python代码判断点是否在cad块内

你可以使用以下代码来判断一个点是否在CAD块内：

def is_point_inside_block(point, block):
    # point: (x, y, z) coordinates of point
    # block: list of (x, y, z) coordinates of block vertices
    
    # Create a point in infinity
    inf_point = (float("inf"), float("inf"), float("inf"))
    
    # Initialize a counter for intersections
    count = 0
    
    # Loop through each pair of vertices in the block
    for i in range(len(block)):
        v1 = block[i]
        if i == len(block) - 1:
            v2 = block[0]
        else:
            v2 = block[i+1]
        
        # Check if the point is on the same plane as the block face
        if (v1[0] == v2[0] == point[0] and 
            min(v1[1], v2[1]) <= point[1] <= max(v1[1], v2[1]) and 
            min(v1[2], v2[2]) <= point[2] <= max(v1[2], v2[2])):
            return True
        
        # Check if the point is on the same line as the block edge
        if (v1[0] == v2[0] and v1[0] == point[0] and 
            min(v1[1], v2[1]) <= point[1] <= max(v1[1], v2[1]) and
            min(v1[2], v2[2]) <= point[2] <= max(v1[2], v2[2])):
            return True
        
        # Check if the point intersects the block edge
        if (v1[0] < point[0] <= v2[0] or v2[0] < point[0] <= v1[0]):
            if v1[1] == v2[1]:
                x = point[0]
                y = v1[1]
            else:
                slope = (v2[1] - v1[1]) / (v2[0] - v1[0])
                x = point[0]
                y = slope * (x - v1[0]) + v1[1]
            if y == point[1] and point[2] < max(v1[2], v2[2]):
                count += 1
        
        # Check if the block edge intersects the ray from the point to infinity
        if ((v1[2] > point[2]) != (v2[2] > point[2])):
            if v1[2] == v2[2]:
                x = v1[0]
                y = v1[1]
            else:
                slope = (v2[1] - v1[1]) / (v2[2] - v1[2])
                x = v1[0]
                y = slope * (x - v1[2]) + v1[1]
            if y > point[1]:
                count += 1
                
    return count % 2 == 1

该函数接受两个参数：点的坐标和块的坐标列表。如果点在块内，函数将返回True。否则，函数将返回False。

请注意，该函数只适用于无限平面上的块，并且假定点和块都不在无限远处。如果点或块在无限远处，则需要使用其他方法来解决这个问题。