languageR (version 1.5.0)

periphrasticDo: The development of periphrastic do in English

Description

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.

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))

# }