Question

我有两个表格，数据非常相似，我想要一个模型：

###first data
x1 <- c(0.271237802,0.253595465,0.299072793,0.355537802,  
0.335295465,0.365922793,0.476437802,0.464095465,0.482172793)  

y1 <- c(0.039937,0.044174,0.062574,0.124286,  
0.108702,0.131217,0.213418,0.216699,0.253712)

####second data
x2 <- c(0.285180641,0.289818303,0.27255962,0.373530641,  
0.356768303,0.34930962,0.463880641,0.471668303,0.46330962)

y2 <- c(0.0499,0.063764,0.05343,0.147753,  
0.14148,0.135757,0.220635,0.245013,0.236258)

####nls.model
fo1 = nls(y1~A*(x1-v)^k, start=list(A=1, v=0.15, k=1))
coef(fo1)
summary(fo1)

fo2 = nls(y2~A*(x2-v)^k, start=list(A=1, v=0.15, k=1))
coef(fo2)
summary(fo2)

####plotting data
s <- seq(from = 0, to = 1, length = 50)
plot(y1~x1, ylab = "D/D0", xlab = "LP(%)", pch = 16, xlim=c(0,0.6), ylim=c(0,0.4), col="blue")
lines(s, predict(fo1, list(x1 = s)), col = "blue")

par(new=T)

plot(y2~x2, ylab = "D/D0", xlab = "LP(%)", pch = 16, xlim=c(0,0.6), ylim=c(0,0.4), axes=F,col="red")

在第一种情况下，该模型运作良好，但在第二种情况下，它很脆弱并给出了消息：

 fo2 = nls(y2~A*((x2-v)^(k)), start=list(A=1, v=0.15, k=1))
Error in numericDeriv(form[[3L]], names(ind), env) : 
  Missing value or an Infinity produced when evaluating the model

虽然数据显示几乎是线性关系，但我想使用nls和建议的函数，因为线性并不总是正确的。
我知道，它可能与起始值有关，但是我无法解决这个问题。有人有线索吗？

Answer 1

x2-v必须是非负面的，所以只要无约束的最小值满足v小于min(x2)，请尝试避免错误。

nls(y2 ~ A * pmax(x2-v, 0)^k, start = list(A = 1, v = 0.15, k = 1))

如果v在最佳状态下不小于min(x2)，那么问题是修订后的模型是否仍然可以接受。

另一种可能性是将v限制为小于min(x2)。例如：

nls(y2 ~ A * pmax(x2-v, 0)^k + 1000 * (v > min(x2)), start = list(A = 1, v = 0.15, k = 1))

或使用约束优化例程。

r中的nls函数出错，几乎相同的数据

1 个答案: