加权数据的威布尔分布 [英] Weibull distribution with weighted data

问题描述

我有一些时间来事件数据，我需要为模拟模型的子组生成大约200个形状/比例参数.我已经分析了数据，并且最好遵循魏布尔分布.通常，我会使用fitdistrplus包和 fitdist(x，"weibull")来执行此操作，但是此数据已使用内核匹配进行了匹配，并且我有一个加权值变量，称为 km，因此需要合并权重，据我所知， fitdist 不能做到这一点.使用我的伽马分布式数据而不是使用 fitdist 时，我使用中的 wtd.mean 和 wtd.var 函数手动进行了计算hsmisc 软件包，效果很好.但是，为韦伯找到一个类似的公式使我难以理解.

我一直在测试一些选项，并将它们与fitdist结果进行比较:

  test_data<-rweibull(100，0.676，946)fitweibull<-fitdist(test_data，"weibull"，method ="mle"，lower = c(0,0))fitweibull $估计形状比例0.6981165 935.0907482 

我首先对此进行了测试: R中的Weibull分布(ExtDist)

 库(bbmle)m1<-mle2(y〜dweibull(shape = exp(lshape)，scale = exp(lscale))，data = data.frame(y = test_data)，start = list(lshape = 0，lscale = 0)) 

这给了我 lshape = -0.3919991 和 lscale = 6.852033

我尝试过的另一件事是 EnvStats 包中的 eweibull .

  eweibull<-eweibull(测试数据)eweibull $参数形状比例0.698091 935.239277 

但是，尽管这些都可以提供结果，但我仍然认为我无法将我的数据与权重相适应.

我也尝试过使用 ExtDist 包中的类似名称的 eWeibull (我不确定100％仍然可以使用，但是确实具有weibull函数，负重！).我收到很多关于输入不可计算(NA或无限)的错误消息.如果我使用 map 进行操作，那么 map(test_data，test_km，eWeibull)的全部100个值都为[[NULL].如果仅使用test_data进行尝试，则会出现一连串与optimx相关的错误.

我还尝试过 propagate 中的 fitDistr ，该错误给出了 weights 应该是特定长度的错误.例如，如果将两者都设置为100，则会出现一个错误，提示 weights 的长度应为94.如果将其设置为94，则表明长度必须为132.>

我需要能够将一组预加权的均值/var/sd等数据传递到计算中，或者要有一个可以获取数据和权重并在计算中使用它们的函数.

解决方案

经过反复试验，我从 EnvStats 包中将 eweibull 函数编辑为而不是使用 mean(x)和 sd(x)，改为使用 wtd.mean(x，w)和 sqrt(wtd.var(x，w)).现在，它将运行并输出加权值.

I have some time to event data that I need to generate around 200 shape/scale parameters for subgroups for a simulation model. I have analysed the data, and it best follows a weibull distribution. Normally, I would use the fitdistrplus package and fitdist(x, "weibull") to do so, however this data has been matched using kernel matching and I have a variable of weighting values called km and so needs to incorporate a weight, which isn't something fitdist can do as far as I can tell. With my gamma distributed data instead of using fitdist I did the calculation manually using the wtd.mean and wtd.var functions from the hsmisc package, which worked well. However, finding a similar formula for the weibull is eluding me.

I've been testing a few options and comparing them against the fitdist results:

test_data <- rweibull(100, 0.676, 946)
fitweibull <- fitdist(test_data, "weibull", method = "mle", lower = c(0,0))
fitweibull$estimate
shape          scale
0.6981165      935.0907482

I first tested this: The Weibull distribution in R (ExtDist)

library(bbmle)
m1 <- mle2(y~dweibull(shape=exp(lshape),scale=exp(lscale)),
 data=data.frame(y=test_data),
 start=list(lshape=0,lscale=0)) 

which gave me lshape = -0.3919991 and lscale = 6.852033

The other thing I've tried is eweibull from the EnvStats package.

eweibull <- eweibull(test_data)
eweibull$parameters
shape       scale
0.698091    935.239277

However, while these are giving results, I still don't think I can fit my data with the weights into any of these.

Edit: I have also tried the similarly named eWeibull from the ExtDist package (which I'm not 100% sure still works, but does have a weibull function that takes a weight!). I get a lot of error messages about the inputs being non-computable (NA or infinite). If I do it with map, so map(test_data, test_km, eWeibull) I get [[NULL] for all 100 values. If I try it just with test_data, I get a long string of errors associated with optimx.

I have also tried fitDistr from propagate which gives errors that weights should be a specific length. For example, if both are set to be 100, I get an error that weights should be length 94. If I set it to 94, it tells me it has to be length of 132.

I need to be able to pass either a set of pre-weighted mean/var/sd etc data into the calculation, or have a function that can take data and weights and use them both in the calculation.

解决方案

After much trial and error, I edited the eweibull function from the EnvStats package to instead of using mean(x) and sd(x), to instead use wtd.mean(x,w) and sqrt(wtd.var(x, w)). This now runs and outputs weighted values.

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