計算列表中的高斯曲線擬合

Question

我有一個像下面這樣的列表數據。我想執行非線性迴歸高斯曲線中頻和計數爲

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., 
     0., 0.0246913580246914, 0.0308641975308642, 
     0.0864197530864197, 0.135802469135802, 0.679, 
     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")) 

我已閱讀本 Fitting a density curve to a histogram in R 我的列表，並報告平均值和標準偏差的每一個元素，但是這是如何適應的嵌合曲線轉換爲直方圖。我要的是最佳擬合值」

‘中庸’ ‘SD’

如果我使用PRISM做到這一點，我應該得到以下結果 對於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 

進行非線性迴歸高斯曲線擬合，我得到

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

第二組 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 

和第三個

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 

我得到這個

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

Answer 1

爲了直方圖轉換回平均值和標準偏差的估計。首先將bin計數的結果轉換爲bin。這將是原始數據的近似值。

根據你上面的例子：

#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 

如果你的數據是整數，然後利用休息時間爲箱將提供更好的估計。取決於直方圖的函數（即right = TRUE/FALSE）可能會將結果移動一個。

編輯

我認爲這將是一件容易的事情。我回顧了視頻，顯示的示例數據爲：

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) 

視頻結果爲mean = -0.7359和sd = 0.4571。提供我發現溶液中的最接近的結果是使用「fitdistrplus」包：

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

使用「最大化擬合優度估計」導致平均= -0.7597280和SD = 0.8320465。
在這一點上，上述方法提供了一個接近的估計，但不完全匹配。我不知道用什麼技術來計算視頻的適合度。

編輯＃2

將上述溶液涉案重新創建原始數據和擬合，使用任一的平均/ SD或使用fitdistrplus包。這種嘗試是嘗試使用高斯分佈執行最小二乘擬合。

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]])}) 

該解決方案對PRISM結果提供了一個更接近的答案，但仍然不盡相同。以下是所有4種解決方案的比較。

從表中，最小二乘擬合（上面的那個）提供了最接近的近似值。也許調整中點dnorm函數可能會有所幫助。但情況B的數據距離正態分佈最遠，但PRISM軟件仍然產生一個小的標準偏差，而其他方法是相似的。 PRISM軟件有可能執行某種類型的數據過濾，以在合適之前移除異常值。