sac*_*cvf 7 regression r linear-regression
这段代码将返回系数:intercept,slop1,slop2
set.seed(1)
n=10
y=rnorm(n)
x1=rnorm(n)
x2=rnorm(n)
lm.ft=function(y,x1,x2)
return(lm(y~x1+x2)$coef)
res=list();
for(i in 1:n){
x1.bar=x1-x1[i]
x2.bar=x2-x2[i]
res[[i]]=lm.ft(y,x1.bar,x2.bar)
}
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如果我输入:
> res[[1]]
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我明白了:
(Intercept) x1 x2
-0.44803887 0.06398476 -0.62798646
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我们如何返回预测值,残差,R square,..等?
我需要一些通用的东西从摘要中提取我需要的东西?
jlh*_*ard 12
这里有几件事情.
首先,最好将变量组合到data.frame中:
df <- data.frame(y=rnorm(10), x1=rnorm(10), x2 = rnorm(10))
fit <- lm(y~x1+x2, data=df)
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如果这样做,那么使用模型进行预测并使用新数据集会更容易.
其次,可以从模型本身访问一些拟合的统计数据,有些可以从中获取summary(fit).
coef <- coefficients(fit) # coefficients
resid <- residuals(fit) # residuals
pred <- predict(fit) # fitted values
rsq <- summary(fit)$r.squared # R-sq for the fit
se <- summary(fit)$sigma # se of the fit
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要获得系数的统计信息,您需要使用摘要:
stat.coef <- summary(fit)$coefficients
coef <- stat.coef[,1] # 1st column: coefficients (same as above)
se.coef <- stat.coef[,2] # 2nd column: se for each coef
t.coef <- stat.coef[,3] # 3rd column: t-value for each coef
p.coef <- stat.coef[,4] # 4th column: p-value for each coefficient
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在您的函数中,您仅返回系数。尝试返回整个模型:
\n\nlm.ft=function(y,x1,x2) lm(y~x1+x2) # You don't need the return statement.\nRun Code Online (Sandbox Code Playgroud)\n\n现在尝试您的代码,然后运行:
\n\nsummary(res[[1]])\n\n# Call:\n# lm(formula = y ~ x1 + x2)\n# \n# Residuals:\n# Min 1Q Median 3Q Max \n# -0.88518 -0.25311 0.03868 0.43110 0.61753 \n# \n# Coefficients:\n# Estimate Std. Error t value Pr(>|t|) \n# (Intercept) -0.44804 0.32615 -1.374 0.2119 \n# x1 0.06398 0.24048 0.266 0.7979 \n# x2 -0.62799 0.26915 -2.333 0.0524 .\n# ---\n# Signif. codes: 0 \xe2\x80\x98***\xe2\x80\x99 0.001 \xe2\x80\x98**\xe2\x80\x99 0.01 \xe2\x80\x98*\xe2\x80\x99 0.05 \xe2\x80\x98.\xe2\x80\x99 0.1 \xe2\x80\x98 \xe2\x80\x99 1\n# \n# Residual standard error: 0.6149 on 7 degrees of freedom\n# Multiple R-squared: 0.5173, Adjusted R-squared: 0.3794 \n# F-statistic: 3.751 on 2 and 7 DF, p-value: 0.07814\nRun Code Online (Sandbox Code Playgroud)\n