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I am trying to fit a GAM model to data under two constraints simultatenously: (1) the fit is monotonic (increasing), (2) the fit goes through a fixed point, say, (x0,y0).
So far, I managed to have these two constraints work separately:
For (1), based on mgcv::pcls() documentation examples, by using mgcv::mono.con() to get linear constraints sufficient for monotonicity, and estimate model coefs via mgcv::pcls(), using the constraints.
For (2), based on this post, by setting the value of spline at knot location x0 to 0 + using offset term in the model formula.
However, I struggle to combine these two constraints simultaneously. I guess a way to go is mgcv::pcls(), but I could work out neither (a) doing a similar trick of setting the value of spline at knot location x0 to 0 + using offset nor (b) setting equality constraint(s) (which I think could yield my (2) constraint setup).
I also note that the approach for setting the value of spline at knot location x0 to 0 for my constrain condition (2) yields weirdly wiggly outcome (as compared to unconstrained GAM fit) -- as showed below.
Attempt so far: fit a smooth function to data under two constraints separately
Simulate some data
library(mgcv)
set.seed(1)
x <- sort(runif(100) * 4 - 1)
f <- exp(4*x)/(1+exp(4*x))
y <- f + rnorm(100) * 0.1
dat <- data.frame(x=x, y=y)
GAM unconstrained (for comparison)
k <- 13
fit0 <- gam(y ~ s(x, k = k, bs = "cr"), data = dat)
# predict from unconstrained GAM fit
newdata <- data.frame(x = seq(-1, 3, length.out = 1000))
newdata$y_pred_fit0 <- predict(fit0, newdata = newdata)
GAM constrained: (1) the fit is monotonic (increasing)
k <- 13
# Show regular spline fit (and save fitted object)
f.ug <- gam(y~s(x,k=k,bs="cr"))
# explicitly construct smooth term's design matrix
sm <- smoothCon(s(x,k=k,bs="cr"),dat,knots=NULL)[[1]]
# find linear constraints sufficient for monotonicity of a cubic regression spline
# it assumes "cr" is the basis and its knots are provided as input
F <- mono.con(sm$xp)
G <- list(
X=sm$X,
C=matrix(0,0,0), # [0 x 0] matrix (no equality constraints)
sp=f.ug$sp, # smoothing parameter estimates (taken from unconstrained model)
p=sm$xp, # array of feasible initial parameter estimates
y=y,
w= dat$y * 0 + 1 # weights for data
)
G$Ain <- F$A # matrix for the inequality constraints
G$bin <- F$b # vector for the inequality constraints
G$S <- sm$S # list of penalty matrices; The first parameter it penalizes is given by off[i]+1
G$off <- 0 # Offset values locating the elements of M$S in the correct location within each penalty coefficient matrix. (Zero offset implies starting in first location)
p <- pcls(G); # fit spline (using smoothing parameter estimates from unconstrained fit)
# predict
newdata$y_pred_fit2 <- Predict.matrix(sm, data.frame(x = newdata$x)) %*% p
# plot
plot(y ~ x, data = dat)
lines(y_pred_fit0 ~ x, data = newdata, col = 2, lwd = 2)
lines(y_pred_fit2 ~ x, data = newdata, col = 4, lwd = 2)
Blue line: constrained; red line: unconstrained
GAM constrained: (2) fitted go through (x0,y0)=(-1, -0.1)
k <- 13
## Create a spline basis and penalty
## Make sure there is a knot at the constraint point (here: -1)
knots <- data.frame(x = seq(-1,3,length=k))
# explicit construction of a smooth term in a GAM
sm <- smoothCon(s(x,k=k,bs="cr"), dat, knots=knots)[[1]]
## 1st parameter is value of spline at knot location -1, set it to 0 by dropping
knot_which <- which(knots$x == -1)
X <- sm$X[, -knot_which] ## spline basis
S <- sm$S[[1]][-knot_which, -knot_which] ## spline penalty
off <- dat$y * 0 + (-0.1) ## offset term to force curve through (x0, y0)
## fit spline constrained through (x0, y0)
gam_1 <- gam(y ~ X - 1 + offset(off), paraPen = list(X = list(S)))
# predict (add offset of -0.1)
newdata_tmp <- Predict.matrix(sm, data.frame(x = newdata$x))
newdata_tmp <- newdata_tmp[, -knot_which]
newdata$y_pred_fit1 <- (newdata_tmp %*% coef(gam_1))[, 1] + (-0.1)
# plot
plot(y ~ x, data = dat)
lines(y_pred_fit0 ~ x, data = newdata, col = 2, lwd = 2)
lines(y_pred_fit1 ~ x, data = newdata, col = 3, lwd = 2)
# lines at cross of which the plot should go throught
abline(v=-1, col = 3); abline(h=-0.1, col = 3)
Green line: constrained; red line: unconstrained
I think you could augment the data vectors x and y with (x0, y0) and then put a (really) high weight on the first observation (i.e. add a weight vector to your G list).
Alternatively to the simple weighting strategy, we can write the quadratic programming problem starting from the results of the preliminary smoothing. This is illustrated in the second R-code below (in this case I used p-spline smoothers, see Eilers and Marx 1991).
Hope this helps a bit (a similar problem is discussed here).
Rcode example 1 (weight strategy)
set.seed(123)
N = 100
x <- sort(runif(N) * 4 - 1)
f <- exp(4*x)/(1+exp(4*x))
y <- f + rnorm(N) * 0.1
x = c(-1, x)
y = c(-0.1, y)
dat = data.frame(x = x, y= y)
k <- 13
fit0 <- gam(y ~ s(x, k = k, bs = "cr"), data = dat)
# predict from unconstrained GAM fit
newdata <- data.frame(x = seq(-1, 3, length.out = 1000))
newdata$y_pred_fit0 <- predict(fit0, newdata = newdata)
k <- 13
# Show regular spline fit (and save fitted object)
f.ug <- gam(y~s(x,k=k,bs="cr"))
# explicitly construct smooth term's design matrix
sm <- smoothCon(s(x,k=k,bs="cr"),dat,knots=NULL)[[1]]
# find linear constraints sufficient for monotonicity of a cubic regression spline
# it assumes "cr" is the basis and its knots are provided as input
F <- mono.con(sm$xp)
G <- list(
X=sm$X,
C=matrix(0,0,0), # [0 x 0] matrix (no equality constraints)
sp=f.ug$sp, # smoothing parameter estimates (taken from unconstrained model)
p=sm$xp, # array of feasible initial parameter estimates
y=y,
w= c(1e8, 1:N * 0 + 1) # weights for data
)
G$Ain <- F$A # matrix for the inequality constraints
G$bin <- F$b # vector for the inequality constraints
G$S <- sm$S # list of penalty matrices; The first parameter it penalizes is given by off[i]+1
G$off <- 0 # Offset values locating the elements of M$S in the correct location within each penalty coefficient matrix. (Zero offset implies starting in first location)
p <- pcls(G); # fit spline (using smoothing parameter estimates from unconstrained fit)
# predict
newdata$y_pred_fit2 <- Predict.matrix(sm, data.frame(x = newdata$x)) %*% p
# plot
plot(y ~ x, data = dat)
lines(y_pred_fit0 ~ x, data = newdata, col = 2, lwd = 2)
lines(y_pred_fit2 ~ x, data = newdata, col = 4, lwd = 2)
abline(v = -1)
abline(h = -0.1)
rm(list = ls())
library(mgcv)
library(pracma)
library(colorout)
set.seed(123)
N = 100
x = sort(runif(N) * 4 - 1)
f = exp(4*x)/(1+exp(4*x))
y = f + rnorm(N) * 0.1
x0 = -1
y0 = -0.1
dat = data.frame(x = x, y= y)
k = 50
# Show regular spline fit (and save fitted object)
f.ug = gam(y~s(x,k=k,bs="ps"))
# explicitly construct smooth term's design matrix
sm = smoothCon(s(x,k=k,bs="ps"), dat,knots=NULL)[[1]]
# Build quadprog to estimate the coefficients
scf = sapply(f.ug$smooth, '[[', 'S.scale')
lam = f.ug$sp / scf
Xp = rbind(sm$X, sqrt(lam) * f.ug$smooth[[1]]$D)
yp = c(dat$y, rep(0, k - 2))
X0 = Predict.matrix(sm, data.frame(x = x0))
sm$deriv = 1
X1 = Predict.matrix(sm, data.frame(x = dat$x))
coef_mono = pracma::lsqlincon(Xp, yp, Aeq = X0, beq = y0, A = -X1, b = rep(0, N))
# fitted values
fit = sm$X %*% coef_mono
sm$deriv = 0
xf = seq(-1, 3, len = 1000)
Xf = Predict.matrix(sm, data.frame(x = xf))
fine_fit = Xf %*% coef_mono
# plot
par(mfrow = c(2, 1), mar = c(3,3,3,3))
plot(dat$x, dat$y, pch = 1, main= 'Data and fit')
lines(dat$x, f.ug$fitted, lwd = 2, col = 2)
lines(dat$x, fit, col = 4, lty = 1, lwd = 2)
lines(xf, fine_fit, col = 3, lwd = 2, lty = 2)
abline(h = -0.1)
abline(v = -1)
plot(dat$x, X1 %*% coef_mono, type = 'l', main = 'Derivative of the fit', lwd = 2)
abline(h = 0.0)
The following package seems to implement what you are looking for:
The proposed shape constrained smoothing has been incorporated into generalized
additive models with a mixture of unconstrained and shape restricted smooth terms
(mono-GAM). [...]
The proposed modelling approach has been implemented in an R package monogam.
The model setup is the same as in mgcv(gam) with the addition of shape constrained
smooths. In order to be consistent with the unconstrained GAM, the package provides
key functions similar to those associated with mgcv(gam).
Additive models with shape constraints
My question is about the typical feed-forward single-hidden-layer backprop neural network, as implemented in package nnet, and trained with train() in package caret. This is related to this question but in the context of the nnet and caret packages in R.
I demonstrate the problem with a simple regression example where Y = sin(X) + small error:
raw Y ~ raw X: predicted outputs are uniformly zero where raw Y < 0.
scaled Y (to 0-1) ~ raw X: solution looks great; see code below.
The code is as follows
library(nnet)
X <- t(t(runif(200, -pi, pi)))
Y <- t(t(sin(X))) # Y ~ sin(X)
Y <- Y + rnorm(200, 0, .05) # Add a little noise
Y_01 <- (Y - min(Y))/diff(range(Y)) # Y linearly transformed to have range 0-1.
plot(X,Y)
plot(X, Y_01)
dat <- data.frame(cbind(X, Y, Y_01)); names(dat) <- c("X", "Y", "Y_01")
head(dat)
plot(dat)
nnfit1 <- nnet(formula = Y ~ X, data = dat, maxit = 2000, size = 8, decay = 1e-4)
nnpred1 <- predict(nnfit1, dat)
plot(X, nnpred1)
nnfit2 <- nnet(formula = Y_01 ~ X, data = dat, maxit = 2000, size = 8, decay = 1e-4)
nnpred2 <- predict(nnfit2, dat)
plot(X, nnpred2)
When using train() in caret, there is a preProcess option but it only scales the inputs. train(..., method = "nnet", ...) appears to be using the raw Y values; see code below.
library(caret)
ctrl <- trainControl(method = "cv", number = 10)
nnet_grid <- expand.grid(.decay = 10^seq(-4, -1, 1), .size = c(8))
nnfit3 <- train(Y ~ X, dat, method = "nnet", maxit = 2000,
trControl = ctrl, tuneGrid = nnet_grid, preProcess = "range")
nnfit3
nnpred3 <- predict(nnfit3, dat)
plot(X, nnpred3)
Of course, I could linearly transform the Y variable(s) to have a positive range, but then my predictions will be on the wrong scale. Though this is only a minor headache, I'm wondering if there is a better solution for training nnet or avNNet models with caret when the output has negative values.
This was answered on cross validated here by user topepo
The relevant part of their answer is:
Since Y is roughly between -1 and 1 you should also use linout = TRUE in your nnet and train calls.
I can create simple graphs. I would like to have observed and predicted values (from a linear regression) on the same graph. I am plotting say Yvariable vs Xvariable. There is only 1 predictor and only 1 response. How could I also add linear regression curve to the same graph?
So to conclude need help with:
plotting actuals and predicted both
plotting regression line
Here is one option for the observed and predicted values in a single plot as points. It is easier to get the regression line on the observed points, which I illustrate second
First some dummy data
set.seed(1)
x <- runif(50)
y <- 2.5 + (3 * x) + rnorm(50, mean = 2.5, sd = 2)
dat <- data.frame(x = x, y = y)
Fit our model
mod <- lm(y ~ x, data = dat)
Combine the model output and observed x into a single object for plott
res <- stack(data.frame(Observed = dat$y, Predicted = fitted(mod)))
res <- cbind(res, x = rep(dat$x, 2))
head(res)
Load lattice and plot
require("lattice")
xyplot(values ~ x, data = res, group = ind, auto.key = TRUE)
The resulting plot should look similar to this
To get just the regression line on the observed data, and the regression model is a simple straight line model as per the one I show then you can circumvent most of this and just plot using
xyplot(y ~ x, data = dat, type = c("p","r"), col.line = "red")
(i.e. you don't even need to fit the model or make new data for plotting)
The resulting plot should look like this
An alternative to the first example which can be used with anything that will give coefficients for the regression line is to write your own panel functions - not as scary as it seems
xyplot(y ~ x, data = dat, col.line = "red",
panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.abline(coef = coef(mod), ...) ## using mod from earlier
}
)
That gives a plot from Figure 2 above, but by hand.
Assuming you've done this with caret then
mod <- train(y ~ x, data = dat, method = "lm",
trControl = trainControl(method = "cv"))
xyplot(y ~ x, data = dat, col.line = "red",
panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.abline(coef = coef(mod$finalModel), ...) ## using mod from caret
}
)
Will produce a plot the same as Figure 2 above.
Another option is to use panel.lmlineq from latticeExtra.
library(latticeExtra)
set.seed(0)
xsim <- rnorm(50, mean = 3)
ysim <- (0 + 2 * xsim) * (1 + rnorm(50, sd = 0.3))
## basic use as a panel function
xyplot(ysim ~ xsim, panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.lmlineq(x, y, adj = c(1,0), lty = 1,xol.text='red',
col.line = "blue", digits = 1,r.squared =TRUE)
})
Is it there a way to adjust a quadratic spline (instead of a cubic one) to some data?
I have this data and I don't seem to find the appropiate function in R to do this.
Expanding just a bit on the comments above, you can use a B-spline basis (implemented in function splines::bs()), setting degree=2 rather than the default degree=3:
library(splines)
## Some example data
set.seed(1)
x <- 1:10
y <- rnorm(10)
## Fit a couple of quadratic splines with different degrees of freedom
f1 <- lm(y ~ bs(x, degree = 2)) # Defaults to 2 - 1 = 1 degree of freedom
f9 <- lm(y ~ bs(x, degree = 2, df=9))
## Plot the splines
x0 <- seq(1, 10, by = 0.1)
plot(x, y, pch = 16)
lines(x0, predict(f1, data.frame(x = x0)), col = "blue")
lines(x0, predict(f9, data.frame(x = x0)), col = "red")
I want to carry out a linear regression in R for data in a normal and in a double logarithmic plot.
For normal data the dataset might be the follwing:
lin <- data.frame(x = c(0:6), y = c(0.3, 0.1, 0.9, 3.1, 5, 4.9, 6.2))
plot (lin$x, lin$y)
There I want to calculate draw a line for the linear regression only of the datapoints 2, 3 and 4.
For double logarithmic data the dataset might be the following:
data = data.frame(
x=c(1:15),
y=c(
1.000, 0.742, 0.623, 0.550, 0.500, 0.462, 0.433,
0.051, 0.043, 0.037, 0.032, 0.028, 0.025, 0.022, 0.020
)
)
plot (data$x, data$y, log="xy")
Here I want to draw the regression line for the datasets 1:7 and for 8:15.
Ho can I calculate the slope and the y-offset als well as parameters for the fit (R^2, p-value)?
How is it done for normal and for logarithmic data?
Thanks for you help,
Sven
In R, linear least squares models are fitted via the lm() function. Using the formula interface we can use the subset argument to select the data points used to fit the actual model, for example:
lin <- data.frame(x = c(0:6), y = c(0.3, 0.1, 0.9, 3.1, 5, 4.9, 6.2))
linm <- lm(y ~ x, data = lin, subset = 2:4)
giving:
R> linm
Call:
lm(formula = y ~ x, data = lin, subset = 2:4)
Coefficients:
(Intercept) x
-1.633 1.500
R> fitted(linm)
2 3 4
-0.1333333 1.3666667 2.8666667
As for the double log, you have two choices I guess; i) estimate two separate models as we did above, or ii) estimate via ANCOVA. The log transformation is done in the formula using log().
Via two separate models:
logm1 <- lm(log(y) ~ log(x), data = dat, subset = 1:7)
logm2 <- lm(log(y) ~ log(x), data = dat, subset = 8:15)
Or via ANCOVA, where we need an indicator variable
dat <- transform(dat, ind = factor(1:15 <= 7))
logm3 <- lm(log(y) ~ log(x) * ind, data = dat)
You might ask if these two approaches are equivalent? Well they are and we can show this via the model coefficients.
R> coef(logm1)
(Intercept) log(x)
-0.0001487042 -0.4305802355
R> coef(logm2)
(Intercept) log(x)
0.1428293 -1.4966954
So the two slopes are -0.4306 and -1.4967 for the separate models. The coefficients for the ANCOVA model are:
R> coef(logm3)
(Intercept) log(x) indTRUE log(x):indTRUE
0.1428293 -1.4966954 -0.1429780 1.0661152
How do we reconcile the two? Well the way I set up ind, logm3 is parametrised to give more directly values estimated from logm2; the intercepts of logm2 and logm3 are the same, as are the coefficients for log(x). To get the values equivalent to the coefficients
of logm1, we need to do a manipulation, first for the intercept:
R> coefs[1] + coefs[3]
(Intercept)
-0.0001487042
where the coefficient for indTRUE is the difference in the mean of group 1 over the mean of group 2. And for the slope:
R> coefs[2] + coefs[4]
log(x)
-0.4305802
which is the same as we got for logm1 and is based on the slope for group 2 (coefs[2]) modified by the difference in slope for group 1 (coefs[4]).
As for plotting, an easy way is via abline() for simple models. E.g. for the normal data example:
plot(y ~ x, data = lin)
abline(linm)
For the log data we might need to be a bit more creative, and the general solution here is to predict over the range of data and plot the predictions:
pdat <- with(dat, data.frame(x = seq(from = head(x, 1), to = tail(x,1),
by = 0.1))
pdat <- transform(pdat, yhat = c(predict(logm1, pdat[1:70,, drop = FALSE]),
predict(logm2, pdat[71:141,, drop = FALSE])))
Which can plot on the original scale, by exponentiating yhat
plot(y ~ x, data = dat)
lines(exp(yhat) ~ x, dat = pdat, subset = 1:70, col = "red")
lines(exp(yhat) ~ x, dat = pdat, subset = 71:141, col = "blue")
or on the log scale:
plot(log(y) ~ log(x), data = dat)
lines(yhat ~ log(x), dat = pdat, subset = 1:70, col = "red")
lines(yhat ~ log(x), dat = pdat, subset = 71:141, col = "blue")
For example...
This general solution works well for the more complex ANCOVA model too. Here I create a new pdat as before and add in an indicator
pdat <- with(dat, data.frame(x = seq(from = head(x, 1), to = tail(x,1),
by = 0.1)[1:140],
ind = factor(rep(c(TRUE, FALSE), each = 70))))
pdat <- transform(pdat, yhat = predict(logm3, pdat))
Notice how we get all the predictions we want from the single call to predict() because of the use of ANCOVA to fit logm3. We can now plot as before:
plot(y ~ x, data = dat)
lines(exp(yhat) ~ x, dat = pdat, subset = 1:70, col = "red")
lines(exp(yhat) ~ x, dat = pdat, subset = 71:141, col = "blue")
#Split the data into two groups
data1 <- data[1:7, ]
data2 <- data[8:15, ]
#Perform the regression
model1 <- lm(log(y) ~ log(x), data1)
model2 <- lm(log(y) ~ log(x), data2)
summary(model1)
summary(model2)
#Plot it
with(data, plot(x, y, log="xy"))
lines(1:7, exp(predict(model1, data.frame(x = 1:7))))
lines(8:15, exp(predict(model2, data.frame(x = 8:15))))
In general, splitting the data into different groups and running different models on different subsets is unusual, and probably bad form. You may want to consider adding a grouping variable
data$group <- factor(rep(letters[1:2], times = 7:8))
and running some sort of model on the whole dataset, e.g.,
model_all <- lm(log(y) ~ log(x) * group, data)
summary(model_all)