| periphrasticDo | R Documentation |
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.
data(periphrasticDo)
A data frame with 44 observations on the following 5 variables.
begina numeric vector with beginnings of the time periods used by Ellegard.
enda numeric vector with ends of these time periods.
typea factor for sentence type, with levels
affdecl (affirmative declarative),
affquest (affirmative question),
negdecl (negative declarative) and
negquest (negative question).
doa numeric vector with the count of sentences with do.
othera numeric vector with the count of sentences without do.
Ellegard, A. (1953) The auxiliary do: The establishment and regulation of its use in English, Stockholm: Almquist & Wiksell.
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.
## 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)
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