Python 将Numpy Lstsq残值转换为R^2
我正在进行如下的最小二乘回归(单变量)。我想用R^2来表示结果的重要性。Numpy返回未缩放残差的值,这是一种合理的归一化方法Python 将Numpy Lstsq残值转换为R^2,python,numpy,linear-regression,Python,Numpy,Linear Regression,我正在进行如下的最小二乘回归(单变量)。我想用R^2来表示结果的重要性。Numpy返回未缩放残差的值,这是一种合理的归一化方法 field_clean,back_clean = rid_zeros(backscatter,field_data) num_vals = len(field_clean) x = field_clean[:,row:row+1] y = 10*log10(back_clean) A = hstack([x, ones((num_vals,1))]) soln = l
field_clean,back_clean = rid_zeros(backscatter,field_data)
num_vals = len(field_clean)
x = field_clean[:,row:row+1]
y = 10*log10(back_clean)
A = hstack([x, ones((num_vals,1))])
soln = lstsq(A, y )
m, c = soln [0]
residues = soln [1]
print residues
看
你的R2值=
1 - residual / sum((y - y.mean())**2)
这相当于
1 - residual / (n * y.var())
例如:
import numpy as np
# Make some data...
n = 10
x = np.arange(n)
y = 3 * x + 5 + np.random.random(n)
# Note that polyfit is an easier way to do this...
# It would just be "model, resid = np.polyfit(x,y,1,full=True)[:2]"
A = np.vstack((x, np.ones(n))).T
model, resid = np.linalg.lstsq(A, y)[:2]
r2 = 1 - resid / (y.size * y.var())
print r2
如果出现
scipy.linalg.lstsq
,我们可以执行此操作吗!?您给出的示例如何处理3D数据!?无法得到它。。