使用 numba 计算向量和矩阵中的行之间的余弦相似度

找到了这个gist https://gist.github.com/FedericoV/0e7d6d8c8794a99a7a42使用 numba 快速计算余弦相似度。

import numba

@numba.jit(target='cpu', nopython=True)
def fast_cosine(u, v):
    m = u.shape[0]
    udotv = 0
    u_norm = 0
    v_norm = 0
    for i in range(m):
        if (np.isnan(u[i])) or (np.isnan(v[i])):
            continue

        udotv += u[i] * v[i]
        u_norm += u[i] * u[i]
        v_norm += v[i] * v[i]

    u_norm = np.sqrt(u_norm)
    v_norm = np.sqrt(v_norm)

    if (u_norm == 0) or (v_norm == 0):
        ratio = 1.0
    else:
        ratio = udotv / (u_norm * v_norm)
    return ratio

结果看起来很有希望（我的机器中没有 jit 装饰器时为 500ns，而只有 200us）。

我想用 numba 来并行化向量之间的计算u和一个候选矩阵M-- 即每行的余弦。

Example:

def fast_cosine_matrix(u, M):
    """
    Return array of cosine similarity between u and rows in M
    >>> import numpy as np
    >>> u = np.random.rand(100)
    >>> M = np.random.rand(10, 100)
    >>> fast_cosine_matrix(u, M)
    """

一种方法是用第二个输入重写矩阵。但我得到一个NotImplementedError如果我尝试迭代矩阵的行。将尝试仅使用切片。

我想过使用vectorize但我无法让它发挥作用。

解决方案重写一下：

import numpy as np
import numba

@numba.jit(target='cpu', nopython=True, parallel=True)
def fast_cosine_matrix(u, M):
    scores = np.zeros(M.shape[0])
    for i in numba.prange(M.shape[0]):
        v = M[i]
        m = u.shape[0]
        udotv = 0
        u_norm = 0
        v_norm = 0
        for j in range(m):
            if (np.isnan(u[j])) or (np.isnan(v[j])):
                continue

            udotv += u[j] * v[j]
            u_norm += u[j] * u[j]
            v_norm += v[j] * v[j]

        u_norm = np.sqrt(u_norm)
        v_norm = np.sqrt(v_norm)

        if (u_norm == 0) or (v_norm == 0):
            ratio = 1.0
        else:
            ratio = udotv / (u_norm * v_norm)
        scores[i] = ratio
    return scores


u = np.random.rand(100)
M = np.random.rand(100000, 100)

fast_cosine_matrix(u, M)