计算列表上的高斯曲线拟合

Question

我有如下所示的列表数据。我想对列表中的每个元素进行、mids、和计数之间的非线性回归高斯曲线拟合，并报告均值和标准差。

mylist<- structure(list(A = structure(list(breaks = c(-10, -9, 
-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4), counts = c(1L, 
0L, 1L, 5L, 9L, 38L, 56L, 105L, 529L, 2858L, 17L, 2L, 0L, 2L), 
    density = c(0.000276014352746343, 0, 0.000276014352746343, 
    0.00138007176373171, 0.00248412917471709, 0.010488545404361, 
    0.0154568037537952, 0.028981507038366, 0.146011592602815, 
    0.788849020149048, 0.00469224399668783, 0.000552028705492686, 
    0, 0.000552028705492686), mids = c(-9.5, -8.5, -7.5, -6.5, 
    -5.5, -4.5, -3.5, -2.5, -1.5, -0.5, 0.5, 1.5, 2.5, 3.5), 
    xname = "x", equidist = TRUE), .Names = c("breaks", "counts", 
"density", "mids", "xname", "equidist"), class = "histogram"), 
    B = structure(list(breaks = c(-7, -6, -5, 
    -4, -3, -2, -1, 0), counts = c(2L, 0L, 6L, 2L, 2L, 1L, 3L
    ), density = c(0.125, 0, 0.375, 0.125, 0.125, 0.0625, 0.1875
    ), mids = c(-6.5, -5.5, -4.5, -3.5, -2.5, -1.5, -0.5), xname = "x", 
        equidist = TRUE), .Names = c("breaks", "counts", "density", 
    "mids", "xname", "equidist"), class = "histogram"), C = structure(list(
        breaks = c(-7, -6, -5, -4, -3, -2, -1, 0, 1), counts = c(2L, 
        2L, 4L, 5L, 14L, 22L, 110L, 3L), density = c(0.0123456790123457, 
        0.0123456790123457, 0.0246913580246914, 0.0308641975308642, 
        0.0864197530864197, 0.135802469135802, 0.679012345679012, 
        0.0185185185185185), mids = c(-6.5, -5.5, -4.5, -3.5, 
        -2.5, -1.5, -0.5, 0.5), xname = "x", equidist = TRUE), .Names = c("breaks", 
    "counts", "density", "mids", "xname", "equidist"), class = "histogram")), .Names = c("A", 
"B", "C"))

我读过这个Fitting a density curve to a histogram in R，但这是如何将曲线拟合成直方图的方法。我想要的是最好的价值观“

“指”“SD”

如果我使用PRISM来做这件事，我应该得到以下A的结果

Mids   Counts
-9.5    1
-8.5    0
-7.5    1
-6.5    5
-5.5    9
-4.5    38
-3.5    56
-2.5    105
-1.5    529
-0.5    2858
0.5     17
1.5     2
2.5     0
3.5     2

进行非线性回归高斯曲线拟合，得到

"Best-fit values"   
"     Amplitude"    3537
"     Mean"       -0.751
"     SD"         0.3842

第二组B

Mids   Counts
-6.5    2
-5.5    0
-4.5    6
-3.5    2
-2.5    2
-1.5    1
-0.5    3



"Best-fit values"   
"     Amplitude"    7.672
"     Mean"         -4.2
"     SD"          0.4275

还有第三个

Mids   Counts
-6.5    2
-5.5    2
-4.5    4
-3.5    5
-2.5    14
-1.5    22
-0.5    110
0.5      3

我明白了

"Best-fit values"   
"     Amplitude"    120.7
"     Mean"       -0.6893
"     SD"        0.4397

Answer 1

为了将直方图转换为均值和标准差的估计。首先，转换bin计数的结果乘以bin。这将是对原始数据的近似。

根据上面的例子：

#extract the mid points and create list of simulated data
simdata<-lapply(mylist, function(x){rep(x$mids, x$counts)})
#if the original data were integers then this may give a better estimate
#simdata<-lapply(mylist, function(x){rep(x$breaks[-1], x$counts)})

#find the mean and sd of simulated data
means<-lapply(simdata, mean)
sds<-lapply(simdata, sd)
#or use sapply in the above 2 lines depending on future process needs

如果您的数据是整数，那么使用中断作为回收箱将提供一个更好的估计。根据直方图的功能(即right=TRUE/FALSE)，可以将结果移动一个。

编辑

我以为这会很容易的。我查看了视频，显示的样本数据如下：

mids<-seq(-7, 7)
counts<-c(7, 1, 2, 2, 2, 5, 217, 70, 18, 0, 2, 1, 2, 0, 1)
simdata<-rep(mids, counts)

视频结果均值为-0.7359，sd=为0.4571。我找到的解决方案提供了最接近的结果，就是使用"fitdistrplus“软件包：

fitdist(simdata, "norm", "mge")

使用“最大拟合优度估计”得到均值= -0.7597280和sd= 0.8320465.

在这一点上，上述方法提供了一个密切的估计，但不完全匹配。我不知道用什么技术从视频中计算出合适的值。

编辑#2

以上解决方案涉及重新创建原始数据，并使用平均值/sd或适配包进行拟合。这种尝试是试图使用高斯分布来执行最小二乘拟合.

simdata<-lapply(mylist, function(x){rep(x$mids, x$counts)})
means<-sapply(simdata, mean)
sds<-sapply(simdata, sd)

#Data from video
#mids<-seq(-7, 7)
#counts<-c(7, 1, 2, 2, 2, 5, 217, 70, 18, 0, 2, 1, 2, 0, 1)

#make list of the bins and distribution in each bin
mids<-lapply(mylist, function(x){x$mids})
dis<-lapply(mylist, function(x) {x$counts/sum(x$counts)})

#function to perform the least square fit
nnorm<-function(values, mids, dis) {
  means<-values[1]
  sds<-values[2]
  #print(paste(means, sds))
  #calculate out the Gaussian distribution for each bin
  modeld<-dnorm(mids, means, sds)  
  #sum of the squares
  diff<-sum( (modeld-dis)^2)
  diff
}

#use optim function with the mean and sd as initial guesses
#find the mininium with the mean and SD as fit parameters
lapply(1:3, function(i) {optim(c(means[[i]], sds[[i]]), nnorm, mids=mids[[i]], dis=dis[[i]])})

这个解决方案为PRISM的结果提供了一个更接近的答案，但仍然不一样。以下是对所有4种解决方案的比较。

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从表中，最小二乘拟合(就在上面)提供了最接近的近似。也许调整中间点的dnorm函数可能会有所帮助。但情况B数据离正态分布最远，但PRISM软件仍产生较小的标准差，而其他方法相似。PRISM软件可能会执行某种类型的数据过滤，以便在拟合之前删除异常值。