使用自然样条拟合进行预测 [英] Prediction using a natural spline fit

问题描述

我有一个简单的自然样条曲线(df = 3)模型，我正在尝试从样本观察中预测一些.使用功能predict()，我能够获得样本中观测值的拟合值，但无法获得新观测值的预测值.

I have a fitted a simple natural spline (df = 3) model and I'm trying to predict for some out of sample observations. Using the function predict(), I'm able to get fitted values for in-sample observations but I've not been able to get the predicted value for new observations.

这是我的代码:

library(splines)

set.seed(12345)
x <- seq(0, 2, by = 0.01)
y <- rnorm(length(x)) + 2*sin(2*pi*(x-1/4))

# My n.s fit:
fit.temp <- lm(y ~ ns(x, knots = seq(0.01, 2, by = 0.1)))

# Getting fitted values:
fit.temp.values <- predict(fit.temp,interval="prediction", level = 1 - 0.05)

# Plotting the data, the fit, and the 95% CI:
plot(x, y, ylim = c(-6, +6))
lines(x, fit.temp.values[,1], col = "darkred")
lines(x, fit.temp.values[,2], col = "darkblue", lty = 2)
lines(x, fit.temp.values[,3], col = "darkblue", lty = 2)

# Consider the points for which we want to get the predicted values:
x.new <- c(0.275, 0.375, 0.475, 0.575, 1.345)

如何获取x.new的预测值?

How can I get the predicted values for x.new?

非常感谢您的帮助，

p.s.我在SO上搜索了所有相关问题，但没有找到答案.

p.s. I searched all related questions on SO and I didn't find the answer.

推荐答案

使用称为x的列创建数据框，并将其作为newdata参数传递给predict:

Create a data frame with a column called x, and pass it as the newdata argument to predict:

predict(fit.temp, newdata=data.frame(x=x.new))

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