plot.etm: Plot method for an etm object

In etm: Empirical Transition Matrix

Description

Plot method for an object of class 'etm'. It draws the estimated transition probabilities in a basic scatterplot.

Usage

## S3 method for class 'etm'
plot(x, tr.choice, xlab = "Time",
     ylab = "Transition Probability", col = 1, lty, xlim, ylim,
     conf.int = FALSE, level = 0.95, ci.fun = "linear",
     ci.col = col, ci.lty = 3,
     legend = TRUE, legend.pos, curvlab, legend.bty = "n", ...)

Arguments

x	An object of class 'etm'
tr.choice	character vector of the form 'c("from to","from to")' specifying which transitions should be plotted. Default, all the transition probabilities are plotted
xlab	x-axis label. Default is "Time"
ylab	y-axis label. Default is "Transition Probability"
col	Vector of colour. Default is black
lty	Vector of line type. Default is 1:number of transitions
xlim	Limits of x-axis for the plot
ylim	Limits of y-axis for the plot
conf.int	Logical. Whether to display pointwise confidence intervals. Default is FALSE.
level	Level of the conficence intervals. Default is 0.95.
ci.fun	Transformation applied to the confidence intervals. It could be different for all transition probabilities, though if length(ci.fun) != number of transitions, only ci.fun[1] will be used. Possible choices are "linear", "log", "log-log" and "cloglog". Default is "linear".
ci.col	Colour of the confidence intervals. Default is col.
ci.lty	Line type of the confidence intervals. Default is 3.
legend	A logical specifying if a legend should be added
legend.pos	A vector giving the legend's position. See legend for further details
curvlab	A character or expression vector to appear in the legend. Default is the name of the transitions
legend.bty	Box type for the legend
...	Further arguments for plot

Details

By default, if the argument strata was used for creating the etm object, the first transition probability for all strata will be plotted. If there is no strata, all transition probabilities are plotted by default.

In any case, a legend with be created by the labels are likely to be ugly. Please use the curvlab argument to control the test or use legend = FALSE and build your own legend.

Value

No value returned

Author(s)

Arthur Allignol, [e-mail]

See Also

plot.default, legend, etm

Examples

data(sir.cont)

# Modification for patients entering and leaving a state
# at the same date
sir.cont <- sir.cont[order(sir.cont$id, sir.cont$time), ]
for (i in 2:nrow(sir.cont)) {
  if (sir.cont$id[i]==sir.cont$id[i-1]) {
    if (sir.cont$time[i]==sir.cont$time[i-1]) {
      sir.cont$time[i-1] <- sir.cont$time[i-1] - 0.5
    }
  }
}

tra <- matrix(ncol=3,nrow=3,FALSE)
tra[1, 2:3] <- TRUE
tra[2, c(1, 3)] <- TRUE

my.etm <- etm(sir.cont,c("0","1","2"),tra,"cens", s = 0)

plot(my.etm, tr.choice = c("0 0"))

Example output

etm documentation built on Sept. 8, 2020, 5:06 p.m.