如何使用numpy在2d数组上执行最大/平均池化
如何使用numpy在2d数组上执行最大/平均池化
提问于 2017-02-26 08:06:16
给定一个2D(M X N)矩阵和一个2D Kernel(K X L)，如何返回一个矩阵，该矩阵是在图像上使用给定内核进行最大或平均池化的结果？
如果可能的话，我想使用numpy。
注: M，N，K，L可以是偶数，也可以是奇数，它们不需要彼此完全整除，例如: 7x5矩阵和2x2核。
最大池化的示例：
matrix:
array([[ 20, 200, -5, 23],
[ -13, 134, 119, 100],
[ 120, 32, 49, 25],
[-120, 12, 09, 23]])
kernel: 2 x 2
soln:
array([[ 200, 119],
[ 120, 49]])回答 5
Stack Overflow用户
发布于 2017-02-26 08:54:10
您可以使用scikit image block_reduce
import numpy as np
import skimage.measure
a = np.array([
[ 20, 200, -5, 23],
[ -13, 134, 119, 100],
[ 120, 32, 49, 25],
[-120, 12, 9, 23]
])
skimage.measure.block_reduce(a, (2,2), np.max)提供：
array([[200, 119],
[120, 49]])Stack Overflow用户
发布于 2017-02-26 08:57:40
如果图像大小可以被内核大小整除，则可以调整数组的形状，并根据需要使用max或mean
import numpy as np
mat = np.array([[ 20, 200, -5, 23],
[ -13, 134, 119, 100],
[ 120, 32, 49, 25],
[-120, 12, 9, 23]])
M, N = mat.shape
K = 2
L = 2
MK = M // K
NL = N // L
print(mat[:MK*K, :NL*L].reshape(MK, K, NL, L).max(axis=(1, 3)))
# [[200, 119], [120, 49]] 如果没有偶数个内核，则必须单独处理边界。(正如注释中指出的，这会导致矩阵被复制，这将影响性能)。
mat = np.array([[20, 200, -5, 23, 7],
[-13, 134, 119, 100, 8],
[120, 32, 49, 25, 12],
[-120, 12, 9, 23, 15],
[-57, 84, 19, 17, 82],
])
# soln
# [200, 119, 8]
# [120, 49, 15]
# [84, 19, 82]
M, N = mat.shape
K = 2
L = 2
MK = M // K
NL = N // L
# split the matrix into 'quadrants'
Q1 = mat[:MK * K, :NL * L].reshape(MK, K, NL, L).max(axis=(1, 3))
Q2 = mat[MK * K:, :NL * L].reshape(-1, NL, L).max(axis=2)
Q3 = mat[:MK * K, NL * L:].reshape(MK, K, -1).max(axis=1)
Q4 = mat[MK * K:, NL * L:].max()
# compose the individual quadrants into one new matrix
soln = np.vstack([np.c_[Q1, Q3], np.c_[Q2, Q4]])
print(soln)
# [[200 119 8]
# [120 49 15]
# [ 84 19 82]]Stack Overflow用户
发布于 2018-03-16 17:50:26
而不是像Elliot的答案所示的那样创建“象限”，我们可以填充它，使其可均匀整除，然后执行最大池或均值池。
由于在CNN中经常使用池化，因此输入数组通常是3D的。所以我做了一个可以在2D或3D数组上工作的函数。
def pooling(mat,ksize,method='max',pad=False):
'''Non-overlapping pooling on 2D or 3D data.
<mat>: ndarray, input array to pool.
<ksize>: tuple of 2, kernel size in (ky, kx).
<method>: str, 'max for max-pooling,
'mean' for mean-pooling.
<pad>: bool, pad <mat> or not. If no pad, output has size
n//f, n being <mat> size, f being kernel size.
if pad, output has size ceil(n/f).
Return <result>: pooled matrix.
'''
m, n = mat.shape[:2]
ky,kx=ksize
_ceil=lambda x,y: int(numpy.ceil(x/float(y)))
if pad:
ny=_ceil(m,ky)
nx=_ceil(n,kx)
size=(ny*ky, nx*kx)+mat.shape[2:]
mat_pad=numpy.full(size,numpy.nan)
mat_pad[:m,:n,...]=mat
else:
ny=m//ky
nx=n//kx
mat_pad=mat[:ny*ky, :nx*kx, ...]
new_shape=(ny,ky,nx,kx)+mat.shape[2:]
if method=='max':
result=numpy.nanmax(mat_pad.reshape(new_shape),axis=(1,3))
else:
result=numpy.nanmean(mat_pad.reshape(new_shape),axis=(1,3))
return result有时，您可能希望以不等于内核大小的步幅执行重叠池化。下面是一个可以做到这一点的函数，无论是否使用填充：
def asStride(arr,sub_shape,stride):
'''Get a strided sub-matrices view of an ndarray.
See also skimage.util.shape.view_as_windows()
'''
s0,s1=arr.strides[:2]
m1,n1=arr.shape[:2]
m2,n2=sub_shape
view_shape=(1+(m1-m2)//stride[0],1+(n1-n2)//stride[1],m2,n2)+arr.shape[2:]
strides=(stride[0]*s0,stride[1]*s1,s0,s1)+arr.strides[2:]
subs=numpy.lib.stride_tricks.as_strided(arr,view_shape,strides=strides)
return subs
def poolingOverlap(mat,ksize,stride=None,method='max',pad=False):
'''Overlapping pooling on 2D or 3D data.
<mat>: ndarray, input array to pool.
<ksize>: tuple of 2, kernel size in (ky, kx).
<stride>: tuple of 2 or None, stride of pooling window.
If None, same as <ksize> (non-overlapping pooling).
<method>: str, 'max for max-pooling,
'mean' for mean-pooling.
<pad>: bool, pad <mat> or not. If no pad, output has size
(n-f)//s+1, n being <mat> size, f being kernel size, s stride.
if pad, output has size ceil(n/s).
Return <result>: pooled matrix.
'''
m, n = mat.shape[:2]
ky,kx=ksize
if stride is None:
stride=(ky,kx)
sy,sx=stride
_ceil=lambda x,y: int(numpy.ceil(x/float(y)))
if pad:
ny=_ceil(m,sy)
nx=_ceil(n,sx)
size=((ny-1)*sy+ky, (nx-1)*sx+kx) + mat.shape[2:]
mat_pad=numpy.full(size,numpy.nan)
mat_pad[:m,:n,...]=mat
else:
mat_pad=mat[:(m-ky)//sy*sy+ky, :(n-kx)//sx*sx+kx, ...]
view=asStride(mat_pad,ksize,stride)
if method=='max':
result=numpy.nanmax(view,axis=(2,3))
else:
result=numpy.nanmean(view,axis=(2,3))
return result页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/42463172
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