如何实现定长空间金字塔池化层? [英] How to implement the fixed length spatial pyramid pooling layer?

问题描述

我想实现本文中介绍的空间金字塔池化层.

I would like to implement the spatial pyramid pooling layer as introduced in this paper.

作为论文设置，重点是定义max_pooling层的variant kernel size和stride size，即:

As the paper setting, the keypoint is to define variant kernel size and stride size of max_pooling layer, which is:

kernel_size = ceil(a/n)
stride_size = floor(a/n)

其中 a 是输入张量空间大小，n 是金字塔级别，即池化输出的空间区间.

where a is the input tensor spatial size, and n is the pyramid level, i.e. spatial bins of the pooling output.

我尝试用 tensorflow 实现这一层:

I try to implement this layer with tensorflow:

import numpy as np
import tensorflow as tf


def spp_layer(input_, name='SPP_layer'):
    """
    4 level SPP layer.

    spatial bins: [6_6, 3_3, 2_2, 1_1]

    Parameters
    ----------
    input_ : tensor
    name : str

    Returns
    -------
    tensor
    """
    shape = input_.get_shape().as_list()

    with tf.variable_scope(name):

        spp_6_6_pool = tf.nn.max_pool(input_,
                                      ksize=[1,
                                             np.ceil(shape[1]/6).astype(np.int32),
                                             np.ceil(shape[2]/6).astype(np.int32),
                                             1],
                                      strides=[1, shape[1]//6, shape[2]//6, 1],
                                      padding='SAME')
        print('SPP layer level 6:', spp_6_6_pool.get_shape().as_list())

        spp_3_3_pool = tf.nn.max_pool(input_,
                                      ksize=[1,
                                             np.ceil(shape[1]/3).astype(np.int32),
                                             np.ceil(shape[2]/3).astype(np.int32),
                                             1],
                                      strides=[1, shape[1]//3, shape[2]//3, 1],
                                      padding='SAME')
        print('SPP layer level 3:', spp_3_3_pool.get_shape().as_list())

        spp_2_2_pool = tf.nn.max_pool(input_,
                                      ksize=[1,
                                             np.ceil(shape[1]/2).astype(np.int32),
                                             np.ceil(shape[2]/2).astype(np.int32),
                                             1],
                                      strides=[1, shape[1]//2, shape[2]//2, 1],
                                      padding='SAME')
        print('SPP layer level 2:', spp_2_2_pool.get_shape().as_list())

        spp_1_1_pool = tf.nn.max_pool(input_,
                                      ksize=[1,
                                             np.ceil(shape[1]/1).astype(np.int32),
                                             np.ceil(shape[2]/1).astype(np.int32),
                                             1],
                                      strides=[1, shape[1]//1, shape[2]//1, 1],
                                      padding='SAME')
        print('SPP layer level 1:', spp_1_1_pool.get_shape().as_list())

        spp_6_6_pool_flat = tf.reshape(spp_6_6_pool, [shape[0], -1])
        spp_3_3_pool_flat = tf.reshape(spp_3_3_pool, [shape[0], -1])
        spp_2_2_pool_flat = tf.reshape(spp_2_2_pool, [shape[0], -1])
        spp_1_1_pool_flat = tf.reshape(spp_1_1_pool, [shape[0], -1])

        spp_pool = tf.concat(1, [spp_6_6_pool_flat,
                                 spp_3_3_pool_flat,
                                 spp_2_2_pool_flat,
                                 spp_1_1_pool_flat])

    return spp_pool

但是当输入大小不同时，它不能保证相同长度的池化输出.

But it cannot gurantee the same length pooling output, when the input sizes are different.

如何解决这个问题?

推荐答案

我认为论文作者有误，公式应该是:

I believe the authors of the paper are wrong, the formula should be:

stride_size = floor(a/n)
kernel_size = floor(a/n) + (a mod n)

请注意，对于 n < 两个公式都给出相同的结果.4.你可以通过用 n 对 a 进行欧几里德除法来证明这个结果.

Notice that both formula give the same result for n < 4. You can prove this result by doing the euclidian division of a by n.

我修改了在 https://github.com/tensorflow/tensorflow/找到的代码问题/6011，这里是:

I modified the code I found at https://github.com/tensorflow/tensorflow/issues/6011 and here it is:

def spp_layer(input_, levels=(6, 3, 2, 1), name='SPP_layer'):
    shape = input_.get_shape().as_list()
    with tf.variable_scope(name):
        pyramid = []
        for n in levels:

            stride_1 = np.floor(float(shape[1] / n)).astype(np.int32)
            stride_2 = np.floor(float(shape[2] / n)).astype(np.int32)
            ksize_1 = stride_1 + (shape[1] % n)
            ksize_2 = stride_2 + (shape[2] % n)
            pool = tf.nn.max_pool(input_,
                                  ksize=[1, ksize_1, ksize_2, 1],
                                  strides=[1, stride_1, stride_2, 1],
                                  padding='VALID')

            # print("Pool Level {}: shape {}".format(n, pool.get_shape().as_list()))
            pyramid.append(tf.reshape(pool, [shape[0], -1]))
        spp_pool = tf.concat(1, pyramid)
    return spp_pool

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