The development of periphrastic do in English: Ellegard's counts
for the use of do across four sentence types in 11 consecutive
time periods between 1390 and 1710.

Usage

data(periphrasticDo)

Format

A data frame with 44 observations on the following 5 variables.

begin

a numeric vector with beginnings of the time periods
used by Ellegard.

end

a numeric vector with ends of these time periods.

type

a factor for sentence type, with levels
affdecl (affirmative declarative),
affquest (affirmative question),
negdecl (negative declarative) and
negquest (negative question).

do

a numeric vector with the count of sentences with
do.

other

a numeric vector with the count of sentences without
do.

Source

Ellegard, A. (1953) The auxiliary do: The establishment and regulation
of its use in English, Stockholm: Almquist & Wiksell.

References

Vulanovic, R. and Baayen, R. H. (2006) Fitting the development of periphrastic
do in all sentence types, in Grzybek, P. and Koehler, R. (eds.), Festschrift
fuer Gabriel Altmann, Berlin: Walter de Gruyter, p. 679-688.

Examples

## Not run:
data(periphrasticDo)
# add midpoints of time periods
periphrasticDo$year = periphrasticDo$begin +
(periphrasticDo$end-periphrasticDo$begin)/2
# and ad an indicator variable distinguishing the first three time periods
# from the others
periphrasticDo$Indicator = rep(c(rep(0, 3), rep(1, 8)), 4)
# fit a logistic regression model
periphrasticDo.glm = glm(cbind(do, other) ~
(year + I(year^2) + I(year^3)) * type + Indicator * type +
Indicator * year, data = periphrasticDo, family = "binomial")
anova(periphrasticDo.glm, test = "F")
# visualization of data and model predictions
periphrasticDo$predict = predict(periphrasticDo.glm, type = "response")
par(mfrow=c(2, 2))
for (i in 1:nlevels(periphrasticDo$type)) {
subset = periphrasticDo[periphrasticDo$type ==
levels(periphrasticDo$type)[i], ]
plot(subset$year,
subset$do/(subset$do + subset$other),
type = "p", ylab = "proportion", xlab = "year",
ylim = c(0, 1), xlim = c(1400, 1700))
mtext(levels(periphrasticDo$type)[i], line = 2)
lines(subset$year, subset$predict, lty = 1)
}
par(mfrow=c(1, 1))
## End(Not run)