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我的要求是在 Java 中提供一个函数，该函数使用线性代数在三维坐标系中找到“校准”的 xyz 位置。以下是我正在尝试实现的接口。如果需要，可以更改参数以适应特定的库。 

/**
 * Given a nominal target xyz point within a three dimensional robotic coordinate system, calculate the theoretical x'y'z'
 * transformation using input arrays that represent the nominal coordinates and actual measured coordinates corresponding to
 * each nominal point.
 *
 * @param nominalPoints - a sampling of nominal values (per engineering specs) for a 3-axis robotic system
 *    e.g.     double[][] nominalPoints = new double[][]{
 *                 new double[]{30.0d, 68.0d, 1.0d},
 *                 new double[]{50.0d, 6.0d, 1.0d},
 *                 new double[]{110.d, 20.0d, 1.0d},
 *                 new double[]{35.0d, 15.0d, 1.0d},
 *                 new double[]{45.0d, 97.0d, 1.0d}
 *              };
 * @param actualPoints - actual/measured values corresponding to the nominal data points.  These points represent the
 *                     variance from nominal due to manufacturing tolerances.
 *    e.g     double[][] measuredPoints = new double[][]{
 *                 new double[]{30.5d, 68.1d, 1.01d},
 *                 new double[]{50.4d, 6.2d, 1.02d},
 *                 new double[]{110.3d, 20.3d, 1.03d},
 *                 new double[]{35.2d, 15.4d, 1.04d},
 *                 new double[]{45.1d, 97.5d, 1.05d}
 *             };
 * @param targetPoint - an x/y/z point in system to transform into an estimated value based on a least-squared evaluation of the
 *                    nominal and actual arrays
 *
 *     e.g.    float[] targetPoint = new float[]{75.21f, 17.56f, 2.765f};
 *
 * @return
 */
public float[] getCalibratedPoint(float[][] nominalPoints, float[][] actualPoints, float[] targetPoint);

下面的代码是我被告知可以在 Python 中使用 numpy 工作的一种解决方案，但我很擅长线性代数，并且很难找到 Java 中的实现来做到这一点。具体来说，我还没有找到一个等效于 np.hstack(..)、np.ones(..) 和一个最小二乘函数，它接受 2 个数组参数，如 np.linalg.lstsq(..)。到目前为止，我已经看过 Apache Commons 和 EJML

import numpy as np

# These are the points in the Tray Coordinate System.
primary = np.array([[40., 1160., 0.],
                    [40., 40., 0.],
                    [260., 40., 0.],
                    [260., 1160., 0.]])

# These are points in the Stage Coordinate System.
secondary = np.array([[610., 560., 0.],
                      [610., -560., 0.],
                      [390., -560., 0.],
                      [390., 560., 0.]])

# Pad the data with ones, so that our transformation can do translations too
def pad(x): return np.hstack([x, np.ones((x.shape[0], 1))])
def unpad(x): return x[:, :-1]

# This is the transform function. Pass Tray Coordinates to this.
def transform(x): return unpad(np.dot(pad(x), A))

X = pad(primary)
Y = pad(secondary)

# Solve the least squares problem X * A = Y
# to find our transformation matrix A
A, res, rank, s = np.linalg.lstsq(X, Y, rcond=None)  # This is where the important work is done.

# Transforming a single point.
print('Transforming point (1, 2, 3) ...')
print(transform(np.array([[1, 2, 3]])))
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