Plotting output of GAM model - r

Question

Edit: following interactions in the responses below, I believe there may be some issues with the plot() or plot.gam() functions when dealing with gam outputs. See responses below.
I am running a non parametric regression model <- gam(y ~ x, bs = "cs", data = data).
My data looks like what follows, where x is in logs. I have 273 observations
y x
[1,] 0.010234756 10.87952
[2,] 0.009165001 10.98407
[3,] 0.001330975 11.26850
[4,] 0.008000957 10.97803
[5,] 0.008579472 10.94924
[6,] 0.009746714 11.01823
I would like to plot the output of the model, basically the fitted curve. When I do
# graph
plot(model)
or
ggplot(data = data, mapping = aes(x = x y = y)) +
geom_point(size = 0.5, alpha = 0.5) +
geom_smooth(method="gam", formula= y~s(x, bs = "cs") )
I get the desired output graphs (apologies for the original labels):
[
However, the two plotted curves are not exactly the same and I did not manage to find the parameters to tweak to remove the differences. Therefore I would like to plot the curve manually.
Here it's my current attempt.
model <- gam(y~ s(x), bs = "cs", data = data)
names(model)
# summary(model)
model_fit <- as.data.frame(cbind(model$y, model$fitted.values,
model$linear.predictors, data$x,
model$residuals))
names(model_fit) <- c("y", "y_fit", "linear_pred", "x", "res")
### here the plotting
ggplot(model_fit) +
geom_point(aes(x = x, y = y_fit), size = 0.5, alpha = 0.5) +
geom_line(aes(x = x, y = y_fit))
However I get the following warning
geom_path: Each group consists of only one observation. Do you need to adjust the group aesthetic?
and wrong output graph
I do not seem to be able to fix the last graph (it seems the error is in geom_point() ) and add the confidence intervals, nor to find where to tweak the first two to make them exactly the same.

Answer 1

The difference is likely due to you using different fitting algorithms. The default in gam() is (currently) method = "GCV.Cp" even through the recommended option is to use method = "REML". stat_smooth() uses method = "REML". GCV-based smoothness selection is known to undersmooth in some circumstances and this seems to be the case here with the REML solution being a much smoother curve.
If you change to method = "REML" in your gam() call, the differences should disappear.
That said, you really shouldn't be ripping things out of model objects like that - for a set off $residuals is not what you think it is - it's not useful in this context as those are the working residuals for PIRLS algorithm. Use the extractor functions like fitted(), residuals() etc.
The easiest way to plot your own version of that drawn by plot.gam() is to capture the object returned by plot.gam() and then use that object to draw what you need.
Via plot.gam()
df <- data_sim("eg1", seed = 2)
m <- gam(y ~ s(x2), data = df, method = "REML")
p_obj <- plot(m, residuals = TRUE)
p_obj <- p_obj[[1]] # just one smooth so select the first component
sm_df <- as.data.frame(p_obj[c("x", "se", "fit")])
data_df <- as.data.frame(p_obj[c("raw", "p.resid")])
## plot
ggplot(sm_df, aes(x = x, y = fit)) +
geom_rug(data = data_df, mapping = aes(x = raw, y = NULL),
sides = "b") +
geom_point(data = data_df, mapping = aes(x = raw, y = p.resid)) +
geom_ribbon(aes(ymin = fit - se, ymax = fit + se, y = NULL),
alpha = 0.3) +
geom_line() +
labs(x = p_obj$xlab, y = p_obj$ylab)
Which produces
Alternatively, you might look at my {gratia} package or the {mgcViz} package of Matteo Fasiolo as options that will do this all for you.
{gratia} example
For example with {gratia}
library('gratia')
draw(m, residuals = TRUE)
which produces

Answer 2

The solution provided by #Gavin Simpson here partially solves the issue, meaning that to make the two curves equal, one needs to add the method = "REML". The two curves then have the same slope.
However, for some reason, when plotting the output of a gam() model using either plot() or plot.gam(), the curve does not fit properly the original data as it should. The same happens by manually plotting the graph by extracting the elements from the object returned by plot.gam(). I am not sure why this happens. In my case, the fitted curve is shifted downwards, clearly "missing" the data points it is supposed to fit. Below the code and the corresponding output graph, the latter being the same you get in plot() or plot.gam() with the addition of the original data points to the plots.
plot(model_1)
# or plot.gam(model_1)
data.plot = as.data.frame(cbind(b[[1]]$x, b[[1]]$fit, b[[1]]$se))
ggplot(data=data.plot, mapping = aes(x= data.plot$V1, y= data.plot$V2)) +
geom_line(aes(x = V1, y = V2)) +
geom_line(aes(x= V1, y = V2 + V3 ), linetype="dashed") +
geom_line(aes(x= V1, y = V2 - V3 ), linetype ="dashed") +
geom_point(data= df_abs, aes(x= log(prd_l_1999), y=prd_gr), size = 0.5, alpha = 0.5)
Misplaced graphs
To note that the ggplot function makes the plot properly. Therefore, my ignorant guess is that this may be an issue with the plotting method.
Working solution
I am not able to prove that the issue is with the plotting functions, but it turns out that this is the same issue as in this question and the partial solution provided by the OP fixes the plotting while still using the gam() function. Below (his) code adapted to my case and the corresponding output graph. As you can see, the graph is plotted properly and the curve fits the data as it is supposed to do. I'd say this may corroborate my hypothesis even though I cannot prove it as I am not knowledgeable enough.
library(data.table)
model_1 <- gam(prd_gr ~ s(log(prd_l_1999)), bs = "cs", data = df_abs, method = "REML")
preds <- predict(model_1,se.fit=TRUE)
my_data <- data.frame(mu=preds$fit, low =(preds$fit - 1.96 * preds$se.fit), high = (preds$fit + 1.96 * preds$se.fit))
ggplot()+
geom_line(data = my_data, aes(x=log(df_abs$prd_l_1999), y=mu), size=1, col="blue")+
geom_smooth(data=my_data,aes(ymin = low, ymax = high, x=log(df_abs$prd_l_1999), y = mu), stat = "identity", col="green")