that continuous distributions are the sort of ethereal mathematical entities you can't get in real life. Computers fake a continuous uniform distribution with a discrete distribution that covers a large number of floating-point values. It's close enough for many applied purposes, but, e.g., a random float will always be rational, whereas a random sample from a continuous uniform distribution will be almost surely irrational. – Kodiologist Dec 4 '18 at 21:58

The formula $$x' = \frac{x - \min{x}}{\max{x} - \min{x}}$$ will normalize the values in $$[0,1]$$.

I am not sure of why you want to exclude $$0$$ and $$1$$, anyway one way would be to choose a new minimum and maximum values for the transformed variable, e.g. $$[0+\epsilon,1-\epsilon]$$. You can then transform the variable using $$x' = \epsilon + (1-2\epsilon) \cdot \left(\frac{x - \min{x}}{\max{x} - \min{x}} \right)$$

Another way could be, as suggested by Sycorax in his comment, to use a logistic transform $$x' = \frac{1}{1 + \exp(-x)}$$ This ensures that $$\forall x \in \mathbb{R} \implies x' \in (0,1)$$. However, depending on the original distribution of $$x$$, $$x'$$ might span only a limited range of the interval $$(0,1)$$, so you might want to try e.g. to standardize $$x$$ before applying the logistic transform.