John Chambers

Mentioned 2

John Chambers turns his attention to R, the enormously successful open-source system based on the S language. His book guides the reader through programming with R, beginning with simple interactive use and progressing by gradual stages, starting with simple functions. More advanced programming techniques can be added as needed, allowing users to grow into software contributors, benefiting their careers and the community. R packages provide a powerful mechanism for contributions to be organized and communicated. This is the only advanced programming book on R, written by the author of the S language from which R evolved.

Mentioned in questions and answers.

This is related to Looping over a Date object result in a numeric iterator

```
> dates <- as.Date(c("2013-01-01", "2013-01-02"))
> class(dates)
[1] "Date"
> for(d in dates) print(class(d))
[1] "numeric"
[1] "numeric"
```

I have two questions:

- What is the preferred way to iterate over a list of Date objects?
- I don't understand Joshua's answer (accepted answer from the question linked above), I'll quote it here: "So your
`Date`

vector is being coerced to`numeric`

because`Date`

objects aren't strictly vectors". So how is it determined that`Date`

should be coerced to`numeric`

?

There are two issues here. One is whether the input gets coerced from `Date`

to `numeric`

. The other is whether the output gets coerced to `numeric`

.

**Input**

For loops coerce `Date`

inputs to `numeric`

, because as @DWin and @JoshuaUlrich point out, `for`

loops take `vectors`

, and `Date`

s are technically not vectors.

```
> for(d in dates) print(class(d))
[1] "numeric"
[1] "numeric"
```

On the other hand, `lapply`

and its simplifier offspring `sapply`

have no such restrictions.

```
> sapply( dates, function(day) class(day) )
[1] "Date" "Date"
```

**Output**

However! The output of `class()`

above is a character. If you try actually returning a date object, `sapply`

is not what you want.

`lapply`

does not coerce to a vector, but `sapply`

does:

```
> lapply( dates, identity )
[[1]]
[1] "2013-01-01"
[[2]]
[1] "2013-01-02"
> sapply( dates, identity )
[1] 15706 15707
```

That's because `sapply`

's simplification function coerces output to a vector.

**Summary**

So: If you have a `Date`

object and want to return a non-`Date`

object, you can use `lapply`

or `sapply`

. If you have a non-`Date`

object, and want to return a `Date`

object, you can use a `for`

loop or `lapply`

. If you have a `Date`

object and want to return a `Date`

object, use `lapply`

.

**Resources for learning more**

If you want to dig deeper into vectors, you can start with John Cook's notes, continue with the R Inferno, and continue with SDA.

**Context of application**

I have a model with random slopes and intercepts. There are numerous levels of the random effects. The new data (to be predicted) may or may not have all of these levels.

To make this more concrete, I am working with music revenue at the album level (`title`

). Each album may come in multiple types `format2`

(CD, vinyl, e-audio, etc). I have measurements for revenue for each album at each type of album. The model is specified as:

```
lmer(physical~ format2+ (0+format2|title))
```

The problem is that future data may not have all the levels of either `title`

or `format2`

. For random intercepts, this is easily resolved with `predict(..., allow.new.levels= TRUE)`

. But it is problematic for the fixed effects and random slopes. I am therefore trying to write a function to do flexible predictions of `merMod`

objects, similar to `lme4::predict.merMod`

; but that will handle the differences between the training data and the prediction data. This is a question asked as much out of ignorance to the exact details of `lme4::predict.merMod`

as anything else.

**Description of problem**

The crux of the problem is getting the correct `model.matrix()`

with fixed and random effects to calculate both predictions and SE's. The S3 method for class `merMod`

returns **only the fixed effects**.

The base `stats::model.matrix()`

function has very limited documentation. Unfortunately, I do not own either Statistical Models in S or Software for Data Analysis, which appear to have the details behind these functions.

`model.matrix()`

is supposed to take a model formula and new data frame and produce a design matrix. But I'm getting an error. Any help you can provide would be much appreciated.

**Example Data**

```
dat1 <- structure(list(dt_scale = c(16, 16, 16, 16, 16, 16, 16, 16, 16,
16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16), title = c("Bahia",
"Jazz Moods: Brazilian Romance", "Quintessence", "Amadeus: The Complete Soundtrack Recording (Bicentennial Edition)",
"Live In Europe", "We'll Play The Blues For You", "The Complete Village Vanguard Recordings, 1961",
"The Isaac Hayes Movement", "Jazz Moods: Jazz At Week's End",
"Blue In Green: The Concert In Canada", "The English Patient - Original Motion Picture Soundtrack",
"The Unique Thelonious Monk", "Since We Met", "You're Gonna Hear From Me",
"The Colors Of Latin Jazz: Cubop!", "The Colors Of Latin Jazz: Samba!",
"Homecoming", "Consecration: The Final Recordings Part 2 - Live At Keystone Korner, September 1980", "More Creedence Gold", "The Stardust Session"), format2 = c("CD", "CD",
"CD", "CD", "CD", "CD", "CD", "SuperAudio", "SuperAudio", "CD", "E Audio", "CD",
"Vinyl", "CD", "E Audio", "CD", "CD", "CD", "CD", "CD"), mf_day = c(TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE), xmas = c(FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE), vday = c(FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE), yr_since_rel = c(16.9050969937038,
8.41815617876864, 9.2991404674865, 25.0870296783559, 39.1267038232812,
27.9156764326061, 9.11596751812513, 23.3052837112449, 14.3123922258974,
30.5208152866414, 5.83025071417496, 21.3090003877291, 7.75022155568392,
11.3601605287827, 0.849006673421519, 31.9918631305662, 13.8861905547041,
12.8342695062012, 29.6916671402534, 13.5912612705038), physical = c(1327.17849171096,
-110.2265302258, -795.37376268564, 355.06192702004, -1357.3492884345,
-1254.93442612023, -816.713683621225, 881.201935773452, -3092.02845691036,
-2268.6304275652, 907.347941142021, -699.130275178185, 377.867849132077,
-1047.50531157311, 1460.25978951805, 1376.84579069304, 3619.03629114089,
962.888173535704, 2514.77880599199, 2539.14958588771)), .Names = c("dt_scale",
"title", "format2", "mf_day", "xmas", "vday", "yr_since_rel",
"physical"), row.names = c(1L, 2L, 5L, 6L, 7L, 8L, 9L, 11L, 12L,
13L, 14L, 15L, 20L, 22L, 23L, 25L, 27L, 32L, 35L, 36L), class = "data.frame")
```

**formula**:

```
f1 <- as.formula(~1 + dt_scale + yr_since_rel + format2 + (0 + format2 + mf_day +
xmas + vday | title))
```

**execution / error**

```
library(lme4)
model.matrix(f1, data= dat1)
Error in 0 + format2 : non-numeric argument to binary operator
```

**Note**
I've also tried this with the `Orthodont`

data; but, I get a different error.

```
library(lme4)
data("Orthodont",package="MEMSS")
fm1 <- lmer(formula = distance ~ age*Sex + (1+age|Subject), data = Orthodont)
newdat <- expand.grid(
age=c(8,10,12,14)
, Sex=c("Male","Female")
, distance = 0
, Subject= c("F01", "F02")
)
f1 <- formula(fm1)[-2] # simpler code via Ben Bolker below
mm <- model.matrix(f1, newdat) # attempt to use model.matrix
Warning message
In Ops.factor(1 + age, Subject) : | not meaningful for factors
# use lme4:::mkNewReTrms as suggested in comments
mm <- lme4:::mkNewReTrms(f1, newdat)
Error in lme4:::mkNewReTrms(f1, newdat) : object 'ReTrms' not found
In addition: Warning message:
In Ops.factor(1 + age, Subject) : | not meaningful for factors
# check if different syntax would fix this
mm <- lme4::mkNewReTrms(f1, newdat)
Error: 'mkNewReTrms' is not an exported object from 'namespace:lme4'
mm <- mkNewReTrms(f1, newdat)
Error: could not find function "mkNewReTrms"
```

**Editted 8/12/15**: see changes on Github and GitHub Repo

**Editted, 10/15/2014:** This answer isn't yet perfect. There are still a couple of use-cases with errors (see comment chain below). But it works in most cases. I'll get around to finalizing it at some point.

I believe this function will solve the more important problem, accurate predictions for merMod objects. Dr Bolker, there are still some issues here (such as sparsity and efficiency); but I believe the method works:

```
data("Orthodont",package="MEMSS")
fm1 <- lmer(formula = distance ~ age*Sex + (1+age|Subject), data = Orthodont)
newdat <- expand.grid(
age=c(8,10,12,14)
, Sex=c("Male","Female")
, distance = 0
, Subject= c("F01", "F02")
)
predict.merMod2 <- function(object, newdat=NULL) {
# 01. get formula and build model matrix
# current problem--model matrix is not sparse, as would be ideal
f1 <- formula(object)[-2]
z.fe <- model.matrix(terms(object), newdat)
z.re <- t(lme4:::mkReTrms(findbars(f1), newdat)$Zt)
mm <- cbind(z.fe,
matrix(z.re, nrow= dim(z.re)[1], ncol= dim(z.re)[2],
dimnames= dimnames(z.re)))
# 02. extract random effect coefficients needed for the new data
# (a) - determine number of coef
len <- length(ranef(object))
re.grp.len <- vector(mode= "integer", length= len)
for (i in 1:len) { # for each random group
re.grp.len[i] <- dim(ranef(object)[[i]])[2] # number of columns (slope and intercept terms)
}
# (b) - create beta vector
fe.names <- unique(colnames(mm)[1:length(fixef(object)) - 1])
re.names <- unique(colnames(mm)[-c(1:length(fixef(object)) - 1)])
beta.re <- as.vector(rep(NA, length= sum(re.grp.len) * length(re.names)), mode= "numeric")
for (i in 1:len) {
re.beta <- ranef(object)[[i]][rownames(ranef(object)[[i]]) %in% re.names,]
ind.i <- sum(!is.na(beta.re)) + 1; ind.j <- length(as.vector(t(re.beta)))
beta.re[ind.i:ind.j] <- as.vector(t(re.beta))
}
beta <- c(fixef(object)[names(fixef(object)) %in% fe.names], beta.re)
# 03. execute prediction
return(mm %*% beta)
}
predict.merMod2(fm1, newdat)
```