pysdic.PointCloud.all_close#

PointCloud.all_close(other, rtol=1e-05, atol=1e-08, equal_nan=True, ordered=True)[source]#

Check if all points in the current point cloud are approximately equal to the points in another PointCloud instance within a tolerance.

This method compares the points of the current point cloud with those of another PointCloud instance and returns True if all corresponding points are approximately equal within the specified relative and absolute tolerances.

Note

If the number of points in both point clouds differ, the method will return False.

See also

Parameters:

  • other (Union[PointCloud, ArrayLike]) – Another PointCloud instance or a NumPy array with shape \((N_p, 3)\) to compare with the current point cloud.

  • rtol (Real, optional) – The relative tolerance parameter (default is 1e-05).

  • atol (Real, optional) – The absolute tolerance parameter (default is 1e-08).

  • equal_nan (bool, optional) – If True, NaN values are considered equal (default is True).

  • ordered (bool, optional) – If True, the points are compared in order. If False, the points are compared without considering the order (default is True).

Returns:

True if all points are approximately equal within the specified tolerances, False otherwise.

Return type:

bool

Raises:

ValueError – If the input is not an instance of PointCloud.

Examples

Creating a PointCloud object from a random NumPy array.

1import numpy as np
2from pysdic import PointCloud
3
4# Create a random point cloud with 100 points
5random_points = np.random.rand(100, 3)  # shape (100, 3)
6point_cloud1 = PointCloud.from_array(random_points)

Compare the point cloud with another point cloud that is slightly modified.

1# Create a second point cloud by adding small noise to the first one
2noise = np.random.normal(scale=1e-8, size=random_points.shape)
3point_cloud2 = PointCloud.from_array(random_points + noise)
4
5# Check if the two point clouds are approximately equal
6are_close = point_cloud1.all_close(point_cloud2, rtol=1e-5, atol=1e-8)
7print(are_close)
8# Output: True (most likely, depending on the noise)

Compare with a point cloud that is significantly different.

1# Create a third point cloud that is significantly different
2different_points = np.random.rand(100, 3) + 1.0  # Shifted by 1.0
3point_cloud3 = PointCloud.from_array(different_points)
4are_close = point_cloud1.all_close(point_cloud3, rtol=1e-5, atol=1e-8)
5print(are_close)
6# Output: False
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