我正在尝试在一个绘图中的相同比例轴上绘制3个不同的绘图。该图进展顺利,但y轴刻度数彼此重叠。 这是我的情节。
h1<-hazard.plot.w2p(beta=beta.spreda,eta=eta.spreda,time=exa1.dat$time,line.colour="orange")
h2<-hazard.plot.w2p(beta=1.007629,eta=32.56836,time=exa1.dat$time,line.colour="red")
h3<-hazard.plot.w2p(beta=1.104483,eta=36.53923,time=exa1.dat$time,line.colour="green")
用于运行此代码的函数:
hazard.plot.w2p <- function(beta, eta, time, line.colour, nincr = 500) {
max.time <- max(time, na.rm = F)
t <- seq(0, max.time, length.out = nincr)
r <- numeric(length(t))
for (i in 1:length(t)) {
r[i] <- failure.rate.w2p(beta, eta, t[i])
}
plot(t, r, type = 'l', bty = 'l',
col = line.colour, lwd = 2,
main = "", xlab = "Time",
ylab = "Failure rate",
las = 1, adj = 0.5,
cex.axis = 0.85, cex.lab = 1.2)
par(new=TRUE)
}
样本数据集:
[fail time
a 4.55
a 4.65
a 5.21
b 3.21
a 1.21
a 5.65
a 7.12][1]
这是我得到的输出:
答案 0 :(得分:0)
这是一个测试,尽管我不知道它是否有效(缺少一些函数/变量):
hazard.plot.w2p <- function(beta, eta, time, line.colour, nincr = 500,
add = FALSE) {
max.time <- max(time, na.rm = F)
t <- seq(0, max.time, length.out = nincr)
r <- failure.rate.w2p(beta, eta, t)
if (!add) {
plot(NA, type = 'n', bty = 'l', xlim=range(t), ylim=range(r),
main = "", xlab = "Time", ylab = "Failure rate",
las = 1, adj = 0.5, cex.axis = 0.85, cex.lab = 1.2)
}
lines(t, r, col = line.colour, lwd = 2)
}
failure.rate.w2p <- function(beta,eta,time) (beta/eta) * (time/eta)^(beta-1)
h1<-hazard.plot.w2p(beta=1.002,eta=30,time=exa1.dat$time,line.colour="orange")
h2<-hazard.plot.w2p(beta=1.007629,eta=32.56836,time=exa1.dat$time,line.colour="red",add=T)
h3<-hazard.plot.w2p(beta=1.104483,eta=36.53923,time=exa1.dat$time,line.colour="green",add=T)