def DSM_grid_sorting_masking_check(DSM,grid_size,threshold_angle): width = DSM.RasterXSize height = DSM.RasterYSize #计算网格数量 grid_num_y =int(np.ceil(height/grid_size)) grid_num_x =int(np.ceil(width/grid_size)) #初始化遮蔽检测结果矩阵 result = np.ones((grid_num_y,grid_num_x),dtype=bool) #计算每个格网进行遮蔽检测 for i in range(grid_num_y): for j in range(grid_num_x): #当前格网内的点坐标 y_min = i*grid_size y_max = min((i+1)*grid_size,height) x_min = j * grid_size x_max = min((j+1)*grid_size,width) coords = np.argwhere(DSM.ReadAsArray(x_min, y_min, x_max - x_min, y_max - y_min) > 0) coords[:, 0] += y_min coords[:, 1] += x_min # 构建KD树 tree = cKDTree(coords) # 查询每个点的最邻近点 k = 2 dist, ind = tree.query(coords, k=k) # 计算每个点的法向量 normals = np.zeros(coords.shape) for l in range(coords.shape[0]): if k == 2: p1 = coords[l, :] p2 = coords[ind[l, 1], :] else: p1 = coords[l, :] p2 = coords[ind[l, 1], :] normals[l, :] = np.cross(p1 - p2, p1 - DSM.ReadAsArray(p1[1], p1[0], 1, 1)) # 计算每个点的可见性 visibilities = np.zeros(coords.shape[0]) for l in range(coords.shape[0]): if k == 2: p1 = coords[l, :] p2 = coords[ind[l, 1], :] else: p1 = coords[l, :] p2 = coords[ind[l, 1], :] angle = np.cross(np.dot(normals[l, :], (p2 - p1) / dist[l, 1])) * 180 / np.pi if angle <= threshold_angle: visibilities[l] = 1 # 判断当前格网是否遮蔽 if np.sum(visibilities) == 0: result[i, j] = False else: result[i, j] = True return result dsm_path = 'C:/yingxiang/output.tif' DSM = gdal.Open(dsm_path) result = DSM_grid_sorting_masking_check(DSM,grid_size=10,threshold_angle=40) print(result)这段代码怎么改可以没有以下错误in method 'BandRasterIONumPy', argument 3 of type 'double'
时间: 2024-03-05 08:52:57 浏览: 332
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这个错误通常是因为传入的参数类型和函数参数类型不匹配导致的。在这段代码中,可能是以下这一行代码引起的问题:
```
coords = np.argwhere(DSM.ReadAsArray(x_min, y_min, x_max - x_min, y_max - y_min) > 0)
```
这里传入到ReadAsArray函数中的参数是x_min, y_min, x_max - x_min, y_max - y_min,这些参数的类型需要是int型的像素坐标。如果这些参数的类型不正确,就会导致函数调用出错,进而导致BandRasterIONumPy函数出错。
要解决这个问题,可以先将输入的像素坐标转换为int类型,代码如下:
```
x_min = int(x_min)
y_min = int(y_min)
```
需要在x_min和y_min的计算后立即进行类型转换。
同样地,还需要检查其他传入函数的参数类型是否正确,确保类型匹配。
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