mat*_*h11 5 regression r lasso-regression lars
我试图使用lars包进行套索回归,但似乎无法使lars位工作.我输入了代码:
diabetes<-read.table("diabetes.txt", header=TRUE)
diabetes
library(lars)
diabetes.lasso = lars(diabetes$x, diabetes$y, type = "lasso")
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但是,我收到一条错误消息:rep(1,n)中的错误:无效的'times'参数.
我试过这样输入:
diabetes<-read.table("diabetes.txt", header=TRUE)
library(lars)
data(diabetes)
diabetes.lasso = lars(age+sex+bmi+map+td+ldl+hdl+tch+ltg+glu, y, type = "lasso")
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但后来我收到错误消息:'lars中的错误(年龄+性别+ bmi +地图+ td + ldl + hdl + tch + ltg + glu,y,type ="lasso"):找不到对象'age'
我哪里错了?
编辑:数据 - 如下所示,但另外5列.
ldl hdl tch ltg glu
1 -0.034820763 -0.043400846 -0.002592262 0.019908421 -0.017646125
2 -0.019163340 0.074411564 -0.039493383 -0.068329744 -0.092204050
3 -0.034194466 -0.032355932 -0.002592262 0.002863771 -0.025930339
4 0.024990593 -0.036037570 0.034308859 0.022692023 -0.009361911
5 0.015596140 0.008142084 -0.002592262 -0.031991445 -0.046640874
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lars::lars似乎没有公式接口,这意味着您不能对列名称使用公式规范(而且它不接受“data=”参数)。有关此主题和其他“数据挖掘”主题的更多信息,您可能需要获取经典文本的副本:“统计学习的要素”。尝试这个:
# this obviously assumes require(lars) and data(diabetes) have been executed.
> diabetes.lasso = with( diabetes, lars(x, y, type = "lasso"))
> summary(diabetes.lasso)
LARS/LASSO
Call: lars(x = x, y = y, type = "lasso")
Df Rss Cp
0 1 2621009 453.7263
1 2 2510465 418.0322
2 3 1700369 143.8012
3 4 1527165 86.7411
4 5 1365734 33.6957
5 6 1324118 21.5052
6 7 1308932 18.3270
7 8 1275355 8.8775
8 9 1270233 9.1311
9 10 1269390 10.8435
10 11 1264977 11.3390
11 10 1264765 9.2668
12 11 1263983 11.0000
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