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New answers tagged normalization

4

Rescaling the input features is just a linear transformation. There's no right or wrong way of rescaling outside a problem context. If you want to map the range 1 - 100 to the range 1 - 10 linearly you should do: $$x \leftarrow \frac{x - 1}{99} \times 9 + 1$$ This maps 1 to 1 and 100 to 10 and it will make the durations have the same range as the other ...

0

One way of standardizing variables is to turn each value into a z-score, by taking $\frac{x - \bar{x}}{sd}$ Doing this, you would only have to do it once; however, this will not result in a range of -1 to 1, the result can be any number. But most values will be between -1 and 1.

1

What do you want to standardize, a binary random variable, or a proportion? It makes no sense to standardize a binary random variable. A random variable is a function that assigns a real value to an event $Y:S\rightarrow \mathbb{R}$. In this case 0 for failure and 1 to success, i.e. $Y\in \lbrace 0,1\rbrace$. In the case of a proportion, this is not a ...

3

Standardizing binary variables does not make any sense. The values are arbitrary; they don't mean anything in and of themselves. There may be a rationale for choosing some values like 0 & 1, with respect to numerical stability issues, but that's it.

1

A binary variable with values 0, 1 can (usually) be scaled to (value - mean) / SD, which is presumably your z-score. The most obvious constraint on that is that if you happen to get all zeros or all ones then plugging in SD blindly would mean that the z-score is indeterminate. There is a case for assigning zero too in so far as value - mean is identically ...

0

apply(m,2,norm<-function(x){return (x/sum(x)}) ?

2

There is simply no single “mathematically” correct way to derive a grade from several variables or dimensions. Without more information on the characteristics of your variables and what you are trying to achieve, it's difficult to provide any useful advice. One important thing to note is that dividing by the maximum value is almost certainly a bad idea ...

1

K-means is inherently distance based, so any normalization implicitly includes assumptions about the relationship between the different variables. Also I notice that your dataset only have 3 continuous variables (and one binary), so you can actually visualize the data relatively easily. Any clusters that are so robust to be meaningful to your client should ...

0

It depends on the structure of a neural network and (even more) on the properties of data. There is no best normalization algorithm because if there would be one, it would be used everywhere by default... In theory, there is no requirement for the data to be normalized at all. This is a purely practical thing because in practice convergence could take ...

3

If you know $m$ and $\sigma$ and your random variable is $X$, then let $Z = m^ \prime + {\frac {\sigma^\prime} \sigma}(X - m)$. It's called an affine transformation, although you'll probably see it more often called a linear transformation. Technically it's a linear transformation followed by a translation (according to Wikipedia).

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