我在编写一个k最近邻分类器时遇到了问题，数组操作花费的时间太长了。

def classify(k, train_data, target):    num_rows = train_data.shape[0]    num_cols = train_data.shape[1]    distances = []    candidates = [0] * 10    for i, row in enumerate(train_data):        dist = euclidean_dist(target[:num_cols - 1], row[:num_cols - 1]) # 速度慢        distances.append((dist, row[num_cols - 1]))    distances.sort(key=lambda tup: tup[0])    distances = distances[:k]    for i, d in enumerate(distances):        candidates[d[1]] += 1    return candidates.index(max(candidates))def euclidean_dist(x1, x2):    assert(len(x1) == len(x2))    result = 0    pdb.set_trace()    for i in range(len(x1)): # 罪魁祸首，x1 和 x2 都是长度为256的列表        result += math.pow(x1[i] - x2[i], 2)    result = math.sqrt(result)    return result

我在上面的代码中标注了问题发生的地方，欢迎提供任何加速的建议。

回答：

看起来你只需要欧几里得距离/2范数，可以通过numpy（导入为np）高效地获得：

def euclidean_dist2(x1, x2):    assert(len(x1) == len(x2))    x1 = np.array(x1)    x2 = np.array(x2)    norm = np.linalg.norm(x1-x2)    return normprint euclidean_dist2([1,2],[4,7])

这将给你5.83095189485，与你原来的函数结果相同，但已被向量化

分解来看，你只是在进行元素-wise的差运算，然后将结果向量乘以自身（平方），求和，最后对和进行开方运算：

def euclidean_dist3(x1, x2):    assert(len(x1) == len(x2))    x1 = np.array(x1)    x2 = np.array(x2)    diff = x1 - x2    squared = np.transpose(diff) * diff    summed = sum(squared)    norm = np.sqrt(summed)    return norm

换句话说，你只是在计算差向量与自身的点积：

def euclidean_dist4(x1, x2):    assert(len(x1) == len(x2))    x1 = np.array(x1)    x2 = np.array(x2)    diff = x1 - x2    dot = np.dot(diff, diff)    norm = np.sqrt(dot)    return norm

不同的方法可以达到同样的效果