9

Python 或其任何模块是否具有等效于 MATLAB 的conv2函数？更具体地说，我对conv2(A, B, 'same')在 MATLAB 中进行相同计算的东西感兴趣。

4

4 回答 4

9

虽然其他答案已经提到scipy.signal.convolve2d作为等价物，但我发现使用mode='same'.

虽然 Matlabconv2会在图像的底部和右侧产生伪影，但在图像scipy.signal.convolve2d的顶部和左侧具有相同的伪影。

有关显示行为的图表，请参阅这些链接（没有足够的声誉直接发布图像）：

曲折的芭芭拉左上角

曲折的芭芭拉右下角

以下包装器可能不是很有效，但在我的情况下通过将输入数组和输出数组都旋转 180 度解决了这个问题：

import numpy as np
from scipy.signal import convolve2d

def conv2(x, y, mode='same'):
    return np.rot90(convolve2d(np.rot90(x, 2), np.rot90(y, 2), mode=mode), 2)
于 2016-07-13T15:27:15.583 回答
5

Looks like scipy.signal.convolve2d is what you're looking for.

于 2010-09-16T23:25:30.390 回答
1 
scipy.ndimage.convolve

does it in n dimensions.

于 2010-09-16T21:55:38.327 回答
1

您必须为每个非单一维度提供偏移量才能重现 Matlab 的 conv2 的结果。仅支持“相同”选项的简单实现可以这样制作

import numpy as np
from scipy.ndimage.filters import convolve

def conv2(x,y,mode='same'):
    """
    Emulate the function conv2 from Mathworks.

    Usage:

    z = conv2(x,y,mode='same')

    TODO: 
     - Support other modes than 'same' (see conv2.m)
    """

    if not(mode == 'same'):
        raise Exception("Mode not supported")

    # Add singleton dimensions
    if (len(x.shape) < len(y.shape)):
        dim = x.shape
        for i in range(len(x.shape),len(y.shape)):
            dim = (1,) + dim
        x = x.reshape(dim)
    elif (len(y.shape) < len(x.shape)):
        dim = y.shape
        for i in range(len(y.shape),len(x.shape)):
            dim = (1,) + dim
        y = y.reshape(dim)

    origin = ()

    # Apparently, the origin must be set in a special way to reproduce
    # the results of scipy.signal.convolve and Matlab
    for i in range(len(x.shape)):
        if ( (x.shape[i] - y.shape[i]) % 2 == 0 and
             x.shape[i] > 1 and
             y.shape[i] > 1):
            origin = origin + (-1,)
        else:
            origin = origin + (0,)

    z = convolve(x,y, mode='constant', origin=origin)

    return z
于 2015-10-14T12:38:07.150 回答