计算列表上的高斯曲线拟合

Question

我有一个如下所示的列表数据。我想为列表中的每个元素在 mids 和 counts 之间执行非线性回归高斯曲线拟合，并报告平均值和标准差

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到直方图。我想要的是最合适的价值观”

“意思” “标清”

如果我使用 PRISM 来做，我应该得到以下结果 对于A

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，我得到

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

第二组 乙

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

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

为了将直方图转换回均值和标准差的估计值。首先将 bin 计数的结果与 bin 相乘。这将是原始数据的近似值。

根据您上面的示例：

#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，则使用间隔作为 bin 将提供更好的估计。取决于直方图的函数（即 right=TRUE/FALSE），结果可能会移动一位。

编辑

我认为这将是一件容易的事。我查看了视频，显示的示例数据是：

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)

视频结果的平均值 = -0.7359 和 sd = 0.4571。我发现提供最接近结果的解决方案是使用“fitdistrplus”包：

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

使用“最大化拟合优度估计”得到平均值 = -0.7597280 和 sd= 0.8320465。
在这一点上，上面的方法提供了一个接近的估计，但并不完全匹配。我不知道用什么技术从视频中计算出适合度。

编辑#2

上述解决方案涉及重新创建原始数据并使用 mean/sd 或使用 fitdistrplus 包进行拟合。此尝试是尝试使用高斯分布执行最小二乘拟合。

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]])})

此解决方案提供了更接近 PRISM 结果的答案，但仍然不一样。这是所有 4 种解决方案的比较。

从表中可以看出，最小二乘拟合（上面的那个）提供了最接近的近似值。也许调整中点 dnorm 函数可能会有所帮助。但是案例 B 的数据离正态分布最远，但 PRISM 软件仍然会产生一个小的标准偏差，而其他方法类似。 PRISM 软件可能会执行某种类型的数据过滤以在拟合之前去除异常值。