如果你试图从另外两个值中预测一个值，那么你应该使用lstsq和a参数作为自变量（加上一列1来估计截距），b作为因变量。
另一方面，如果你只是想得到数据的最佳拟合线，即。如果你把数据投影到这条线上，这条线会使真实的点和它的投影之间的平方距离最小，那么你需要的是第一个主成分。
定义它的一种方法是其方向向量是协方差矩阵的特征向量的直线，该特征向量对应于最大特征值，该特征值通过数据的均值。也就是说，eig(cov(data))是一种非常糟糕的计算方法，因为它做了很多不必要的计算和复制，并且可能比使用svd更不准确。见下文：

import numpy as np

# Generate some data that lies along a line

x = np.mgrid[-2:5:120j]
y = np.mgrid[1:9:120j]
z = np.mgrid[-5:3:120j]

data = np.concatenate((x[:, np.newaxis], 
                       y[:, np.newaxis], 
                       z[:, np.newaxis]), 
                      axis=1)

# Perturb with some Gaussian noise
data += np.random.normal(size=data.shape) * 0.4

# Calculate the mean of the points, i.e. the 'center' of the cloud
datamean = data.mean(axis=0)

# Do an SVD on the mean-centered data.
uu, dd, vv = np.linalg.svd(data - datamean)

# Now vv[0] contains the first principal component, i.e. the direction
# vector of the 'best fit' line in the least squares sense.

# Now generate some points along this best fit line, for plotting.

# I use -7, 7 since the spread of the data is roughly 14
# and we want it to have mean 0 (like the points we did
# the svd on). Also, it's a straight line, so we only need 2 points.
linepts = vv[0] * np.mgrid[-7:7:2j][:, np.newaxis]

# shift by the mean to get the line in the right place
linepts += datamean

# Verify that everything looks right.

import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d as m3d

ax = m3d.Axes3D(plt.figure())
ax.scatter3D(*data.T)
ax.plot3D(*linepts.T)
plt.show()

它看起来是这样的：

如果你的数据表现得相当好，那么它应该足以找到分量距离的最小二乘和。然后你可以找到线性回归，z独立于x，然后再次独立于y。
以documentation为例：

import numpy as np

pts = np.add.accumulate(np.random.random((10,3)))
x,y,z = pts.T

# this will find the slope and x-intercept of a plane
# parallel to the y-axis that best fits the data
A_xz = np.vstack((x, np.ones(len(x)))).T
m_xz, c_xz = np.linalg.lstsq(A_xz, z)[0]

# again for a plane parallel to the x-axis
A_yz = np.vstack((y, np.ones(len(y)))).T
m_yz, c_yz = np.linalg.lstsq(A_yz, z)[0]

# the intersection of those two planes and
# the function for the line would be:
# z = m_yz * y + c_yz
# z = m_xz * x + c_xz
# or:
def lin(z):
    x = (z - c_xz)/m_xz
    y = (z - c_yz)/m_yz
    return x,y

#verifying:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt

fig = plt.figure()
ax = Axes3D(fig)
zz = np.linspace(0,5)
xx,yy = lin(zz)
ax.scatter(x, y, z)
ax.plot(xx,yy,zz)
plt.savefig('test.png')
plt.show()

如果你想最小化从直线（与直线正交）到三维空间中的点的实际正交距离（我不确定这是否被称为线性回归）。然后我将构建一个计算RSS的函数，并使用scipy.optimize最小化函数来解决它。