# Generalized Linear Models and Mixed-Effects in Agriculture

July 10, 2017
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After publishing my previous post, I realized that it was way too long and so I decided to split it in 2-3 parts. If you think something is missing in the explanation here it may be related to the fact that this was originally part of the previous post (http://r-video-tutorial.blogspot.co.uk/2017/06/linear-models-anova-glms-and-mixed.html), so please look there first (otherwise please post your question in the comment section and I will try to answer).

## Dealing with non-normal data – Generalized Linear Models

As you remember, when we first introduced the simple linear model () we defined a set of assumptions that need to be met to apply this model. We already talked about methods to deal with deviations from the assumption of independence, equality of variances and balanced designs and the fact that, particularly if our dataset is large, we may reach robust results even if our data are not perfectly normal. However, there are datasets for which the target variable has a completely different distribution from the normal, this means that also the error terms will not be normally distributed. In these cases we need to change our modelling method and employ generalized linear models. Common scenarios where this model should be considered are for example researchers where the variable of interest is binary, for example presence or absence of a species, or where we are interested in modelling counts, for example the number of insects present in a particular location. In these cases, where the target variable is not continuous but rather discrete or categorical, the assumption of normality is usually not met. In this section we will focus on the two scenarios mentioned above, but GLM can be used to deal with data distributed in many different ways, and we will introduce how to deal with more general cases.

### Count Data

Data of this type, i.e. counts or rates, are characterized by the fact that their lower bound is always zero. This does not fit well with a normal linear model, where the regression line may well estimate negative values. For this type of variable we can employ a Poisson Regression, which fits the following model:
As you can see the equation is very similar to the standard linear model, the difference is that to insure that all Y are positive (since we cannot have negative values for count data) we are estimating the log of Y. This is the link function, meaning the transformation of y that we need to make to insure the linearity of the response. For the model we are going to look at below, which are probably the most common, the link function is implicitly called, meaning that glm call the right function for us and we do not have to specify it explicitly. However, we can do that if needed.

From this equation you may think that instead of using glm we could log transform y and run a normal lm. The problem with that would be that lm is assuming an error term with constant variance (as we saw with the plot fitted versus residuals), but for this model the variance is not constant so that assumption would be violated. That is why we need to use glm.

In R fitting this model is very easy. For this example we are going to use another dataset available in the package agridat, named beall.webworms, which represents counts of webworms in a beet field, with insecticide treatments:
`` > dat = beall.webworms   > str(dat)   'data.frame':  1300 obs. of 7 variables:    \$ row : int 1 2 3 4 5 6 7 8 9 10 ...    \$ col : int 1 1 1 1 1 1 1 1 1 1 ...    \$ y  : int 1 0 1 3 6 0 2 2 1 3 ...    \$ block: Factor w/ 13 levels "B1","B10","B11",..: 1 1 1 1 1 6 6 6 6 6 ...    \$ trt : Factor w/ 4 levels "T1","T2","T3",..: 1 1 1 1 1 1 1 1 1 1 ...    \$ spray: Factor w/ 2 levels "N","Y": 1 1 1 1 1 1 1 1 1 1 ...    \$ lead : Factor w/ 2 levels "N","Y": 1 1 1 1 1 1 1 1 1 1 ...     ``

We can check the distribution of our data with the function hist:
`` hist(dat\$y, main="Histogram of Worm Count", xlab="Number of Worms")  ``

We are going to fit a simple model first to see how to interpret its results, and then compare it with a more complex model:
`` pois.mod = glm(y ~ trt, data=dat, family=c("poisson"))  ``

As you can see the model features a new option called family, here you specify the distribution of the error term, in this case a poisson distribution. We should also specify a log link function as we saw before, but this is the default setting so there is no need to include it.

`` > summary(pois.mod)      Call:   glm(formula = y ~ trt, family = c("poisson"), data = dat)      Deviance Residuals:      Min    1Q  Median    3Q   Max    -1.6733 -1.0046 -0.9081  0.6141  4.2771       Coefficients:         Estimate Std. Error z value Pr(>|z|)     (Intercept) 0.33647  0.04688  7.177 7.12e-13 ***   trtT2    -1.02043  0.09108 -11.204 < 2e-16 ***   trtT3    -0.49628  0.07621 -6.512 7.41e-11 ***   trtT4    -1.22246  0.09829 -12.438 < 2e-16 ***   ---   Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1      (Dispersion parameter for poisson family taken to be 1)        Null deviance: 1955.9 on 1299 degrees of freedom   Residual deviance: 1720.4 on 1296 degrees of freedom   AIC: 3125.5      Number of Fisher Scoring iterations: 6     ``

The first valuable information is related to the residuals of the model, which should be symmetrical as for any normal linear model. From this output we can see that minimum and maximum, as well as the first and third quartiles, are similar, so this assumption is confirmed. Then we can see that the variable trt (i.e. treatment factor) is highly significant for the model, with very low p-values. As for lm, glm compare the variables using non-parametric alternatives to the t-test, and that is why it is computing z-values.

We can use the function plot to obtain more info about how the model fits our data:

`` par(mfrow=c(2,2))   plot(pois.mod)  ``

This creates the following plot, where the four outputs are included in the same image:

This plots tell a lot about the goodness of fit of the model. The first image in top left corner is the same we created for lm (i.e. residuals versus fitted values). This again does not show any trend, just a general underestimation. Then we have the normal QQ plot, where we see that the residuals are not normal, which violates one of the assumptions of the model. Even though we are talking about non-linear error term, we are “linearizing” the model with the link function and by specifying a different family for the error term. Therefore, we still need to obtain normal residuals.

The effects of the treatments are all negative and referred to the first level T1, meaning for example that a change from T1 to T2 will decrease the count by 1.02. We can check this effect by estimating changes between T1 and T2 with the function predict, and the option newdata:

`` > predict(pois.mod, newdata=data.frame(trt=c("T1","T2")))        1     2     0.3364722 -0.6839588     ``

Another important piece of information are the Null and Residuals deviances, which allow us to compute the probability that this model is better than the Null hypothesis, which states that a constant (with no variables) model would be better.
We can compute the p-value of the model with the following line:
`` > 1-pchisq(deviance(pois.mod), df.residual(pois.mod))   [1] 1.709743e-14     ``

This p-value is very low, meaning that this model fits the data well. However, it may not be the best possible model, and we can use the AIC parameter to compare it to other models. For example, we could include more variables:
`` pois.mod2 = glm(y ~ block + spray*lead, data=dat, family=c("poisson"))  ``

How does this new model compare with the previous?
`` > AIC(pois.mod, pois.mod2)        df   AIC   pois.mod  4 3125.478   pois.mod2 16 3027.438     ``

As you can see the second model has a lower AIC, meaning that fits the data better than the first.
One of the assumptions of the Poisson distribution is that its mean and variance have the same value. We can check by simply comparing mean and variance of our data:
`` > mean(dat\$y)   [1] 0.7923077   > var(dat\$y)   [1] 1.290164     ``

In cases such as this when the variance is larger than the mean (in this case we talk about overdispersed count data) we should employ different methods, for example a quasipoisson distribution:
`` pois.mod3 = glm(y ~ trt, data=dat, family=c("quasipoisson"))  ``

The summary function provides us with the dispersion parameter, which for a Poisson distribution should be 1:
`` > summary(pois.mod3)      Call:   glm(formula = y ~ trt, family = c("quasipoisson"), data = dat)      Deviance Residuals:      Min    1Q  Median    3Q   Max    -1.6733 -1.0046 -0.9081  0.6141  4.2771       Coefficients:         Estimate Std. Error t value Pr(>|t|)     (Intercept) 0.33647  0.05457  6.166 9.32e-10 ***   trtT2    -1.02043  0.10601 -9.626 < 2e-16 ***   trtT3    -0.49628  0.08870 -5.595 2.69e-08 ***   trtT4    -1.22246  0.11440 -10.686 < 2e-16 ***   ---   Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1      (Dispersion parameter for quasipoisson family taken to be 1.35472)        Null deviance: 1955.9 on 1299 degrees of freedom   Residual deviance: 1720.4 on 1296 degrees of freedom   AIC: NA      Number of Fisher Scoring iterations: 6     ``

Since the dispersion parameter is 1.35, we can conclude that our data are not terrible dispersed, so maybe a Poisson regression would still be appropriate for this dataset.

Another way of directly comparing the two models is with the analysis of deviance, which can be performed with the function anova:

`` > anova(pois.mod, pois.mod2, test="Chisq")   Analysis of Deviance Table   Model 1: y ~ trt   Model 2: y ~ block + spray * lead    Resid. Df Resid. Dev Df Deviance Pr(>Chi)     1   1296   1720.4                2   1284   1598.4 12  122.04 < 2.2e-16 ***   ---   Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  ``

This test compares the residual deviance of the two models to see whether they are difference and calculates a p-values. In this case the p-value is highly significant, meaning that the models are different. Since we already compared the AIC, we can conclude that pois.mod2 is significantly (low p-value) better (lower AIC) than pois.mod.

However, there are cases where the data are very overdispersed. In those cases, when we see that the distribution has lots of peaks we need to employ the negative binomial regression, with the function glm.nb available in the package MASS:
`` library(MASS)      NB.mod1 = glm.nb(y ~ trt, data=dat)  ``

NOTE:
For GLM it is possible to also compute pseudo R-Squared to ease the interpretation of their accuracy. This can be done with the function pR2 from the package pscl. Please read below (Logistic Regression section) for an example on the use of this function.

### Logistic Regression

Another popular for of regression that can be tackled with GLM is the logistic regression, where the variable of interest is binary (0 and 1, presence and absence or any other binary outcome). In this case the regression model takes the following equation:
Again, the equation is identical to the standard linear model, but what we are computing from this model is the log of the probability that one of the two outcomes will occur.
For this example we are going to use another dataset available in the package agridatcalled johnson.blight, where the binary variable of interest is the presence or absence of blight (either 0 or 1) in potatoes:
`` > dat = johnson.blight   > str(dat)   'data.frame':  25 obs. of 6 variables:    \$ year  : int 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 ...    \$ area  : int 0 0 0 0 50 810 120 40 0 0 ...    \$ blight : int 0 0 0 0 1 1 1 1 0 0 ...    \$ rain.am : int 8 9 9 6 16 10 12 10 11 8 ...    \$ rain.ja : int 1 4 6 1 6 7 12 4 10 9 ...    \$ precip.m: num 5.84 6.86 47.29 8.89 7.37 ...     ``

In R fitting this model is very easy. In this case we are trying to see if the presence of blight is related to the number of rainy days in April and May (column rain.am):
`` mod9 = glm(blight ~ rain.am, data=dat, family=binomial)  ``

we are now using the binomial distribution for a logistic regression. To check the model we can rely again on summary:
`` > summary(mod9)      Call:   glm(formula = blight ~ rain.am, family = binomial, data = dat)      Deviance Residuals:      Min    1Q  Median    3Q   Max    -1.9395 -0.6605 -0.3517  1.0228  1.6048       Coefficients:         Estimate Std. Error z value Pr(>|z|)    (Intercept) -4.9854   2.0720 -2.406  0.0161 *   rain.am    0.4467   0.1860  2.402  0.0163 *   ---   Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1      (Dispersion parameter for binomial family taken to be 1)        Null deviance: 34.617 on 24 degrees of freedom   Residual deviance: 24.782 on 23 degrees of freedom   AIC: 28.782      Number of Fisher Scoring iterations: 5     ``

This table is very similar to the one created for count data, so a lot of the discussion above can be used here. The main difference is in the way we can interpret the coefficients, because we need to remember that here we are calculating the logit function of the probability, so 0.4467 (coefficient for rain.am) is not the actual probability associated with an increase in rain. However, what we can say by just looking at the coefficients is that rain has a positive effect on blight, meaning that more rain increases the chances of finding blight in potatoes.
To estimate probabilities we need to use the function predict:
`` > predict(mod9, type="response")        1     2     3     4     5     6     7    0.19598032 0.27590141 0.27590141 0.09070472 0.89680283 0.37328295 0.59273722         8     9     10     11     12     13     14    0.37328295 0.48214935 0.19598032 0.69466455 0.19598032 0.84754431 0.27590141        15     16     17     18     19     20     21    0.93143346 0.05998586 0.19598032 0.05998586 0.84754431 0.59273722 0.59273722        22     23     24     25    0.48214935 0.59273722 0.98109229 0.89680283     ``

This calculates the probability associated with the values of rain in the dataset. To know the probability associated with new values of rain we can again use predict with the option newdata:
`` >predict(mod9,newdata=data.frame(rain.am=c(15)),type="response")        1    0.8475443     ``

This tells us that when rain is equal to 15 we have 84% chances of finding blight (i.e. chances of finding 1) in potatoes.

We could use the same method to compute probabilities for a series of value of rain to see what is the threshold of rain that increases the probability of blight above 50%:

`` prob.NEW = predict(mod9,newdata=data.frame(rain.am=1:30),type="response")    plot(1:30, prob.NEW, xlab="Rain", ylab="Probability of Blight")   abline(h=0.5)  ``

As you can see we are using once again the function predict, but in this case we are estimating the probabilities for increasing values of rain. Then we are plotting the results:

From this plot it is clear that we reach a 50% probability at around 12 rainy days between April and May.

To assess the accuracy of the model we can use two approaches, the first is based on the deviances listed in the summary. The Residual deviance compares this model with the one that fits the data perfectly. So if we calculate the following p-value (using the deviance and df in the summary table for residuals):
`` > 1-pchisq(24.782, 23)   [1] 0.3616226     ``

We see that because it is higher than 0.05 we cannot reject that this model fits the data as well as the perfect model, therefore our model seems to be good.
We can repeat the same procedure for the Null hypothesis, which again tests whether this model fits the data well:
`` > 1-pchisq(34.617, 24)   [1] 0.07428544     ``

Since this is again not significant it suggests (contrary to what we obtained before) that this model is not very good.
An additional and probably easier to understand way to assess the accuracy of a logistic model is calculating the pseudo R2, which can be done by installing the package pscl:
`` install.packages("pscl")   library(pscl)     ``

Now we can run the following function:
`` > pR2(mod9)       llh   llhNull     G2  McFadden    r2ML    r2CU      -12.3910108 -17.3086742  9.8353268  0.2841155  0.3252500  0.4338984     ``

From this we can see that our model explains around 30-40% of the variation in blight, which is not particularly good. We can use this index to compare models, as we did for AIC.
Each of these R2 is computed in a different way and you can read the documentation to know more. In general, one of the most commonly reported is the McFadden, which however tends to be conservative, and the r2ML. In this paper you can find a complete overview/comparison of various pseudo R-Squared: http://www.glmj.org/archives/articles/Smith_v39n2.pdf

### Dealing with other distributions and transformation

As mentioned, GLM can be used for fitting linear models not only in the two scenarios we described above, but in any occasion where data do not comply with the normality assumption. For example, we can look at another dataset available in agridat, where the variable of interest is slightly non-normal:
`` > dat = hughes.grapes   > str(dat)   'data.frame':  270 obs. of 6 variables:    \$ block  : Factor w/ 3 levels "B1","B2","B3": 1 1 1 1 1 1 1 1 1 1 ...    \$ trt   : Factor w/ 6 levels "T1","T2","T3",..: 1 2 3 4 5 6 1 2 3 4 ...    \$ vine  : Factor w/ 3 levels "V1","V2","V3": 1 1 1 1 1 1 1 1 1 1 ...    \$ shoot  : Factor w/ 5 levels "S1","S2","S3",..: 1 1 1 1 1 1 2 2 2 2 ...    \$ diseased: int 1 2 0 0 3 0 7 0 1 0 ...    \$ total  : int 14 12 12 13 8 9 8 10 14 10 ...     ``

The variable total presents a skewness of 0.73, which means that probably with a transformation it should fit with a normal distribution. However, for the sake of the discussion we will assume it cannot be transformed. So now our problem is identify the best distribution for our data, to do so we can use the function descdist in the package fitdistrplus we already loaded:
`` descdist(dat\$total, discrete = FALSE)  ``

this returns the following plot:
Where we can see that our data (blue dot) are close to normal and maybe closer to a gamma distribution. So now we can further check this using another function from the same package:
`` plot(fitdist(dat\$total,distr="gamma"))  ``

which creates the following plot:
From this we can see that in fact our data seem to be close to a gamma distribution, so now we can proceed with modelling:
`` mod8 = glm(total ~ trt * vine, data=dat, family=Gamma(link=identity))  ``

in the option family we included the name of the distribution, plus a link function that is used if we want to transform our data (in this case the function identity is for leaving data not transformed).
This is what we do to model other types of data that do not fit with a normal distribution. Other possible families supported by GLM are:
binomial, gaussian, Gamma, inverse.gaussian, poisson, quasi, quasibinomial, quasipoisson
Other possible link functions (which availability depends on the family) are:
logit, probit, cauchit, cloglog, identity, log, sqrt, 1/mu^2, inverse.

### Generalized Linear Mixed Effects models

As linear model, linear mixed effects model need to comply with normality. If our data deviates too much we need to apply the generalized form, which is available in the package lme4:
`` install.packages("lme4")   library(lme4)  ``

For this example we will use again the dataset johnson.blight:
`` dat = johnson.blight  ``

Now we can fit a GLME model with random effects for area, and compare it with a model with only fixed effects:
`` mod10 = glm(blight ~ precip.m, data=dat, family="binomial")      mod11 = glmer(blight ~ precip.m + (1|area), data=dat, family="binomial")       > AIC(mod10, mod11)      df    AIC   mod10 2 37.698821   mod11 3 9.287692     ``

As you can see this new model reduces the AIC substantially.
The same function can be used for Poisson regression, but it does not work for quasipoisson overdispersed data. However, within lme4 there is the function glmer.nb for negative binomial mixed effect. The syntax is the same as glmer, except that in glmer.nb we do not need to include family.

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