中的累积分布拟合到R后创建正态分布在用Gompertz函数成功拟合我的累积数据之后，我需要从拟合函数创建正态分布。将拟合参数

这是迄今为止代码：

 df <- data.frame(x = c(0.01,0.011482,0.013183,0.015136,0.017378,0.019953,0.022909,0.026303,0.0302,0.034674,0.039811,0.045709,0.052481,0.060256,0.069183,0.079433,0.091201,0.104713,0.120226,0.138038,0.158489,0.18197,0.20893,0.239883,0.275423,0.316228,0.363078,0.416869,0.47863,0.549541,0.630957,0.724436,0.831764,0.954993,1.096478,1.258925,1.44544,1.659587,1.905461,2.187762,2.511886,2.884031,3.311311,3.801894,4.365158,5.011872,5.754399,6.606934,7.585776,8.709636,10,11.481536,13.182567,15.135612,17.378008,19.952623,22.908677,26.30268,30.199517,34.673685,39.810717,45.708819,52.480746,60.255959,69.183097,79.432823,91.201084,104.712855,120.226443,138.038426,158.489319,181.970086,208.929613,239.883292,275.42287,316.227766,363.078055,416.869383,478.630092,549.540874,630.957344,724.43596,831.763771,954.992586,1096.478196), 
       y = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.00044816,0.00127554,0.00221488,0.00324858,0.00438312,0.00559138,0.00686054,0.00817179,0.00950625,0.01085188,0.0122145,0.01362578,0.01514366,0.01684314,0.01880564,0.02109756,0.0237676,0.02683182,0.03030649,0.0342276,0.03874555,0.04418374,0.05119304,0.06076553,0.07437854,0.09380666,0.12115065,0.15836926,0.20712933,0.26822017,0.34131335,0.42465413,0.51503564,0.60810697,0.69886817,0.78237651,0.85461023,0.91287236,0.95616228,0.98569093,0.99869001,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999)) 

library(drc) 
fm <- drm(y ~ x, data = df, fct = G.3()) 

options(scipen = 10) #to avoid scientific notation in x axis 

plot(df$x, predict(fm),type = "l", log = "x",col = "blue", main = "Cumulative function distribution",xlab = "x", ylab = "y") 

points(df,col = "red") 

legend("topleft", inset = .05,legend = c("exp","fit") 
     ,lty = c(NA,1), col = c("red", "blue"), pch = c(1,NA), lwd=1, bty = "n") 


summary(fm)

这是下面的情节：现在

我的想法是此累积适合某种程度上转化为正态分布。有什么想法我怎么能这样做？

来源

2017-06-01 numb

我在考虑cumdiff（因为没有更好的术语）。 link帮助了很多。

编辑

plot(df$x[-1], Mod(df$y[-length(df$y)]-df$y[-1]), log = "x", type = "b", 
     main = "Normal distribution for original data", 
     xlab = "x", ylab = "y")

产生：

加成

为了从fitted功能得到高斯：

df$y_pred<-predict(fm) 
plot(df$x[-1], Mod(df$y_pred[-length(df$y_pred)]-df$y_pred[-1]), log = "x", 
    type = "b", main="Normal distribution for fitted function", 
    xlab = "x", lab = "y")

产生：

来源

2017-06-01 16:00:11 amonk

好的，所以这是您从我的初始数据集创建正常绘图而不是从拟合创建正常绘图的方式。现在我会深入研究并了解如何配合eq。 – numb

非常感谢！但是有没有一种方法可以在x轴而不是索引上绘制x值？我试过了，但是最后我遇到了一个问题：Mod（df $ y [-length（df $ y）] - df $ y [-1]）''和'df $ x' ... – numb

对了......我也在想这个。试试'str（fm）'，看看你能否得到一些信息。毕竟，我并不熟悉'drc'软件包。现在我不能潜入，但我保证我会尽快回复你。 – amonk

虽然你的初衷可能是无参数的话，建议使用参数估计方法：矩量法，被广泛用于诸如这样的问题，因为你有一定的参数分布（正态分布）来拟合。这个想法很简单，从拟合的累积分布函数中，可以计算我的代码中的平均值（E1）和方差（我的代码中的SD的平方），然后解决问题，因为正态分布可以完全由均值和方差。

df <- data.frame(x=c(0.01,0.011482,0.013183,0.015136,0.017378,0.019953,0.022909,0.026303,0.0302,0.034674,0.039811,0.045709,0.052481,0.060256,0.069183,0.079433,0.091201,0.104713,0.120226,0.138038,0.158489,0.18197,0.20893,0.239883,0.275423,0.316228,0.363078,0.416869,0.47863,0.549541,0.630957,0.724436,0.831764,0.954993,1.096478,1.258925,1.44544,1.659587,1.905461,2.187762,2.511886,2.884031,3.311311,3.801894,4.365158,5.011872,5.754399,6.606934,7.585776,8.709636,10,11.481536,13.182567,15.135612,17.378008,19.952623,22.908677,26.30268,30.199517,34.673685,39.810717,45.708819,52.480746,60.255959,69.183097,79.432823,91.201084,104.712855,120.226443,138.038426,158.489319,181.970086,208.929613,239.883292,275.42287,316.227766,363.078055,416.869383,478.630092,549.540874,630.957344,724.43596,831.763771,954.992586,1096.478196), 
       y=c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.00044816,0.00127554,0.00221488,0.00324858,0.00438312,0.00559138,0.00686054,0.00817179,0.00950625,0.01085188,0.0122145,0.01362578,0.01514366,0.01684314,0.01880564,0.02109756,0.0237676,0.02683182,0.03030649,0.0342276,0.03874555,0.04418374,0.05119304,0.06076553,0.07437854,0.09380666,0.12115065,0.15836926,0.20712933,0.26822017,0.34131335,0.42465413,0.51503564,0.60810697,0.69886817,0.78237651,0.85461023,0.91287236,0.95616228,0.98569093,0.99869001,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999)) 

library(drc) 
fm <- drm(y ~ x, data = df, fct = G.3()) 

options(scipen = 10) #to avoid scientific notation in x axis 

plot(df$x, predict(fm),type="l", log = "x",col="blue", main="Cumulative distribution function",xlab="x", ylab="y") 

points(df,col="red") 

E1 <- sum((df$x[-1] + df$x[-length(df$x)])/2 * diff(predict(fm))) 
E2 <- sum((df$x[-1] + df$x[-length(df$x)])^2/4 * diff(predict(fm))) 
SD <- sqrt(E2 - E1^2) 
points(df$x, pnorm((df$x - E1)/SD), col = "green") 

legend("topleft", inset = .05,legend= c("exp","fit","method of moment") 
     ,lty = c(NA,1), col = c("red", "blue", "green"), pch = c(1,NA), lwd=1, bty="n") 


summary(fm)

而且估计结果：

## > E1 (mean of fitted normal distribution) 
## [1] 65.78474 
## > E2 (second moment of fitted normal distribution) 
##[1] 5792.767 
## > SD (standard deviation of fitted normal distribution) 
## [1] 38.27707 
## > SD^2 (variance of fitted normal distribution) 
## [1] 1465.134

编辑：更新的方法来计算CDF配合装配drc时刻。下面定义的函数moment使用连续r.v的矩公式计算矩估计。 E(X^k) = k * \int x^{k - 1} (1 - cdf(x)) dx。这些是我能从合适的cdf中得到的最好估计。当x接近零时，由于原始数据集中的原因，正如我在评论中所讨论的那样，该拟合并不是很好。

df <- data.frame(x=c(0.01,0.011482,0.013183,0.015136,0.017378,0.019953,0.022909,0.026303,0.0302,0.034674,0.039811,0.045709,0.052481,0.060256,0.069183,0.079433,0.091201,0.104713,0.120226,0.138038,0.158489,0.18197,0.20893,0.239883,0.275423,0.316228,0.363078,0.416869,0.47863,0.549541,0.630957,0.724436,0.831764,0.954993,1.096478,1.258925,1.44544,1.659587,1.905461,2.187762,2.511886,2.884031,3.311311,3.801894,4.365158,5.011872,5.754399,6.606934,7.585776,8.709636,10,11.481536,13.182567,15.135612,17.378008,19.952623,22.908677,26.30268,30.199517,34.673685,39.810717,45.708819,52.480746,60.255959,69.183097,79.432823,91.201084,104.712855,120.226443,138.038426,158.489319,181.970086,208.929613,239.883292,275.42287,316.227766,363.078055,416.869383,478.630092,549.540874,630.957344,724.43596,831.763771,954.992586,1096.478196), 
       y=c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.00044816,0.00127554,0.00221488,0.00324858,0.00438312,0.00559138,0.00686054,0.00817179,0.00950625,0.01085188,0.0122145,0.01362578,0.01514366,0.01684314,0.01880564,0.02109756,0.0237676,0.02683182,0.03030649,0.0342276,0.03874555,0.04418374,0.05119304,0.06076553,0.07437854,0.09380666,0.12115065,0.15836926,0.20712933,0.26822017,0.34131335,0.42465413,0.51503564,0.60810697,0.69886817,0.78237651,0.85461023,0.91287236,0.95616228,0.98569093,0.99869001,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999,0.99999999)) 

library(drc) 
fm <- drm(y ~ x, data = df, fct = G.3()) 

moment <- function(k){ 
    f <- function(x){ 
     x^(k - 1) * pmax(0, 1 - predict(fm, data.frame(x = x))) 
    } 
    k * integrate(f, lower = min(df$x), upper = max(df$x))$value 
} 

E1 <- moment(1) 
E2 <- moment(2) 
SD <- sqrt(E2 - E1^2)

来源

2017-06-01 18:38:50 Consistency

谢谢你的想法。我会深入研究这个，看看我该如何处理时刻的方法。我有点担心它在图的第一部分不适合（在x = 80之前的某处）。你有什么想法，为什么？ – numb

所以，我试图从你的时刻绘制正态分布，并且在起点上的不合适会导致怪异的正态分布（因为它在第一时期没有达到0）。这是我使用的代码： 'y2 < - dnorm（df $ x，mean = E1，sd = SD）plot（df $ x，y2，type =“b”）' 这是[plot]（ http://imgur.com/a/0ZCWE） – numb

@numb我发现问题是因为SD估计过大。这是因为当x很小时，原来的'df $ x'是稠密的，但当x很大时很稀疏，这就导致了这个问题。我正在寻找方法来获得更好的估计。 – Consistency

将拟合参数

回答

编辑

加成

相关问题