Equivalent of Matlab's 'fit' for Gaussian mixture models in R? - r

Question

I have some time series data that looks like this:
x <- c(0.5833, 0.95041, 1.722, 3.1928, 3.941, 5.1202, 6.2125, 5.8828,
4.3406, 5.1353, 3.8468, 4.233, 5.8468, 6.1872, 6.1245, 7.6262,
8.6887, 7.7549, 6.9805, 4.3217, 3.0347, 2.4026, 1.9317, 1.7305,
1.665, 1.5655, 1.3758, 1.5472, 1.7839, 1.951, 1.864, 1.6638,
1.5624, 1.4922, 0.9406, 0.84512, 0.48423, 0.3919, 0.30773, 0.29264,
0.19015, 0.13312, 0.25226, 0.29403, 0.23901, 0.000213074755156413,
5.96565965097398e-05, 0.086874, 0.000926808687858284, 0.000904641782399267,
0.000513042259030044, 0.40736, 4.53928073402494e-05, 0.000765719624469057,
0.000717419263673946)
I would like to fit a curve to this data, using mixtures of one to five Gaussians. In Matlab, I could do the following:
fits{1} = fit(1:length(x),x,fittype('gauss1'));
fits{2} = fit(1:length(x),x,fittype('gauss2'));
fits{3} = fit(1:length(x),x,fittype('gauss3'));
... and so on.
In R, I am having difficulty identifying a similar method.
dat <- data.frame(time = 1:length(x), x = x)
fits[[1]] <- Mclust(dat, G = 1)
fits[[2]] <- Mclust(dat, G = 2)
fits[[3]] <- Mclust(dat, G = 3)
... but this does not really seem to be doing quite the same thing. For example, I am not sure how to calculate the R^2 between the fit curve and the original data using the Mclust solution.
Is there a simpler alternative in base R to fitting a curve using a mixture of Gaussians?

Answer 1

Function
With the code given below, and with a bit of luck in finding good initial parameters, you should be able to curve-fit Gaussian's to your data.
In the function fit_gauss, aim is to y ~ fit_gauss(x) and the number of Gaussians to use is determined by the length of the initial values for parameters: a, b, d all of which should be equal length
I have demonstrated curve-fitting of OP's data up to three Gaussian's.
Specifying Initial Values
This it pretty much most work I have done with nls (thanks to OP for that). So, I am not quite sure what is the best method select the initial values. Naturally, they depend on height's of peaks (a), mean and standard deviation of x around them (b and d).
One option would be for given number of Gaussian's, try with a number of starting values, and find the one that has best fit based on residual standard error fit$sigma.
I fiddled a bit to find initial parameters, but I dare say the parameters and
the plot with three Gaussian model looks solid.
Fitting one, two and thee Gaussian's to Example data
ind <- 1 : length(x)
# plot original data
plot(ind, x, pch = 21, bg = "blue")
# Gaussian fit
fit_gauss <- function(y, x, a, b, d) {
p_model <- function(x, a, b, d) {
rowSums(sapply(1:length(a),
function(i) a[i] * exp(-((x - b[i])/d[i])^2)))
}
fit <- nls(y ~ p_model(x, a, b, d),
start = list(a=a, b = b, d = d),
trace = FALSE,
control = list(warnOnly = TRUE, minFactor = 1/2048))
fit
}
Single Gaussian
g1 <- fit_gauss(y = x, x = ind, a=1, b = mean(ind), d = sd(ind))
lines(ind, predict(g1), lwd = 2, col = "green")
Two Gaussian's
g2 <- fit_gauss(y = x, x = ind, a = c(coef(g1)[1], 1),
b = c(coef(g1)[2], 30),
d = c(coef(g1)[1], 2))
lines(ind, predict(g2), lwd = 2, col = "red")
Three Gaussian's
g3 <- fit_gauss(y = x, x = ind, a=c(5, 4, 4),
b = c(12, 17, 11), d = c(13, 2, 2))
lines(ind, predict(g3), lwd = 2, col = "black")
Summery of fit with three Gaussian
summary(g3)
# Formula: x ~ p_model(ind, a, b, d)
#
# Parameters:
# Estimate Std. Error t value Pr(>|t|)
# a1 5.9307 0.5588 10.613 5.93e-14 ***
# a2 3.5689 0.7098 5.028 8.00e-06 ***
# a3 -2.2066 0.8901 -2.479 0.016894 *
# b1 12.9545 0.5289 24.495 < 2e-16 ***
# b2 17.4709 0.2708 64.516 < 2e-16 ***
# b3 11.3839 0.3116 36.538 < 2e-16 ***
# d1 11.4351 0.8568 13.347 < 2e-16 ***
# d2 1.8893 0.4897 3.858 0.000355 ***
# d3 1.0848 0.6309 1.719 0.092285 .
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.7476 on 46 degrees of freedom
#
# Number of iterations to convergence: 34
# Achieved convergence tolerance: 8.116e-06