我可能会遗漏一些关于预测的内容 - 但我的多元线性回归似乎按预期工作:
> bigmodel <- lm(score ~ lean + gender + age, data = mydata)
> summary(bigmodel)
Call:
lm(formula = score ~ lean + gender + age, data = mydata)
Residuals:
Min 1Q Median 3Q Max
-25.891 -4.354 0.892 6.240 18.537
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 70.96455 3.85275 18.419 <2e-16 ***
lean 0.62463 0.05938 10.518 <2e-16 ***
genderM -2.24025 1.40362 -1.596 0.1121
age 0.10783 0.06052 1.782 0.0764 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9 on 195 degrees of freedom
Multiple R-squared: 0.4188, Adjusted R-squared: 0.4098
F-statistic: 46.83 on 3 and 195 DF, p-value: < 2.2e-16
> head(predict(bigmodel),20)
1 2 3 4 5 6 7 8 9 10
75.36711 74.43743 77.02533 78.76903 79.95515 79.09251 80.38647 81.65807 80.14846 78.96234
11 12 13 14 15 16 17 18 19 20
82.39052 82.04468 81.05187 81.26753 84.50240 81.80667 80.92169 82.40895 81.76197 82.94809
但在阅读?predict.lm
之后,我无法绕过预测。对于我的原始数据集,此输出看起来不错 - 但是如果我想针对不同的数据集运行预测而不是我用来创建bigmodel
的数据集呢?
例如,如果我将.csv文件导入到名为newmodel
的R中,其中200人完成了倾向,性别和年龄 - 我如何使用bigmodel
中的回归公式来生成预测对于newmodel
?
谢谢!
答案 0 :(得分:3)
如果您阅读了predict.lm
的文档,则会看到以下内容。因此,使用newdata
参数传递您导入的newmodel
数据以获得预测。
predict(object, newdata, se.fit = FALSE, scale = NULL, df = Inf,
interval = c("none", "confidence", "prediction"),
level = 0.95, type = c("response", "terms"),
terms = NULL, na.action = na.pass,
pred.var = res.var/weights, weights = 1, ...)
Arguments
object
Object of class inheriting from "lm"
newdata
An optional data frame in which to look for variables with which to predict.
If omitted, the fitted values are used.
更新。关于使用预测导出数据的问题,以下是您可以执行此操作的方法。
predictions = cbind(newmodel, pred = predict(bigmodel, newdata = newmodel))
write.csv(predictions, 'predictions.csv', row.names = F)
更新2.完全可重复的完整解决方案
bigmodel <- lm(mpg ~ wt, data = mtcars)
newdata = data.frame(wt = runif(20, min = 1.5, max = 6))
cbind(
newdata,
mpg = predict(bigmodel, newdata = newdata)
)