即使xpd = TRUE,我如何绘制垂直和水平线?

时间:2017-12-22 19:56:26

标签: r plot

这是一个简化的图表:

env <- data.frame(site = c('BLK','DUC','WHP','BLK','DUC','WHP','BLK','DUC','WHP'),
                  sal = c(5,6,3,2,4,5,6,8,4),
                  date = c(2013,2013,2013,2015,2015,2015,2017,2017,2017))
sitelist <- c('BLK','DUC','WHP')
par(mar=c(3,5,3,6), xpd = T)
plot(sal~date, data = env, type = 'n', ylim = c(0,10), ylab = 'Salinity',
     bty = 'n', xlab = '')
abline(v=2016, col = 'khaki', lwd = 20)
abline(mean(env$sal), 0, lty = 3)
for (ii in seq_along(sitelist)) {
  i <- sitelist[ii]; lines(sal[site==i] ~ date[site==i], data = env,
                            col = c(4,2,5)[ii],  lwd = 2,
                            lty = c(1,2,3)[ii]);
  points(sal[site==i] ~ date[site==i], data = env,
         pch = c(0,1,2)[ii], col = c(4,2,5)[ii])}
legend('topright', title = 'sites', inset=c(-0.2,0), lty = c(1,2,3), 
       col = c(4,2,5), lwd = 2, sitelist, 
       pch = c(0,1,2))

正如所写,这段代码产生了一个图,其中abline函数创建了超出图的边界的线,这要归功于xpd=T。但是,我不想设置xpd=F,因为我无法在边界之外绘制我的传奇。解决方案必须是使用xpd=F绘制边界外的图例的方法,或者是绘制在边界处停止的线的方法。理想情况下,解决方案将使用基本程序并且是相当标准的,因此我可以将其放入我的~20个图中的每个图中,而无需过多的自定义。

我尝试使用segments但对段的圆角边不满意,因为我的垂直线应该是一种阴影区域来表示某个时间段。

2 个答案:

答案 0 :(得分:0)

这可以解决您的问题。

替换

abline(v=2016, col = 'khaki', lwd = 20)
abline(mean(env$sal), 0, lty = 3)

lines(c(2013, 2017), rep(mean(env$sal), 2), col="black", lwd = 2, lty = 2)
lines(rep(2016, 2), c(0, 10), col="khaki", lwd = 20)

来源:https://stackoverflow.com/a/24741885/5874001

par(mar=c(3,5,3,6), xpd = T)
plot(sal~date, data = env, type = 'n', ylim = c(0,10), ylab = 'Salinity', bty = 'n', xlab = '')
lines(c(2013, 2017), rep(mean(env$sal), 2), col="black", lwd = 2, lty = 2)
lines(rep(2016, 2), c(0, 10), col="khaki", lwd = 20)
for (ii in seq_along(sitelist)) {
  i <- sitelist[ii]; lines(sal[site==i] ~ date[site==i], 
                            data = env,
                            col = c(4,2,5)[ii],  
                            lwd = 2,
                            lty = c(1,2,3)[ii]);
  points(sal[site==i] ~ date[site==i], data = env,
         pch = c(0,1,2)[ii], col = c(4,2,5)[ii])}
legend('topright',  title = 'sites', inset=c(-0.2,0), 
       lty = c(1,2,3), col = c(4,2,5), lwd = 2, 
       sitelist, pch = c(0,1,2))

enter image description here

如果你有20多个情节,我会看看你是否可以编写循环来执行该任务。

答案 1 :(得分:0)

您可以在ngIf调用中将xpd设置为FALSE,并在par调用中插入xpd = TRUE,如下所示:

legend

或者在env <- data.frame(site = c('BLK','DUC','WHP','BLK','DUC','WHP','BLK','DUC','WHP'), sal = c(5,6,3,2,4,5,6,8,4), date = c(2013,2013,2013,2015,2015,2015,2017,2017,2017)) sitelist <- c('BLK','DUC','WHP') par(mar=c(3,5,3,6), xpd = F) plot(sal~date, data = env, type = 'n', ylim = c(0,10), ylab = 'Salinity', bty = 'n', xlab = '') abline(v=2016, col = 'khaki', lwd = 20) abline(mean(env$sal), 0, lty = 3) for (ii in seq_along(sitelist)) { i <- sitelist[ii]; lines(sal[site==i] ~ date[site==i], data = env, col = c(4,2,5)[ii], lwd = 2, lty = c(1,2,3)[ii]); points(sal[site==i] ~ date[site==i], data = env, pch = c(0,1,2)[ii], col = c(4,2,5)[ii])} legend('topright', title = 'sites', inset=c(-0.2,0), lty = c(1,2,3), col = c(4,2,5), lwd = 2, sitelist, pch = c(0,1,2), xpd = T) 调用中保持xpd = TRUE,并在par调用中将xpd设置为FALSE,如下所示:

abline

plot