Home Articles

线性回归系数信息作为数据帧或矩阵

Asked
Viewed 295 times
3

我正在尝试创建一个脚本来优化线性回归分析,我真的想对模型输出进行操作,最具体的是Pr(> | t |)值 . 不幸的是,我不知道如何将模型输出到矩阵或数据表中 .

这是一个例子:在下面的代码中,我创建了七列数据,并使用其他六个数据拟合第七列 . 当我得到模型的摘要时,很明显三个参数比其他三个参数更重要 . 如果我可以用数字方式访问系数输出,我可以创建一个脚本来删除最不重要的参数并重新运行分析......但是就这样,我手动执行此操作 .

做这个的最好方式是什么?

非常感谢您的帮助 .

q = matrix( 
c(2,14,-4,1,10,9,41,8,13,2,0,20,3,27,1,10,-1,0,
10,-6,23,6,13,-8,1,15,-7,55,7,14,10,0,20,-3,6,4,20,
-1,5,19,-2,48,10,19,8,8,10,-2,24,8,13,9,8,14,5,7,7,
12,1,0,16,7,27,7,10,-1,1,15,7,31,2,20,-5,10,12,3,57,
0,19,-8,8,11,-4,63,5,11,7,8,10,-7,6,9,10,-7,2,19,8,
51,2,18,3,3,14,4,30), nrow=15, ncol=7, byrow = TRUE)
#
colnames(q) <- c("A","B","C","D","E","F","Z")
#
q <- as.data.frame(q)
#
qmodel <- lm(Z~.,data=q)
#
summary(qmodel)
#

输出:

Call:
lm(formula = Z ~ ., data = q)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.25098 -0.52655 -0.02931  0.62350  1.26649 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) -2.09303    1.51627  -1.380    0.205    
A            0.91161    0.11719   7.779 5.34e-05 ***
B            1.99503    0.09539  20.914 2.87e-08 ***
C           -2.98252    0.04789 -62.283 4.91e-12 ***
D            0.13458    0.10377   1.297    0.231    
E            0.15191    0.09397   1.617    0.145    
F            0.01417    0.04716   0.300    0.772    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9439 on 8 degrees of freedom
Multiple R-squared:  0.9986,    Adjusted R-squared:  0.9975 
F-statistic: 928.9 on 6 and 8 DF,  p-value: 6.317e-11

现在这是我想看到的:

> coeffs
             Estimate Std. Error t value Pr(>|t|)
 (Intercept) -2.09303    1.51627  -1.380 2.05e-01
 A            0.91161    0.11719   7.779 5.34e-05
 B            1.99503    0.09539  20.914 2.87e-08
 C           -2.98252    0.04789 -62.283 4.91e-12
 D            0.13458    0.10377   1.297 2.31e-01
 E            0.15191    0.09397   1.617 1.45e-01
 F            0.01417    0.04716   0.300 7.72e-01

事实上,我以这种方式得到了......根本不是自动化的......

coeffs = matrix(
c(-2.09303,1.51627,-1.38,0.205,0.91161,0.11719,
7.779,0.0000534,1.99503,0.09539,20.914,0.0000000287,
-2.98252,0.04789,-62.283,0.00000000000491,0.13458,
0.10377,1.297,0.231,0.15191,0.09397,1.617,0.145,
0.01417,0.04716,0.3,0.772), nrow=7, ncol=4, byrow = TRUE)
#
rownames(coeffs) <- c("(Intercept)","A","B","C","D","E","F")
colnames(coeffs) <- c("Estimate","Std. Error","t value","Pr(>|t|)")
#
coeffs <- as.data.frame(coeffs)
#
coeffs

2 Answers

  • 6

    如果我理解正确,您需要摘要返回的矩阵 . 这非常直截了当:

    fit <- lm( formula, data=yourData)
    coeffs <- summary(fit)$coefficients
    

    之后,您可以从 coeffs 中选择符合条件的记录,就像使用任何矩阵一样 . 例:

    coeffs[coeffs[4,] < 1e-12,]
    
  • 2

    你想要的是摘要对象的 coefficients 组件 .

    m <- lm(Z~.,data=q)
    
    summary(m)$coefficients
    

    进一步评论:

    • 使用 step 进行逐步变量选择而不是自己编码;

    • 逐步变量选择具有错误的统计特性;考虑像 glmnet (在同名的包中)做一些规范化的模型构建 .

Related