I want to estimate the parameters of a non linear model.
The model equation is Z = A * exp(- a * X) + B * exp(- b * Y) + C
What I did is to transform the model into a linear problem by doing a exponential transformation before doing a linear regression:
a and b between 0 and 1, I compute exp_x = exp(- a * X) and exp_y = exp(- b * Y)Z ~ exp_x + exp_yIt works very well as we can see in this simulation
x = 1:10
y = 1:10
combination = expand.grid(x = x, y = y)
df = data.frame(
X = combination$x,
Y = combination$y,
Z = 2 * exp(-0.3 * combination$x) +
5 * exp(-0.6 * combination$y) +
rnorm(n = 100, mean = 0, sd = 0.1 )
)
a_hat = 0
b_hat = 0
best_ols = NULL
best_rsquared = 0
for (a in seq(0.01, 1, 0.01)){
for (b in seq(0.01, 1, 0.01)){
df$exp_x = exp(- a * df$X)
df$exp_y = exp(- b *df$Y)
ols = lm(data = df, formula = Z ~ exp_x + exp_y)
r_squared = summary(ols)$r.squared
if (r_squared > best_rsquared){
best_rsquared = r_squared
a_hat = a
b_hat = b
best_ols = ols
}
}
}
a_hat
b_hat
best_ols
best_rsquared
> a_hat
[1] 0.34
> b_hat
[1] 0.63
> best_ols
Call:
lm(formula = Z ~ exp_x + exp_y, data = df)
Coefficients:
(Intercept) exp_x exp_y
0.0686 2.0550 5.1189
> best_rsquared
[1] 0.9898669
Problem: this is slow
It takes around 10 secs and I need to do it thousands times on others data frame.
How could I drastically speed it up?
Perhaps use nls instead. Since you did not set.seed(), cannot see whether our predictions would be similar, but at least I got the a and b estimates "right" after the edit:
nmod <- nls( Z ~ A*exp(-a*X)+B*exp(-b*Y), data=df, start=list(A=0.5, B=0.5, a=.1,b=.1))
> coef(nmod)
A B a b
2.0005670 4.9541553 0.2951589 0.5937909
#--------
> nmod
Nonlinear regression model
model: Z ~ A * exp(-a * X) + B * exp(-b * Y)
data: df
A B a b
2.0006 4.9542 0.2952 0.5938
residual sum-of-squares: 0.9114
Number of iterations to convergence: 9
Achieved convergence tolerance: 5.394e-06
Much faster than your 10 second experience. And this is on an 8 year-old machine.
> system.time( nmod <- nls( Z ~ A*exp(-a*X)+B*exp(-b*Y), data=df, start=list(A=0.5, B=0.5, a=.1,b=.1)) )
user system elapsed
0.036 0.002 0.033
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