Michael Gruenstaeudl - 3 months ago 22

R Question

The documentation for function

`anova.lm`

`anova`

`data(LifeCycleSavings)`

fit0 <- lm(sr ~ 1, data = LifeCycleSavings)

fit1 <- update(fit0, . ~ . + pop15)

fit2 <- update(fit1, . ~ . + pop75)

fit3 <- update(fit2, . ~ . + dpi)

fit4 <- update(fit3, . ~ . + ddpi)

anova(fit0, fit1, fit2, fit3, fit4, test = "F")

You could also use

`lapply`

`anova`

`fit_L = list(fit0, fit1, fit2, fit3, fit4)`

lapply(fit_L, anova)

Analogously, the documentation for function

`find.mle`

`anova`

`library(diversitree)`

#

pars <- c(0.1, 0.2, 0.03, 0.03, 0.01, 0.01)

phy <- tree.bisse(pars, max.t=60, x0=0)

lik <- make.bisse(phy, phy$tip.state)

fit <- find.mle(lik, pars)

lik.l <- constrain(lik, lambda0 ~ lambda1)

fit.l <- find.mle(lik.l, pars[-2])

anova(fit, equal.lambda=fit.l)

Here, however, I cannot use lapply to execute the function

`anova`

`fit_L = list(fit, fit.l)`

lapply(fit_L, anova)

# Error in anova.fit.mle(X[[i]], ...) : Need to specify more than one model

Can anyone think of a way to use

`lapply`

`diversitree`

To clarify my question: The underlying idea of my post is to make

`lapply`

`lapply`

`fit_L`

Answer

`lapply`

iterates over the lements of a list and applies a function to them. This is not what you want. You want to pass all list elements as arguments to a function, which is what `do.call`

does:

```
do.call(anova, c(fit_L, test = "F"))
```

If you look at your examples with `anova.lm`

, you see that the output is different if you use `lapply`

. From the documentation:

Specifying a single object gives a sequential analysis of variance table for that fit. [...] If more than one object is specified, the table has a row for the residual degrees of freedom and sum of squares for each model. For all but the first model, the change in degrees of freedom and sum of squares is also given. ...

`lapply`

passes single objects to `anova.lm`

. This doesn't work for your `mle`

fits because the corresponding `anova`

method only does model comparison.

Source (Stackoverflow)

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