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我一直在尝试使用拉格朗日插值来解决问题，该插值在 R 语言的poly.calc方法（包）中实现。polynom

基本上，我的问题是使用拉格朗日插值法预测某个国家的人口。我有过去几年（1961 - 2014）的人口。csv文件在这里

w1 = read.csv(file="country.csv", sep=",", head=TRUE)
array_x = w1$x
array_y = w1$y

#calls Lagrange Method
p = poly.calc(array_x, array_y)

#create a function to evaluate the polynom
prf <- as.function(p)
#create some points to plot
myx = seq(1961, 2020, 0.5)
#y's to plot
myy = prf(myx)
#plot
plot(myx, myy,col='blue')

之后，绘制的曲线正在下降，y 轴为（非常大）负（134 的幂）。它没有任何意义。但是，如果我使用五点，那是正确的。

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1 回答 1

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This is not really an SO question but rather a numerical analysis question.

R is doing everything you want it to, it's not a programming error. It's just that what you want it to do is notoriously bad. Lagrange polynomials are notorious for being incredibly unstable, especially when a large number of points are fit.

A much more stable alternative is the use of splines, such as B-splines. They can be fit very easily with R's default spline library into any regression model, i.e. you could fit a least squares model with 

library(splines)
x <- sort(runif(500, -3,3) ) #sorting makes for easier plotting ahead
y <- sin(x)
splineFit <- lm(y ~ bs(x, df = 5) )
est_y <- predict(splineFit)
plot(x, y, type = 'l')
lines(x, est_y, col = 'blue')

You can see from the above model that the splines can do a good job of fitting non-linear relations.

于 2015-11-02T05:44:40.510 回答