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If under the hamming weight you understand the number of nonzero components, then something along these lines will do the job:

def random_vector(q,n,w):
    F = GF(q)
    S = Subsets(range(n),w).random_element()
    v = vector(F,n)
    for s in S:
        while True:
            v[s] = F.random_element()
            if v[s]:
                break
    return v

Here random_vector(5^2,10,3) returns a random vector over $GF(5^2)$ with 3 nonzero out of total 10 components, such as (0, 0, 2*z2 + 3, 0, 4*z2 + 1, 0, 0, 4*z2 + 1, 0, 0).

If under the hamming weight you understand the number of nonzero components, then something along these lines will do the job:

def random_vector(q,n,w):
    F F.<b> = GF(q)
GF(q, modulus='primitive')
    S = Subsets(range(n),w).random_element()
    v = vector(F,n)
    return vector([b^randint(0,q-2) if k in S else F.zero() for s k in S:
        while True:
            v[s] = F.random_element()
            if v[s]:
                break
    return v
range(n)])

Here random_vector(5^2,10,3) returns a random vector over $GF(5^2)$ with 3 nonzero out of total 10 components, such as (0, 0, 2*z2 + 3, 0, 4*z2 + 1, 0, 0, 4*z2 + 1, 0, 0).