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This is a very similar question to Plotting GAMs on Response Scale with Multiple Smooth and Linear Terms, but I am specifically interested in the answer from @gavinsimpson.

I am using the code provided in the answer to the above question to plot individual smooths on the scale of the response but the plots it produces seem "off" and I can't figure out why.

Model:

model <- gam(count ~ 
                   s(x1, k = 40) + 
                   s(x2, k = 40) + 
                   s(x3, k = 40) +
                   s(x4, k = 40) +
                   s(x5, k = 40) +
                   s(x6, bs = "re") + 
                   s(Latitude, Longitude, bs = "sos", k = 230),
                 gamma = 1.4, family = nb(), method = "REML", 
                 data = dat)

All predictors are scaled, and I am only interested in plotting the individual smooths of x1, x2, x3, x4, the rest are "nuisance" variables.

Plot of smooth x1:

enter image description here

Two issues:

  1. Fitted trend is much higher than expected,

  2. Credible interval has "jagged" edges.

Both issues are present in all plots of the individual smooths I am interested in.

Code used for above plot:

new_data <- with(dat, expand.grid(x1 = seq(min(x1), max(x1), 
                length = 200),
                    x2 = median(x2),
                    x3 = median(x3),
                    x4 = median(x4),
                    x5 = median(x5),
                    x6 = x6,
                    Latitude = median(Latitude),
                    Longitude = median(Longitude)))

ilink <- family(model)$linkinv
pred  <- predict(model, new_data, newdata.guaranteed=TRUE, 
           type = "link", se.fit = TRUE, unconditional = T)
pred  <- cbind(pred, new_data)
pred  <- transform(pred, lwr_ci = ilink(fit - (2 * se.fit)),
                         upr_ci = ilink(fit + (2 * se.fit)),
                         fitted = ilink(fit))

ggplot(pred, aes(x = x1 , y = fitted)) +
  geom_point(dat, aes(x = x1, y = count)) + 
  geom_ribbon(aes(ymin = lwr_ci, ymax = upr_ci), alpha = 0.2) +
  geom_line(colour = "red", lwd = 1.1) +
  theme_classic() + 
  xlab("x1") + 
  ylab("Count") 

Ps First stack post so please let me know if you require more details or the format of the post is wrong.

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  • $\begingroup$ what about using plot(model) ? $\endgroup$
    – rep_ho
    Sep 29, 2021 at 15:14
  • $\begingroup$ What does "Fitted trend is much higher than expected" mean here? On what basis do you claim this? $\endgroup$
    – Roland
    Oct 1, 2021 at 6:39
  • $\begingroup$ From visual inspection and other models of a similar sort using the same predictors. Furthermore, when I plot using base R commands the fitted trend is more where you would expect it to be given the data. $\endgroup$
    – Jam
    Oct 1, 2021 at 17:31
  • $\begingroup$ What do you inspect visually? The plot in your question? You set four predictors to their median values. You should inspect the distribution of these values in your data. E.g., one of them could be mostly in a narrow range of values with a few more extreme values. I think potting smooths on the predictor scale like this shouldn't be done if your model includes more than one smoother. $\endgroup$
    – Roland
    Oct 4, 2021 at 7:01
  • 1
    $\begingroup$ The "jagged edges" probably result from the random effects smoother. You should predict for the population mean (but consider carefully what you confidence interval is conditioned on). $\endgroup$
    – Roland
    Oct 4, 2021 at 7:04

1 Answer 1

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As @Roland surmised, the jagged edges are due to you predicting multiple values of the response at each value of x1. You get multiple predictions because you plugged in all the values of x6 when you created new_data. That you didn't get a spaghetti plot is likely down to the ordering of the values in the expand.grid call; if x1 were your random effect variable then the plot could have been a visual mess and highlighted the problem more easily.

When predicting from models like this you typically need to provide a data frame of new values for all covariates used to fit the model. You can provide a single value for the covariates you are not interested in, in which case the predicted values are conditioned on those values. You could do this for the random effect also by selecting a single level of the random effect variable to predict for:

new_data <- with(dat,
                 expand.grid(x1 = seq(min(x1), max(x1), length = 200),
                             x2 = median(x2),
                             x3 = median(x3),
                             x4 = median(x4),
                             x5 = median(x5),
                             x6 = factor(x6[1], levels = levels(x6)), # <- here
                             Latitude = median(Latitude),
                             Longitude = median(Longitude)))

In the example I'm choosing a single level from x6 but making sure that new_data$x6 is still a factor with the right levels.

This would condition the predictions on that subject, which may not be what you want.

An alternative is to exclude certain terms from the predictions. Basically this is like setting or forcing the effect of $f(\mathsf{Lon}, \mathsf{Lat}) = 0$, which given how the smooths are set up means the average spatial effect. In terms of the random effects, because these are mean 0 i.i.d. Gaussian random effects, setting the effect of the random effect smooth to 0 is the same as generating population level predictions.

Note: The smooths are centred about the model constant term, so if you have factor terms, then this will mean the predictions are conditioned on the reference level(s) for that/those factor(s); changing the contrasts for those factors will change the interpretation of the model constant term so you should bear that in mind as you might prefer a different set of contrasts.

Regardless, we can exclude those terms that we might consider superfluous or nuisance using the terms or exclude arguments to predict.gam, which have the following description:

  • terms: if type=="terms" or type="iterms" then only results for the terms (smooth or parametric) named in this array will be returned. Otherwise any smooth terms not named in this array will be set to zero. If NULL then all terms are included.

  • exclude: if type=="terms" or type="iterms" then terms (smooth or parametric) named in this array will not be returned. Otherwise any smooth terms named in this array will be set to zero. If NULL then no terms are excluded. Note that this is the term names as it appears in the model summary, see example. You can avoid providing the covariates for the excluded terms by setting newdata.guaranteed=TRUE, which will avoid all checks on newdata.

(emphasis added) In your case it is easier perhaps to exclude the terms you want excluding.

Note the bit in italics above. You have to name the smooths in this vector you pass to terms or exclude exactly as {mgcv} knows them. So read them off the summary() output if you are unsure.

If you want to exclude the effects of s(x5), s(x6), and s(Latitude, Longitude) from your predictions then you would do:

excl <- c("s(x5)", "s(x6)", "s(Latitude,Longitude)")
pred  <- predict(model, new_data, exclude = excl,
                 type = "link", se.fit = TRUE, unconditional = TRUE)

Note how I removed spaces in the label for the spatial smooth.

You probably don't want to use newdata.guaranteed=TRUE here unless you are very sure you have everything perfectly correct in the definition of your new_data; it's very easy to mess up factor coding etc.

As a double check, you should confirm that your new_data has as many rows as you planned to plot. Here you wanted to show how the predicted value of your response changed over the range of x1 and generated 200 values of x1 at which you wanted predictions. Hence your new_data should have 200 rows: if it doesn't then you messed something up and need to check how you are creating new_data and invariably you have an issue in the expand.grid call.

Note also here that you will get different predictions if you predict at the median latitude and longitude than if you exclude those effects with exclude as sum to zero constraint imposed on each smooth will centre the smooths at the mean, and hence the 0 represents the mean spatial effect, not the effect on Y at the median location. I may be glossing over some minor implementation details there, but just bear that in mind when choosing values to predict at if you are conditioning on them as you did in your example.

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  • $\begingroup$ Apologies for the delay this was right before my masters submission, I have accepted your answer which was very useful and gave me much to consider. Thank you. $\endgroup$
    – Jam
    Mar 4, 2022 at 12:29

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