使用numpy实现topk函数操作(并排序)

>>> x = np.array([3, 5, 6, 4, 2, 7, 1])
>>> x[np.argpartition(x, 3)]
array([2, 1, 3, 4, 5, 7, 6])

def naive_arg_topK(matrix, K, axis=0):
    """
    perform topK based on np.argsort
    :param matrix: to be sorted
    :param K: select and sort the top K items
    :param axis: dimension to be sorted.
    :return:
    """
    full_sort = np.argsort(matrix, axis=axis)
    return full_sort.take(np.arange(K), axis=axis)

# Example
>>> dists = np.random.permutation(np.arange(30)).reshape(6, 5)
array([[17, 28,  1, 24, 23,  8],
       [ 9, 21,  3, 22,  4,  5],
       [19, 12, 26, 11, 13, 27],
       [10, 15, 18, 14,  7, 16],
       [ 0, 25, 29,  2,  6, 20]])
>>> naive_arg_topK(dists, 2, axis=0)
array([[4, 2, 0, 4, 1, 1],
       [1, 3, 1, 2, 4, 0]])
>>> naive_arg_topK(dists, 2, axis=1)
array([[2, 5],
       [2, 4],
       [3, 1],
       [4, 0],
       [0, 3]])

def partition_arg_topK(matrix, K, axis=0):
    """
    perform topK based on np.argpartition
    :param matrix: to be sorted
    :param K: select and sort the top K items
    :param axis: 0 or 1. dimension to be sorted.
    :return:
    """
    a_part = np.argpartition(matrix, K, axis=axis)
    if axis == 0:
        row_index = np.arange(matrix.shape[1 - axis])
        a_sec_argsort_K = np.argsort(matrix[a_part[0:K, :], row_index], axis=axis)
        return a_part[0:K, :][a_sec_argsort_K, row_index]
    else:
        column_index = np.arange(matrix.shape[1 - axis])[:, None]
        a_sec_argsort_K = np.argsort(matrix[column_index, a_part[:, 0:K]], axis=axis)
        return a_part[:, 0:K][column_index, a_sec_argsort_K]

# Example
>>> dists = np.random.permutation(np.arange(30)).reshape(6, 5)
array([[17, 28,  1, 24, 23,  8],
       [ 9, 21,  3, 22,  4,  5],
       [19, 12, 26, 11, 13, 27],
       [10, 15, 18, 14,  7, 16],
       [ 0, 25, 29,  2,  6, 20]])
>>> partition_arg_topK(dists, 2, axis=0)
array([[4, 2, 0, 4, 1, 1],
       [1, 3, 1, 2, 4, 0]])
>>> partition_arg_topK(dists, 2, axis=1)
array([[2, 5],
       [2, 4],
       [3, 1],
       [4, 0],
       [0, 3]])

# 构建小顶堆跳转def sift(li, low, higt):
    tmp = li[low]
    i = low
    j = 2 * i + 1
    while j <= higt:  # 情况2：i已经是最后一层
        if j + 1 <= higt and li[j + 1] < li[j]:  # 右孩子存在并且小于左孩子
            j += 1
        if tmp > li[j]:
            li[i] = li[j]
            i = j
            j = 2 * i + 1
        else:
            break  # 情况1：j位置比tmp小
    li[i] = tmp


def top_k(li, k):
    heap = li[0:k]
    # 建堆
    for i in range(k // 2 - 1, -1, -1):
        sift(heap, i, k - 1)
    for i in range(k, len(li)):
        if li[i] > heap[0]:
            heap[0] = li[i]
            sift(heap, 0, k - 1)
    # 挨个输出
    for i in range(k - 1, -1, -1):
        heap[0], heap[i] = heap[i], heap[0]
        sift(heap, 0, i - 1)
    return heap


li = [0, 8, 6, 2, 4, 9, 1, 4, 6]
print(top_k(li, 3))