# How to compare multiple weight vectors of different size?

I am facing a statistical problem that I am not sure whether it's solvable. Simply put, I am given multiple weight vectors and here are two examples:

$$w_1 = [.2, .3, .4, .1]$$ for items A, B, C, D, respectively.

$$w_2 = [.1, .1, .05, .05, .03, .03, .02, .5, .12]$$ for items A, B, C, D, E, F, G, H, I, respectively.

and I have many weight vectors like this (sum of all elements is 1), but of different sizes.

Now I want to construct an "importance vector" for all 26 items (A ~ Z), each entry represents each item's importance.

Since many weight vectors are of different sizes, their weights are of different scale. I don't think it's a good idea to simply add them up and take average because of the reason stated before. Any ideas? Or any pieces of literature that I can refer to? Or what is this kind of problems called?

• So the importance vector would use the weight information for each item across all these vectors, right? – Lucas Farias Feb 2 at 14:18
• @LucasFarias Yes – zcylywde Feb 2 at 15:20

In this sense, given the data in these vectors are probabilities, isn't it the case that the probability of an item that does not belong to a particular vector is $$0$$? Namely, $$w_1 = \left[0.1,0.2,0.7\right]$$ for $$A,B,C$$ imply $$\mathbb{P}(D)=0$$.
If that's so, you can project them all onto $$\mathbb{R}^{26}$$, and this would allow you to aggregate (e.g. sum, average, etc.) them properly. By this I mean that the resulting projection of the $$w_1$$ of my example would be $$w_1^{*} = \left[0.1,0.2,0.7,0,...,0\right]$$.
• I am afraid I cannot simply infer P(D) = 0. In my weight vector, e.g. $w_1 = [0.1, 0.2, 0.7]$ for A, B, C, one piece of information should be "C is more important than B", but I do not know whether C is more important than D or than the rest of letters that are not included in this vector. – zcylywde Feb 4 at 3:22