Question

我试图估计Heaps定律的常数。我有以下数据集novels_colection：

  Number of novels DistinctWords WordOccurrences
1                1         13575          117795
2                1         34224          947652
3                1         40353         1146953
4                1         55392         1661664
5                1         60656         1968274

然后我构建下一个函数：

# Function for Heaps law
heaps <- function(K, n, B){
  K*n^B
}
heaps(2,117795,.7) #Just to test it works

所以n = Word Occurrences，K和B是应该是常量的值，以便找到我对不同词汇的预测。

我尝试了这个，但它给了我一个错误：

fitHeaps <- nls(DistinctWords ~ heaps(K,WordOccurrences,B), 
    data = novels_collection[,2:3], 
    start = list(K = .1, B = .1), trace = T)

错误= Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model

我是如何解决此问题或方法以适应函数并获取K和B的值？

Answer 1

如果您在y = K * n ^ B的两边进行日志转换，则会获得log(y) = log(K) + B * log(n)。这是log(y)和log(n)之间的线性关系，因此您可以使用线性回归模型来查找log(K)和B。

logy <- log(DistinctWords)
logn <- log(WordOccurrences)

fit <- lm(logy ~ logn)

para <- coef(fit)  ## log(K) and B
para[1] <- exp(para[1])    ## K and B

Answer 2

使用minpack.lm我们可以拟合非线性模型，但我想它会比对数变换变量上的线性模型更容易过度拟合（如哲源所做），但我们可以比较残差对某些数据集进行线性/非线性模型得到实证结果，这将是有趣的。

library(minpack.lm)
fitHeaps = nlsLM(DistinctWords ~ heaps(K, WordOccurrences, B),
                     data = novels_collection[,2:3], 
                     start = list(K = .01, B = .01))
coef(fitHeaps)
#        K         B 
# 5.0452566 0.6472176 

plot(novels_collection$WordOccurrences, novels_collection$DistinctWords, pch=19)
lines(novels_collection$WordOccurrences, predict(fitHeaps, newdata = novels_collection[,2:3]), col='red')

`nls`无法估计模型的参数

2 个答案: