I am trying to get the nonlinear formula of y0, a, and b for my curve so I cam plot it on my graph. the summary(nls.mod) output does not show the y0 that I will need to plot the curve and I am not sure why as I have tried everything. The code is below:
# BH this version of plot is used for diagnostic plots for
# BH residuals of a linear model, i.e. using lm.
plot(mdl3 <- lm(ETR ~ wp_Mpa + I(wp_Mpa^2) + I(wp_Mpa^3), data = dat3))
prd <- data.frame(x = seq(-4, 0, by = 0.5))
result <- prd
result$mdl3 <- predict(mdl3, newdat3 = prd)
# BH use nls to fit this model y0+a*exp(b*x)
nls.mod <- nls(ETR ~ y0 + a*exp(b*wp_Mpa), start=c(a = -4, b = 0), data=dat3.no_na)
summary(nls.mod)
and here is the output:
Formula: ETR ~ y0 + a * exp(b * wp_Mpa)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 85.85515 8.62005 9.960 <2e-16 ***
b 0.14157 0.07444 1.902 0.0593 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 58.49 on 134 degrees of freedom
Number of iterations to convergence: 8
Achieved convergence tolerance: 1.515e-06
as you can see for some reason only the a and b show up but the y0 is supposed to be above that
I tried to reassign the variables and that just continued to give me the same output contacted a stats consultant and they just said I needed to change the variables but it still didnt work
Let's consider this code as an example:
a = 10
b = 2
c = 1.05
set.seed(123,kind="Mersenne-Twister",normal.kind="Inversion")
x = runif(100)
data = data.frame(X = x, Y = a + b/c * (((1-x)^-c)-1))
fit_sp <- minpack.lm::nlsLM( formula = Y ~ a + b/c * (((1-X)^-c)-1),
data = data, start = c(a = 0, b = 0.1, c = 0.01),
control = nls.control(maxiter = 1000),
lower = c(0.0001,0.0001,0.0001))
fit_sp
Nonlinear regression model
model: Y ~ a + b/c * (((1 - X)^-c) - 1)
data: data
a b c
10.00 2.00 1.05
residual sum-of-squares: 1.507e-26
Number of iterations to convergence: 13
Achieved convergence tolerance: 1.49e-08
summary(fit_sp)
Formula: Y ~ a + b/c * (((1 - X)^-c) - 1)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 1.000e+01 1.516e-15 6.598e+15 <2e-16 ***
b 2.000e+00 4.372e-16 4.574e+15 <2e-16 ***
c 1.050e+00 5.319e-17 1.974e+16 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.246e-14 on 97 degrees of freedom
Number of iterations to convergence: 13
Achieved convergence tolerance: 1.49e-08
I know that non linear least squares calculates the coefficients that minimize the sum of squared residuals. But how is it possible to obtain by hand the standard errors and the t-statistics for the parameters estimate?
I am using the nls package in R to perform a nonlinear fit. I have specified my independent variable as follows:
t <- seq(1,7)
and my dependent variables as P <- c(0.0246, 0.2735, 0.5697, 0.6715, 0.8655, 0.9614, 1)
I then have tried:
m <- nls(P ~ 1 / (c + q*exp(-b*t))^(1/v)),
but every time I get:
"Error in c + q * exp(-b * t) : non-numeric argument to binary
operator"
Every one of my variables is numeric. Any ideas?
Thanks!
You have more than one problem in your script. The main issue is that you should never use names which are used by R: t is the matrix transpose, c is a common method to create vectors, and q is the quit instruction. nls() will not try to fit them, as they are already defined. I recommend using more meaningful and less dangerous variables such as Coef1, Coef2, …
The second problem is that you are trying to fit a model with 4 variables to a dataset with 7 data... This may yield singularities and other problems.
For the sake of the argument, I have reduced your model to three variables, and changed some names:
Time <- seq(1,7)
Prob <- c(0.0246, 0.2735, 0.5697, 0.6715, 0.8655, 0.9614, 1)
plot(Time, Prob)
And now we perform the nls() fit:
Fit <- nls(Prob ~ 1 / (Coef1 + Coef2 * exp(-Coef3 * Time)))
X <- data.frame(Time = seq(0, 7, length.out = 100))
Y <- predict(object = Fit, newdata = X)
lines(X$Time, Y)
And a summary of the results:
summary(Fit)
# Formula: Prob ~ 1/(Coef1 + Coef2 * exp(-Coef3 * Time))
#
# Parameters:
# Estimate Std. Error t value Pr(>|t|)
# Coef1 1.00778 0.06113 16.487 7.92e-05 ***
# Coef2 23.43349 14.42378 1.625 0.1796
# Coef3 1.04899 0.21892 4.792 0.0087 **
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.06644 on 4 degrees of freedom
#
# Number of iterations to convergence: 12
# Achieved convergence tolerance: 3.04e-06
I know it is not exactly what you wanted, but I hope it helps.
I'm trying to find solution in R that performs similarly to MATLAB's trust region reflective algorithm. This question has been asked before but the author was asked to provide reproducible example. I couldn't comment there so the only solution was to post new question. Here's my example:
x <- c(5000,5000,5000,5000,2500,2500,2500,2500,1250,1250,1250,1250,625,625, 625,625,312,312,312,312,156,156,156,156)
y <- c(0.209065186,0.208338898,0.211886104,0.209638321,0.112064803,0.110535275,0.111748670,0.111208841,0.060416469,0.059098975,0.059274827,0.060859512,0.032178730,0.033190833,0.031621743,0.032345817,0.017983939,0.016632180,0.018468540,0.019513489,0.011490089,0.011076365,0.009282322,0.012309134)
Since initial parameter values are the central issue I tried using 'nls2' package which uses 'brute-force' algorithm to find good starting parameters. Even with that, nls and nls.lm cannot reach convergence. Here's some basic code for this:
library('nls2'); library('minpack.lm')
fo <- y ~ I(A * (x ^ B) + C)
sA <- seq(-2,1,len=10) # range of parameter values
sB <- seq(-1,1,len=10)
sC <- seq(-1,1,len=10)
st1 <- expand.grid(A=sA,B=sB,C=sC)
mod1 <- nls2(fo,start=st1,algorithm="brute-force")
fit_ <- nls(fo,start=coef(mod1)) # basic nls
# or nls.lm
fit_ <- nlsLM(fo, start=coef(mod1),algorithm = "LM")
MATLAB produced:
a = 7.593e-05 (6.451e-05, 8.736e-05)
b = 0.9289 (0.9116, 0.9462)
c = 0.002553 (0.001333, 0.003772)
Goodness of fit:
SSE: 2.173e-05
R-square: 0.9998
Adjusted R-square: 0.9998
RMSE: 0.001017
and yes, using these parameter values, R also produced the solution.
Question: how to obtain this in R without using matlab ?
After looking at a the plotted data, I have no problem guessing suitable starting values:
plot(y ~ x)
The data is almost on a straight line through 0. So good starting value vor B and C should be 1 and 0, respectively. Then you only need to guestimate the slope of the straight line. Of course you could also use lm(y ~ x) to find starting values for A and C.
fo <- y ~ A * (x ^ B) + C
DF <- data.frame(x, y)
fit <- nls(fo, start = list(A = 0.001, B = 1, C = 0), data = DF)
summary(fit)
#Formula: y ~ A * (x^B) + C
#
#Parameters:
# Estimate Std. Error t value Pr(>|t|)
#A 7.593e-05 5.495e-06 13.820 5.17e-12 ***
#B 9.289e-01 8.317e-03 111.692 < 2e-16 ***
#C 2.552e-03 5.866e-04 4.351 0.000281 ***
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#Residual standard error: 0.001017 on 21 degrees of freedom
#
#Number of iterations to convergence: 5
#Achieved convergence tolerance: 9.084e-07
lines(seq(min(x), max(x), length.out = 100),
predict(fit, newdata = data.frame(x = seq(min(x), max(x), length.out = 100))),
col = "blue")
I am trying to fit a two-part line to data.
Here's some sample data:
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)
I am hoping to be able to define the best fitting two part line (hand drawn example shown)
I then define a piecewise function that should find a two part linear function. The definition is based on the gradients of the two lines and their intercept with each other, which should completely define the lines.
# 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) )
However I get the error:
Error in nls(y ~ I(x <= Cx) * Cy - A * (Cx - x) + I(x > Cx) * Cy + B * :
singular gradient
I got the basic method from Finding a curve to match data
Any ideas where I am going wrong?
I don't have an elegant answer, but I do have an answer.
(SEE THE EDIT BELOW FOR A MORE ELEGANT ANSWER)
If Cx is small enough that there are no data points to fit A and Cy to, or if Cx is big enough that there are no data points to fit B and Cy to, the QR decomposition matrix will be singular because there will be many different values of Cx, A and Cy or Cx, B and Cy respectively that will fit the data equally well.
I tested this by preventing Cx from being fitted. If I fix Cx at (say) Cx = mean(x), nls() solves the problem without difficulty:
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))
... gives:
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
That led me to think that if I transformed Cx so that it could never go outside the range [min(x),max(x)], that might solve the problem. In fact, I'd want there to be at least three data points available to fit each of the "A" line and the "B" line, so Cx has to be between the third lowest and the third highest values of x. Using the atan() function with the appropriate arithmetic let me map a range [-inf,+inf] onto [0,1], so I got the code:
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))
Unfortunately, however, I still get the singular gradient matrix at initial parameters error from this code, so the problem is still over-parameterised. As #Henrik has suggested, the difference between the bilinear and single linear fit is not great for these data.
I can nevertheless get an answer for the bilinear fit, however. Since nls() solves the problem when Cx is fixed, I can now find the value of Cx that minimises the residual standard error by simply doing a one-dimensional minimisation using optimize(). Not a particularly elegant solution, but better than nothing:
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$minimum))
... gives output of:
[1] 0.8541683
Nonlinear regression model
model: y ~ ifelse(x < xlo + (xhi - xlo) * f, ya + A * x, yb + B * x)
data: data.frame(x, y)
A B ya yb
-1317.215 -872.002 2.620 1.407
residual sum-of-squares: 0.0414
Number of iterations to convergence: 1
Achieved convergence tolerance: 2.913e-08
Formula: y ~ ifelse(x < xlo + (xhi - xlo) * f, ya + A * x, yb + B * x)
Parameters:
Estimate Std. Error t value Pr(>|t|)
A -1.317e+03 1.792e+01 -73.493 < 2e-16 ***
B -8.720e+02 1.207e+02 -7.222 1.14e-07 ***
ya 2.620e+00 2.791e-02 93.854 < 2e-16 ***
yb 1.407e+00 3.200e-01 4.399 0.000164 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.0399 on 26 degrees of freedom
Number of iterations to convergence: 1
There isn't a huge difference between the values of A and B and ya and yb for the optimum value of f, but there is some difference.
(EDIT -- ELEGANT ANSWER)
Having separated the problem into two steps, it isn't necessary to use nls() any more. lm() works fine, as follows:
function (x,y)
{
f <- function (Cx)
{
lhs <- function(x) ifelse(x < Cx,Cx-x,0)
rhs <- function(x) ifelse(x < Cx,0,x-Cx)
fit <- lm(y ~ lhs(x) + rhs(x))
c(summary(fit)$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
}
... which gives:
If the breakpoint is known it is possible to use linear regression
Broken stick regression from "Practical Regression and Anova using R"
Julian J. Faraway
December 2000
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))
The package segmented was designed for this type of problem.
First, create a regular linear regression with lm:
linmod <- lm(y ~ x)
summary(linmod)
Which gives us:
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
Next, we use the linear model to produce a segmented model with 1 break point:
segmod <- segmented(linmod, seg.Z = ~x)
summary(segmod)
And the segmented model provides a slightly better r-squared:
***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
You can check the plot, intercept and slope:
plot(segmod)
intercept(segmod)
slope(segmod)
Thank to Henrik for putting me on the right path!
Here's a more complete and relatively elegant solution with a simple plot:
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