0
$\begingroup$

I am working with psychology, more precisely, the Big Five personality traits. I have a test which measures 5 variables, that is believed to describe personality traits of human beings. The variables are; openness, conscientiousness, extraversion, agreeableness & neuroticism. I will refer to them as O C E A N.

A test is carried out to measure the variables. It can yield scores from 1.0 to 7.0. Where E = 7.0 would mean that the person has the maximum value of extraversion, suggesting a very social person.

Lets say both Eve and Adam makes the test, and their final scores are:

Eve = [4.0, 2.4, 5.2, 5.1, 6.9]

Adam = [1.1, 2.2, 4.2, 5.6, 6.1]

What would be a good way to measure their similarity? The ordering of the letters does not matter, as long as they both use the same order, e.g instead of O C E A N it could be N E O A C. There might be correlation between the variables, e.g openness O might have a positive correlation with extroversion E.

$\endgroup$
0
$\begingroup$

You are asking about the distance in vector space. There exist many possible distance functions, but most common are ($x_{1\dots5}$ and $y_{1\dots5}$ are 5 your numbers in order):

  1. Euclidian distance (from $L_2$ norm) $d_E = \sqrt{(x_1-y_1)^2 +\dots + (x_5-y_5)^2}$
  2. Manhattan distance (from $L_1$ norm) $d_M = |x_1-y_1|+\dots+|x_5-y_5|$
  3. Chebyshev distance (from $L_\infty$ norm) $d_C=\max(|x_1-y_1|,\dots |x_5-y_5|)$

You are probably good with the first one.

$\endgroup$
  • $\begingroup$ Would those distance functions capture possible correlation of the variables? (Might be a silly question, am not a mathematician) $\endgroup$ – Isbister Jan 31 at 15:53
  • $\begingroup$ They won't capture correlations, they capture how close are those points are in some 5d space. If Alice is (5,5,5,5,5), Bob is (4,4,4,4,4) and Charlie is (5,6,4,4,6), who is more similar to Alice from your point of view, Bob or Charlie? $\endgroup$ – Vasily Mitch Jan 31 at 15:56
  • $\begingroup$ It depends. Because of possible correlation. If the order of the 5 variabels are ”ocean” then Bob is. If the order is ”neoac” Charlie is. $\endgroup$ – Isbister Jan 31 at 16:37
  • $\begingroup$ So are you saying that some variables are more important then others? $\endgroup$ – Vasily Mitch Jan 31 at 16:39
  • $\begingroup$ another question. Lets say I have a dataset with Alice, Bob and 500000 more samples. Could I benefit from creating a k-nn model from this set, and then perform nearest neighbour search to find who is most similar to Alice? (can i benefit for having a big data set when checking similarities between pairs?) $\endgroup$ – Isbister Feb 7 at 18:57
-1
$\begingroup$

There are several approaches you can take

Cosine similarity

Define $|{\rm Eve}| = \sqrt{O_{\rm eve}^2 + C_{\rm eve}^2 + E_{\rm eve}^2 + A_{\rm eve}^2 + N_{\rm eve}^2}$, with a similar expression for Adam and

$$ {\rm Eve}\cdot {\rm Adam} = O_{\rm eve}O_{\rm adam} + C_{\rm eve}C_{\rm adam} + \cdots + N_{\rm eve}N_{\rm adam} $$

The cosine similarity between these two observations is

$$ \text{cosine similarity} = \frac{{\rm Eve}\cdot {\rm Adam}}{|{\rm Eve}||\rm Adam|} $$

You can then calculate the angular distance

$$ \text{angular distance} = \frac{1}{\pi}\arccos(\text{cosine similarity}) $$

and the angular similarity

$$ \text{angular similarity} = 1 - \text{cosine distance} $$

In this scenario, a similarity of $1$ means observations are close, $0$ means they are different

$\endgroup$
  • $\begingroup$ If you take one person with all traits scored 1 and other person with all traits scored 7 (they are as far from each other as they can be), but angular distance will be 0 and angular similarity will be 1 $\endgroup$ – Vasily Mitch Jan 31 at 15:47
  • $\begingroup$ @VasilyMitch I agree to your comment! I was considiring cosine distance aswell, but I thought there might be more sophisticated methods! $\endgroup$ – Isbister Jan 31 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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