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Question

我有一个如下所示的列表数据。我想对列表中的每个元素执行中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到直方图。我想要的是最合适的价值“

“意思是” “SD”

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

执行非线性回归高斯曲线拟合，我得

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

为第二组乙

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

和第三个

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平均值和标准差的估计值。首先将bin计数的结果转换为bin。这将是原始数据的近似值。

根据您的上述示例：

#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，那么使用中断作为二进制文件将提供更好的估计。根据直方图的功能（即right = TRUE / FALSE）可能会将结果移一。

修改

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)

视频结果为平均值= -0.7359，sd = 0.4571。我发现提供最接近结果的解决方案是使用＆＃34; fitdistrplus＆＃34;包：

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

使用＆＃34;最大化拟合优度估计＆＃34;得到平均值= -0.7597280，sd = 0.8320465 此时，上述方法提供了近似估计但不完全匹配。我不知道用什么技术来计算视频的拟合度。

编辑＃2

以上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]])})

此解决方案为PRISM结果提供了更接近的答案，但仍然不尽相同。以下是所有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的数据距离正态分布最远，但PRISM软件仍然会产生较小的标准偏差，而其他方法则相似。 PRISM软件可以执行某种类型的数据过滤，以在拟合之前删除异常值。

计算列表

1 个答案:

修改

编辑＃2