在下面的代码中，我希望两个打印语句打印相同的结果，因为我明确地将参数传递s给两个预测函数。
library(glmnet)
set.seed(1)
x=rnorm(100)
eps=rnorm(100)
y = 1 + x + x^2 + x^3 + eps
xmat=model.matrix(y~poly(x,10,raw=T),data=data.frame(x=x))
grid=10^seq(10,-2,length=100)
lasso.mod = glmnet(xmat,y, alpha=1,lambda=grid)
lasso.coef=predict(lasso.mod,type="coefficients",s=0.01495444)[1:10,]
print(lasso.coef)
lasso.mod = glmnet(xmat,y, alpha=1,lambda=5)
lasso.coef=predict(lasso.mod,type="coefficients",s=0.01495444)[1:10,]
print(lasso.coef)
但是，结果非常不同，我想了解原因。
(Intercept) (Intercept) poly(x, 10, raw = T)1
1.1329454011 0.0000000000 1.3081576745
poly(x, 10, raw = T)2 poly(x, 10, raw = T)3 poly(x, 10, raw = T)4
0.6887020751 0.6576599481 0.0336098492
poly(x, 10, raw = T)5 poly(x, 10, raw = T)6 poly(x, 10, raw = T)7
0.0566899437 0.0002744787 0.0006870169
poly(x, 10, raw = T)8
0.0001053833
(Intercept) (Intercept) poly(x, 10, raw = T)1
2.092266 0.000000 0.000000
poly(x, 10, raw = T)2 poly(x, 10, raw = T)3 poly(x, 10, raw = T)4
0.000000 0.000000 0.000000
poly(x, 10, raw = T)5 poly(x, 10, raw = T)6 poly(x, 10, raw = T)7
0.000000 0.000000 0.000000
poly(x, 10, raw = T)8
0.000000
我做了一个实验，我改成lasso.mod = glmnet(xmat,y, alpha=1,lambda=5)，
lasso.mod = glmnet(xmat,y, alpha=1,lambda=0.015)结果更接近了。
似乎 predict 函数依赖于grid传递给训练函数的 ，但文档似乎表明s参数 onpredict应该覆盖它。是否存在依赖关系，如果是，它是什么以及如何解决它来处理任意系数s？
更新 1：我在glmnet.
Do not
supply a single value for 'lambda' (for predictions after CV
use 'predict()' instead). Supply instead a decreasing
sequence of 'lambda' values. 'glmnet' relies on its warms
starts for speed, and its often faster to fit a whole path
than compute a single fit.
警告仅指性能而不是准确性/稳定性，但我尝试改变不同的递减顺序grid，结果predict仍然不同。