[英]piecewise function fitting with nls() in R
我试图将数据分为两部分。
这是一些示例数据:
x<-c(0.00101959664756622, 0.001929220749155, 0.00165657261751726,
0.00182514724375389, 0.00161532360585458, 0.00126991061099209,
0.00149545009309177, 0.000816386510029308, 0.00164402569283353,
0.00128029006251656, 0.00206892841921455, 0.00132378793976235,
0.000953143467154676, 0.00272964503695939, 0.00169743839571702,
0.00286411493120396, 0.0016464862337286, 0.00155672067449593,
0.000878271561566836, 0.00195872573138819, 0.00255412836538339,
0.00126212428137799, 0.00106206607962734, 0.00169140916371657,
0.000858015581562961, 0.00191955159274793, 0.00243104345247067,
0.000871042201994687, 0.00229814264111745, 0.00226756341241083)
y<-c(1.31893118849162, 0.105150790530179, 0.412732029152914, 0.25589805483046,
0.467147868109498, 0.983984462069833, 0.640007862668818, 1.51429617241365,
0.439777145282391, 0.925550163462951, -0.0555942758921906, 0.870117027565708,
1.38032147826294, -0.96757052387814, 0.346370836378525, -1.08032147826294,
0.426215616848312, 0.55151485221263, 1.41306889485598, 0.0803478641720901,
-0.86654892295057, 1.00422341998656, 1.26214517662281, 0.359512373951839,
1.4835398594013, 0.154967053938309, -0.680501679226447, 1.44740598234453,
-0.512732029152914, -0.359512373951839)
我希望能够定义最合适的两部分线(显示的手绘示例)
然后我定义了一个分段函数,它应该找到一个两部分线性函数。 该定义基于两条线的梯度和它们彼此的截距,它们应该完全定义线。
# A=gradient of first line segment
# B=gradient of second line segment
# Cx=inflection point x coord
# Cy=inflexion point y coord
out_model <- nls(y ~ I(x <= Cx)*Cy-A*(Cx-x)+I(x > Cx)*Cy+B*(x),
data = data.frame(x,y),
start = c(A=-500,B=-500,Cx=0.0001,Cy=-1.5) )
但是我得到错误:
nls中的错误(y~I(x <= Cx)* Cy - A *(Cx - x)+ I(x> Cx)* Cy + B *:奇异梯度
我从查找曲线以匹配数据中获得了基本方法
我出错的任何想法?
我没有一个优雅的答案,但我确实有一个答案。
(请参阅下面的编辑以获得更优雅的答案)
如果Cx
足够小以至于没有数据点适合A
和Cy
,或者如果Cx
足够大以至于没有数据点适合B
和Cy
,则QR分解矩阵将是单数的,因为会有许多不同的分别适合数据的Cx
, A
和Cy
或Cx
, B
和Cy
。
我通过阻止Cx
安装来测试这个。 如果我将Cx
修复为(比方说) Cx = mean(x)
,则nls()
可以毫无困难地解决问题:
nls(y ~ ifelse(x < mean(x),ya+A*x,yb+B*x),
data = data.frame(x,y),
start = c(A=-1000,B=-1000,ya=3,yb=0))
......给出:
Nonlinear regression model
model: y ~ ifelse(x < mean(x), ya + A * x, yb + B * x)
data: data.frame(x, y)
A B ya yb
-1325.537 -1335.918 2.628 2.652
residual sum-of-squares: 0.06614
Number of iterations to convergence: 1
Achieved convergence tolerance: 2.294e-08
这让我想到,如果我改变Cx
使它永远不会超出范围[min(x),max(x)]
,那么这可能会解决问题。 实际上,我希望至少有三个数据点可用于拟合每条“A”线和“B”线,因此Cx必须介于x
的第三个最低值和第三个最高值之间。 使用具有适当算术的atan()
函数让我将范围[-inf,+inf]
映射到[0,1]
,所以我得到了代码:
trans <- function(x) 0.5+atan(x)/pi
xs <- sort(x)
xlo <- xs[3]
xhi <- xs[length(xs)-2]
nls(y ~ ifelse(x < xlo+(xhi-xlo)*trans(f),ya+A*x,yb+B*x),
data = data.frame(x,y),
start = c(A=-1000,B=-1000,ya=3,yb=0,f=0))
不幸的是,我仍然从这段代码得到singular gradient matrix at initial parameters
错误的singular gradient matrix at initial parameters
,所以问题仍然过度参数化。 正如@Henrik建议的那样,双线性和单线性拟合之间的差异对于这些数据来说并不是很好。
不过,我可以得到双线性拟合的答案。 由于nls()
在Cx
修复时解决了问题,我现在可以通过使用optimize()
简单地进行一维最小化来找到最小化残差标准误差的Cx
值。 不是一个特别优雅的解决方案,但总比没有好:
xs <- sort(x)
xlo <- xs[3]
xhi <- xs[length(xs)-2]
nn <- function(f) nls(y ~ ifelse(x < xlo+(xhi-xlo)*f,ya+A*x,yb+B*x),
data = data.frame(x,y),
start = c(A=-1000,B=-1000,ya=3,yb=0))
ssr <- function(f) sum(residuals(nn(f))^2)
f = optimize(ssr,interval=c(0,1))
print (f$minimum)
print (nn(f$minimum))
summary(nn(f$minimum))
...给出的输出:
[1] 0.8541683
Nonlinear regression model
model: y ~ ifelse(x < xlo + (xhi - xlo) * f, ya + A * x, yb + B * x)
data: data.frame(x, y)
A B ya yb
-1317.215 -872.002 2.620 1.407
residual sum-of-squares: 0.0414
Number of iterations to convergence: 1
Achieved convergence tolerance: 2.913e-08
Formula: y ~ ifelse(x < xlo + (xhi - xlo) * f, ya + A * x, yb + B * x)
Parameters:
Estimate Std. Error t value Pr(>|t|)
A -1.317e+03 1.792e+01 -73.493 < 2e-16 ***
B -8.720e+02 1.207e+02 -7.222 1.14e-07 ***
ya 2.620e+00 2.791e-02 93.854 < 2e-16 ***
yb 1.407e+00 3.200e-01 4.399 0.000164 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.0399 on 26 degrees of freedom
Number of iterations to convergence: 1
对于f
的最佳值, A
和B
的值与ya
和yb
之间没有巨大的差异,但是存在一些差异。
(编辑 - 优雅答案)
将问题分成两个步骤后,不再需要使用nls()
了。 lm()
工作正常,如下:
function (x,y)
{
f <- function (Cx)
{
lhs <- function(x) ifelse(x < Cx,Cx-x,0)
rhs <- function(x) ifelse(x < Cx,0,x-Cx)
fit <- lm(y ~ lhs(x) + rhs(x))
c(summary(fit)$r.squared,
summary(fit)$coef[1], summary(fit)$coef[2],
summary(fit)$coef[3])
}
r2 <- function(x) -(f(x)[1])
res <- optimize(r2,interval=c(min(x),max(x)))
res <- c(res$minimum,f(res$minimum))
best_Cx <- res[1]
coef1 <- res[3]
coef2 <- res[4]
coef3 <- res[5]
plot(x,y)
abline(coef1+best_Cx*coef2,-coef2) #lhs
abline(coef1-best_Cx*coef3,coef3) #rs
}
... 这使:
如果断点已知,则可以使用线性回归
k <- 0.0025
lhs <- function(x) ifelse(x < k,k-x,0)
rhs <- function(x) ifelse(x < k,0,x-k)
fit <- lm(y ~ lhs(x) + rhs(x))
segmented
包是为这类问题而设计的。
首先,使用lm
创建常规线性回归:
linmod <- lm(y ~ x)
summary(linmod)
这给了我们:
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-0.108783 -0.025432 -0.006484 0.040092 0.088638
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.630e+00 2.732e-02 96.28 <2e-16 ***
x -1.326e+03 1.567e+01 -84.63 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.04869 on 28 degrees of freedom
Multiple R-squared: 0.9961, Adjusted R-squared: 0.996
F-statistic: 7163 on 1 and 28 DF, p-value: < 2.2e-16
接下来,我们使用线性模型生成具有1个断点的分段模型:
segmod <- segmented(linmod, seg.Z = ~x)
summary(segmod)
分段模型提供了更好的r平方:
***Regression Model with Segmented Relationship(s)***
Call:
segmented.lm(obj = linmod, seg.Z = ~x)
Estimated Break-Point(s):
Est. St.Err
0.003 0.000
Meaningful coefficients of the linear terms:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.659e+00 2.882e-02 92.239 <2e-16 ***
x -1.347e+03 1.756e+01 -76.742 <2e-16 ***
U1.x 5.167e+02 4.822e+02 1.072 NA
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.04582 on 26 degrees of freedom
Multiple R-Squared: 0.9968, Adjusted R-squared: 0.9964
Convergence attained in 3 iterations with relative change 0
你可以检查情节,拦截和斜率:
plot(segmod)
intercept(segmod)
slope(segmod)
感谢Henrik让我走上了正确的道路! 这是一个更完整,相对优雅的解决方案,简单的情节:
range_x<-max(x)-min(x)
intervals=1000
coef1=c()
coef2=c()
coef3=c()
r2=c()
for (i in 1:intervals)
{
Cx<-min(x)+(i-1)*(range_x/intervals)
lhs <- function(x) ifelse(x < Cx,Cx-x,0)
rhs <- function(x) ifelse(x < Cx,0,x-Cx)
fit <- lm(y ~ lhs(x) + rhs(x))
coef1[i]<-summary(fit)$coef[1]
coef2[i]<-summary(fit)$coef[2]
coef3[i]<-summary(fit)$coef[3]
r2[i]<-summary(fit)$r.squared
}
best_r2<-max(r2) # get best r squared
pos<-which.max(r2)
best_Cx<-min(x)+(pos-1)*(range_x/intervals) # get Cx for best r2
plot(x,y)
abline(coef1[pos]+best_Cx*coef2[pos],-coef2[pos]) #lhs
abline(coef1[pos]-best_Cx*coef3[pos],coef3[pos]) #rs
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