如何排除NAs？(fitdist功能)

Question

我有100x2数据帧DFN.fitdist在DFN $ Lret列上运行会出现错误消息"函数mle无法估计参数,错误代码为100".我想是原因是最后一行包含一个NA.因此我运行fitdist排除NAs,现在我得到错误"数据必须是长度大于1的数字向量".有关如何解决此问题的任何想法？非常感谢.

DFN <- structure(list(LRet = c(0.0011, 0, -0.0026, 0, -0.0015, 0.0038, 3e-04, -0.0021, 4e-04, -0.001, 0, 0.0019, -6e-04, -8e-04, -5e-04, -8e-04, 3e-04, -5e-04, -0.0026, 0.0014, 7e-04, 0, -2e-04, 0.0011, -0.0025, 0.0042, 0.0022, -0.0017, -0.0058, 1e-04, 2e-04, 8e-04, -9e-04, -0.0014, -0.0014, -0.001, -0.0032, -0.0015, 6e-04, -8e-04, 0.001, -0.0014, -0.0017, -8e-04, -0.001, 0.0011, 0.0013, -0.001, 5e-04, 9e-04, -8e-04, -0.0025, 0.0027, 6e-04, 2e-04, -6e-04, 9e-04, -3e-04, -7e-04, 3e-04, 0, 2e-04, -6e-04, 1e-04, -1e-04, -7e-04, -8e-04, 7e-04, -1e-04, -7e-04, 7e-04, 8e-04, -8e-04, 8e-04, 0.0058, -1e-04, -5e-04, 0.0027, -0.0012, 7e-04, 7e-04, 0, 3e-04, -1e-04, 2e-04, -2e-04, -0.0013, -1e-04, 1e-04, -0.0011, 0.0013, 2e-04, -3e-04, -7e-04, 0, 0.0015, 1e-04, 3e-04, -0.0012, NA), LRetPct = c("0.11%", "0.00%", "-0.26%", "0.00%", "-0.15%", "0.38%", "0.03%", "-0.21%", "0.04%", "-0.10%", "0.00%", "0.19%", "-0.06%", "-0.08%", "-0.05%", "-0.08%", "0.03%", "-0.05%", "-0.26%", "0.14%", "0.07%", "0.00%", "-0.02%", "0.11%", "-0.25%", "0.42%", "0.22%", "-0.17%", "-0.58%", "0.01%", "0.02%", "0.08%", "-0.09%", "-0.14%", "-0.14%", "-0.10%", "-0.32%", "-0.15%", "0.06%", "-0.08%", "0.10%", "-0.14%", "-0.17%", "-0.08%", "-0.10%", "0.11%", "0.13%", "-0.10%", "0.05%", "0.09%", "-0.08%", "-0.25%", "0.27%", "0.06%", "0.02%", "-0.06%", "0.09%", "-0.03%", "-0.07%", "0.03%", "0.00%", "0.02%", "-0.06%", "0.01%", "-0.01%", "-0.07%", "-0.08%", "0.07%", "-0.01%", "-0.07%", "0.07%", "0.08%", "-0.08%", "0.08%", "0.58%", "-0.01%", "-0.05%", "0.27%", "-0.12%", "0.07%", "0.07%", "0.00%", "0.03%", "-0.01%", "0.02%", "-0.02%", "-0.13%", "-0.01%", "0.01%", "-0.11%", "0.13%", "0.02%", "-0.03%", "-0.07%", "0.00%", "0.15%", "0.01%", "0.03%", "-0.12%", " NA%")), .Names = c("LRet", "LRetPct"), class = "data.frame", row.names = 901:1000)

library(fitdistrplus)

#Following gives error code 100
f1 <- fitdist(DFN$LRet,"norm") 

#Following gives error code 100
f1 <- fitdist(DFN$LRet,"norm", na.rm=T)

#Following gives error data must be a numeric vector of length greater than 1"
f1 <- fitdist(na.exclude(DFN$LRet),"norm")
#Same result using na.omit

请注意,如果删除包含NA的最后一行,则上述代码可以正常工作.fitdist如果可以避免的话,我宁愿不必在运行之前消除最后一行.

编辑/更新:消除NA的最后一行确实首先解决了问题,但我现在无法一致地重现(即在消除最后一行后成功运行了几次代码,但并非总是如此).我想知道为什么.我尝试使用25x2数据帧,100x2和300x2,以及矢量,具有类似的结果.认为数据框或向量的大小可能是问题的一部分,因此具有不同大小的试验.

Answer 1

通过fitdist节目调试

 if (!(is.vector(data) & is.numeric(data) & length(data) > 1)) 
    stop("data must be a numeric vector of length greater than 1")

看?is.vector:

'is.vector'如果'x'是指定模式的向量而没有名称以外的属性,则返回'TRUE' .

na.exclude及其亲属(na.omit等)将有关排除值的信息保存为属性,因此is.vector()变为FALSE......

其中一个副作用c()是删除名称以外的属性,因此is.vector(c(na.exclude(DFN$LRet)))也是TRUE如此

fitdist(c(na.exclude(DFN$LRet)), "norm")

至少没有得到"必须是数字向量"错误 - 但我仍然得到"错误100".进一步调查......

挖掘fitdist更多的内容,似乎(正如@ 42-所建议的)optim()遇到了麻烦.具体来说,它实际上得到了答案,但是当它试图计算解决方案的Hessian时,它会尝试标准差参数和barfs的负值.

作为一个例子,这工作:

nn <- c(na.exclude(DFN$LRet))
fn <- function(x) -sum(dnorm(nn,mean=x[1],sd=x[2],log=TRUE))
optim(fn,par=c(mean(nn),sd(nn)),method="Nelder-Mead")

但这失败了:

optim(fn,par=c(mean(nn),sd(nn)),method="Nelder-Mead",hessian=TRUE)