Question

我需要在不连续的步骤中平滑包含高程的2D numpy数组。这里我Test Execution位于z-value = 1 (elevation)但是，它可以是介于0和1之间的任何值。我的数组如下所示：

(0, 0)

我尝试使用['1.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00'] ['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']应用余弦内核，然后使用scipy。像这样：

convolve2d

这导致数组看起来像这样：

import numpy as np
from scipy import signal

peak_array = np.array([
   [ 1.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

def elevation_grid():
    """Creates a smooth elevation-grid, from an array of peaks."""
    def gkern():
        """Returns a 2D Triangle kernel array."""
        # Decay is a cosine.
        # It is centered, so multiply by two, and add one
        # This way, we cover the entire grid, and assert
        # equal smoothing (due to an odd width)
        gkern1d_c = signal.cosine(peak_array.shape[0]*2 + 1)
        gkern1d_r = signal.cosine(peak_array.shape[1]*2 + 1)

        gkern2d = np.outer(gkern1d_c, gkern1d_r)
        return gkern2d

    kernel = gkern()
    grad = signal.convolve2d(peak_array, kernel, mode='same')

    # Normalize the grid
    grad -= np.amin(grad)
    grad /= np.amax(grad)

    return grad

def print_readable(array):
    """Prints the map to a human-readable format."""
    for row in range(0, array.shape[0]):
        # Round to two decimals
        r = ["%.2f" % array[col][row] for col in range(0, array.shape[1])]
        print(r)

smooth_array = elevation_grid()

print_readable(smooth_array)

预期结果如何。但是，如果我在每个角落放置峰值：

['1.00', '0.99', '0.95', '0.90', '0.82', '0.72', '0.60', '0.47', '0.33', '0.18']
['0.99', '0.98', '0.94', '0.89', '0.81', '0.71', '0.60', '0.47', '0.33', '0.18']
['0.95', '0.94', '0.91', '0.85', '0.78', '0.68', '0.57', '0.45', '0.32', '0.17']
['0.90', '0.89', '0.85', '0.80', '0.73', '0.64', '0.54', '0.42', '0.29', '0.16']
['0.82', '0.81', '0.78', '0.73', '0.67', '0.59', '0.49', '0.38', '0.27', '0.14']
['0.72', '0.71', '0.68', '0.64', '0.59', '0.51', '0.43', '0.33', '0.23', '0.12']
['0.60', '0.60', '0.57', '0.54', '0.49', '0.43', '0.36', '0.28', '0.19', '0.09']
['0.47', '0.47', '0.45', '0.42', '0.38', '0.33', '0.28', '0.21', '0.14', '0.06']
['0.33', '0.33', '0.32', '0.29', '0.27', '0.23', '0.19', '0.14', '0.09', '0.03']
['0.18', '0.18', '0.17', '0.16', '0.14', '0.12', '0.09', '0.06', '0.03', '0.00']

他们在中间加起来导致中心的一个更大的峰值，我期望中间的最低值......

['1.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '1.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00']
['1.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '0.00', '1.00']

如何抚平我的山峰？它需要高效，因为我有大型阵列（高达1000x1000）。

Answer 1

因此，如果k是你的内核，并且*是卷积，并且v1，... v4输入矩阵就像你发布的第一个矩阵一样，你可以将它们相加并获得另一个矩阵作为w

w = v1 + v2 + v3 + v4

您正在应用卷积，这是一个线性操作。

输出= k * w = k *（v1 + v2 + v3 + v4）= k * v1 + k2 * v2 ...

如果你注意到，你在拐角处输入四个1（Inp2）与初始矩阵的四个总和基本相同，只有一个1（Inp1）。

你可以通过简单地旋转90度Inp1和求和（在numpy，transposes和fliprl中）来构建Inp2。因此，您可以旋转第一个计算的结果并将它们相加以找到正确的最终结果。

Inp1的四个旋转版本的总和不太可能与适用于Inp2的卷积不同。

如果是这种情况，则可能是一个数字问题（如果您使用这些数字则不应该发生）。检查如何使用max和min重新缩放数据。你除以max，所以要确保它是非负的，超过eps

平滑的离散2D阵列

1 个答案: