[开发技巧]·AdaptivePooling与Max/AvgPooling相互转换
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1.问题描述
自适应池化Adaptive Pooling是PyTorch的一种池化层,根据1D,2D,3D以及Max与Avg可分为六种形式。
自适应池化Adaptive Pooling与标准的Max/AvgPooling区别在于,自适应池化Adaptive Pooling会根据输入的参数来控制输出output_size,而标准的Max/AvgPooling是通过kernel_size,stride与padding来计算output_size:
output_size = ceil ( (input_size+2∗padding−kernel_size)/stride)+1
Adaptive Pooling仅存在与PyTorch,如果需要将包含Adaptive Pooling的代码移植到Keras或者TensorFlow就会遇到问题。
本文将提供一个公式,可以简便的将AdaptivePooling准换为Max/AvgPooling,便于大家移植使用。
2.原理讲解
我们已经知道了普通Max/AvgPooling计算公式为:output_size = ceil ( (input_size+2∗padding−kernel_size)/stride)+1
当我们使用Adaptive Pooling时,这个问题就变成了由已知量input_size,output_size求解kernel_size与stride
为了简化问题,我们将padding设为0(后面我们可以发现源码里也是这样操作的c++源码部分)
stride = floor ( (input_size / (output_size) )
kernel_size = input_size − (output_size−1) * stride
3.实战演示
下面我们通过一个实战来操作一下,验证公式的正确性
输出结果
tensor([[[ 0.9095, 0.8043, 0.4052, 0.3410, 1.8831, 0.8703, -0.0839],
[ 0.3300, -1.2951, -1.8148, -1.1118, -1.1091, 1.5657, 0.7093],
[-0.6788, -1.2790, -0.6456, 1.9085, 0.8627, 1.1711, 0.5614],
[-0.0129, -0.6447, -0.6685, -1.2087, 0.8535, -1.4802, 0.5274],
[ 0.7347, 0.0374, -1.7286, -0.7225, -0.4257, -0.0819, -0.9878],
[-1.2553, -1.0774, -0.1936, -1.4741, -0.9028, -0.1584, -0.6612]],
[[-0.3473, 1.0599, -1.5744, -0.2023, -0.5336, 0.5512, -0.3200],
[-0.2518, 0.1714, 0.6862, 0.3334, -1.2693, -1.3348, -0.0878],
[ 1.0515, 0.1385, 0.4050, 0.8554, 1.0170, -2.6985, 0.3586],
[-0.1977, 0.8298, 1.6110, -0.9102, 0.7129, 0.2088, 0.9553],
[-0.2218, -0.7234, -0.4407, 1.0369, -0.8884, 0.3684, 1.2134],
[ 0.5812, 1.1974, -0.1584, -0.0903, -0.0628, 3.3684, 2.0330]]])
tensor([[[-0.3627, 0.0799, 0.7145],
[-0.5343, -0.7190, -0.3686]],
[[ 0.1488, -0.0314, -0.4797],
[ 0.2753, 0.0900, 0.8788]]])
tensor([[[-0.3627, 0.0799, 0.7145],
[-0.5343, -0.7190, -0.3686]],
[[ 0.1488, -0.0314, -0.4797],
[ 0.2753, 0.0900, 0.8788]]])
可以发现adp = t.nn.AdaptiveAvgPool2d(list(outputsz))与avg = t.nn.AvgPool2d(kernel_size=list(kernelsz),stride=list(stridesz))结果一致
为了防止这是偶然现象,修改参数,使用AdaptiveAvgPool1d进行试验
输出结果
tensor([[[ 1.3405, 0.3509, -1.5119, -0.1730, 0.6971, 0.3399, -0.0874,
-1.2417, 0.6564],
[ 2.0482, 0.3528, 0.0703, 1.2012, -0.8829, -0.3156, 1.0603,
-0.7722, -0.6086],
[ 1.0470, -0.9374, 0.3594, -0.8068, 0.5126, 1.4135, 0.3538,
-1.0973, 0.3046]],
[[-0.1688, 0.7300, -0.3457, 0.5645, -1.2507, -1.9724, 0.4469,
-0.3362, 0.7910],
[ 0.5676, -0.0614, -0.0243, 0.1529, 0.8276, 0.2452, -0.1783,
0.7460, 0.2577],
[-0.1433, -0.7047, -0.4883, 1.2414, -1.4316, 0.9704, -1.7088,
-0.0094, -0.3739]]])
tensor([[[ 0.0598, -0.3293, 0.3165, -0.2242],
[ 0.8237, 0.1295, -0.0461, -0.1069],
[ 0.1563, 0.0217, 0.7600, -0.1463]],
[[ 0.0718, -0.3440, -0.9254, 0.3006],
[ 0.1606, 0.3187, 0.2982, 0.2751],
[-0.4454, -0.2262, -0.7233, -0.6973]]])
tensor([[[ 0.0598, -0.3293, 0.3165, -0.2242],
[ 0.8237, 0.1295, -0.0461, -0.1069],
[ 0.1563, 0.0217, 0.7600, -0.1463]],
[[ 0.0718, -0.3440, -0.9254, 0.3006],
[ 0.1606, 0.3187, 0.2982, 0.2751],
[-0.4454, -0.2262, -0.7233, -0.6973]]])
可以发现adp = t.nn.AdaptiveAvgPool1d(list(outputsz))与avg = t.nn.AvgPool1d(kernel_size=list(kernelsz),stride=list(stridesz))结果也是相同的。
4.总结分析
在以后遇到别人代码使用Adaptive Pooling,可以通过这两个公式转换为标准的Max/AvgPooling,从而应用到不同的学习框架中
stride = floor ( (input_size / (output_size) )
kernel_size = input_size − (output_size−1) * stride
只需要知道输入的input_size ,就可以推导出stride 与kernel_size ,从而替换为标准的Max/AvgPooling
Hope this helps