Finding x,y,z points from x,y coordinates-CodePudding

I have an array of 3D points called points. Here are the first 20 points of that array:

points = np.array([[-1.33309284,-1.30808319,4.00199986],[-1.33209359,-1.30417236,3.99900007],[-1.33109427,-1.30026584,3.99600005],[-1.35235626,-1.30090892,4.00699997],[-1.34447409,-1.29896096,4.00099993],[-1.33627494,-1.29668835,3.99399996],[-1.35134384,-1.29700102,4.00400019],[-1.34346598,-1.29505737,3.99799991],[-1.33527122,-1.29278989,3.99099994],[-0.43118706,-1.29732492,4.00500011],[-0.42564261,-1.29829663,4.0079999,],[-1.35033125,-1.29309735,4.00099993],[-1.34245787,-1.29115818,3.99499989],[-1.33393295,-1.28857266,3.98699999],[-1.35809791,-1.28919816,3.99799991],[-1.35089223,-1.28790834,3.99399996],[-1.34470857,-1.2875859,3.99300003],[-1.36034349,-1.28562515,3.99600005],[-1.35381569,-1.28498166,3.99399996],[-1.34695627,-1.28401647,3.99099994]])

I have four x,y coordinates, which represent estimated corner points

corner_points = np.array([[-1.33109427,-1.30026584],[-1.33527122,-1.29278989],[-1.35089223,-1.28790834],[-1.33393295,-1.28857266]])

I want to find the four points in the points array that match the x,y coordinates of the corner points. I know that I can do an elaborate for loop where I check the four x,y coordinates agaist all of the points, but I assume that numpy.where can do it more effectively?

I have tried:

points_with_height = np.where(points[0:1] == corner_points[0:1])

But it returns empty

I have searched for ways to compare 2D arrays but they only provides slicing of one element in the array.

I just tested with the for loop:

for point in points:
    if point[:1] == corner_points[0][:1] or point[:1] == corner_points[1][:1] or point[:1] == corner_points[2][:1] or point[:1] == corner_points[3][:1]:
        print(point)

It worked perfectly, but it is just not very pretty and I assume that there must be a better way to do it?

CodePudding user response：

First, I agree with Jérôme Richard that it isn't good practice to assume equality between the 2D/3D points (usually, one would "unproject" the 2D points to do a similar thing here).

Here is an approach using numpy. The relations/mapping aspect of numpy isn't as great as pandas, so there may be alternate approaches if you wanted to expand the dependencies into a more relational library such as it (Pandas uses Numpy behind the scenes).

At this time, I can't think of a way to conditionally reduce the mask_2d array into a 1d mask only when both conditions are true (without pandas), but I am sure there is a way. Anyone else, feel free to expand on this answer.

import numpy as np

points = np.array([[-1.33309284,-1.30808319,4.00199986],[-1.33209359,-1.30417236,3.99900007],[-1.33109427,-1.30026584,3.99600005],[-1.35235626,-1.30090892,4.00699997],[-1.34447409,-1.29896096,4.00099993],[-1.33627494,-1.29668835,3.99399996],[-1.35134384,-1.29700102,4.00400019],[-1.34346598,-1.29505737,3.99799991],[-1.33527122,-1.29278989,3.99099994],[-0.43118706,-1.29732492,4.00500011],[-0.42564261,-1.29829663,4.0079999,],[-1.35033125,-1.29309735,4.00099993],[-1.34245787,-1.29115818,3.99499989],[-1.33393295,-1.28857266,3.98699999],[-1.35809791,-1.28919816,3.99799991],[-1.35089223,-1.28790834,3.99399996],[-1.34470857,-1.2875859,3.99300003],[-1.36034349,-1.28562515,3.99600005],[-1.35381569,-1.28498166,3.99399996],[-1.34695627,-1.28401647,3.99099994]])

# corner_points: removed [-1.26,0.48,3.5999999999999996], and fixed too many ] 
corner_points = np.array([-1.33109427,-1.30026584],[-1.33527122,-1.29278989],,[-1.33393295,-1.28857266],[-1.36034349,-1.28562515]]) 

mask_2d = np.isin(points, corner_points)
""" 
Out[92]: 
array(
       [[False, False, False],
       [False, False, False],
       [ True,  True, False], ....
"""

mask_1d = mask[:,0] # Note that this assumes the first and second value  are equal. Possible hole in code.
"""
Out[93]: 
array([False, False,  True, False, False, False, False, False,  True,
   False, False, False, False,  True, False, False, False,  True,
   False, False])
"""

points[mask_1d]
"""
Out[94]: 
array([[-1.33109427, -1.30026584,  3.99600005],
       [-1.33527122, -1.29278989,  3.99099994],
       [-1.33393295, -1.28857266,  3.98699999],
       [-1.36034349, -1.28562515,  3.99600005]])
"""