Vectors like this
v1 = {0 0 0 1 1 0 0 1 0 1 1}
v2 = {0 1 1 1 1 1 0 1 0 1 0}
v3 = {0 0 0 0 0 0 0 0 0 0 1}
Need to calculate similarity between them. Hamming distance between v1 and v2 is 4 and between v1 and v3 is also 4. But because I am interested in the groups of '1' which are together for me v2 is far more similar to v1 then v3 is.
Is there any distance metrics that can capture this in the data?
The data represent occupancy of house in time, that's why it is important to me. '1' means occupied, '0' means non occupied.
It sounds like you need cosine similarity measure:
similarity = cos(v1, v2) = v1 * v2 / (|v1| |v2|)
where v1 * v2 is dot product between v1 and v2:
v1 * v2 = v1[1]*v2[1] + v1[2]*v2[2] + ... + v1[n]*v2[n]
Essentially, dot product shows how many elements in both vectors have 1 at the same position: if v1[k] == 1 and v2[k] == 1, then final sum (and thus similarity) is increased, otherwise it isn't changed.
You can use dot product itself, but sometimes you would want final similarity to be normalized, e.g. be between 0 and 1. In this case you can divide dot product of v1 and v2 by their lengths - |v1| and |v2|. Essentially, vector length is square root of dot product of the vector with itself:
|v| = sqrt(v[1]*v[1] + v[2]*v[2] + ... + v[n]*v[n])
Having all of these, it's easy to implement cosine distance as follows (example in Python):
from math import sqrt
def dot(v1, v2):
return sum(x*y for x, y in zip(v1, v2))
def length(v):
return sqrt(dot(v, v))
def sim(v1, v2):
return dot(v1, v2) / (length(v1) * length(v2))
Note, that I described similarity (how much two vectors are close to each other), not distance (how far they are). If you need exactly distance, you can calculate it as dist = 1 / sim.