根据,我应该能够根据变量中的颜色值对图上的点进行着色,如下所示(直接引用):
ui-list
我想用一维散点图来做这件事,称为(jDatColor <- merge(jDat, jColors))
## continent country year pop lifeExp gdpPercap color
## 1 Africa Eritrea 2007 4906585 58.04 641.4 #1B9E77
## 2 Africa Chad 2007 10238807 50.65 1704.1 #1B9E77
## 3 Americas Jamaica 2007 2780132 72.57 7320.9 #D95F02
## 4 Americas Cuba 2007 11416987 78.27 8948.1 #D95F02
## 5 Americas Costa Rica 2007 4133884 78.78 9645.1 #D95F02
## 6 Asia Nepal 2007 28901790 63.78 1091.4 #7570B3
## 7 Europe Germany 2007 82400996 79.41 32170.4 #E7298A
## 8 Europe Norway 2007 4627926 80.20 49357.2 #E7298A
#And now we can provide this new color variable as the value of the col = argument.
plot(lifeExp ~ gdpPercap, jDatColor, log = 'x', xlim = jXlim, ylim = jYlim,
col = color,
main = 'custom color scheme based on Dark2', cex = 2)
legend(x = 'bottomright',
legend = as.character(jColors$continent),
col = jColors$color, pch = par("pch"), bty = 'n', xjust = 1)
。我的(MWE)代码如下所示:
stripchart
产生这个:
根据我在total_data <- floor(runif(8000, 1,101))
class_val <- c(rep('one', 1000), rep('two', 1000), rep('three', 1000), rep('four', 1000), rep('five', 1000), rep('six', 1000), rep('seven', 1000), rep('eight', 1000))
colour_vals_pl <- rep(c('red', 'blue', 'yellow', 'dark blue', 'grey', 'pink', 'cornflowerblue', 'brown'), each=1000)
colour_vals_pl <- factor(colour_vals_pl, levels=c('red', 'blue', 'yellow', 'dark blue', 'grey', 'pink', 'cornflowerblue', 'brown'), ordered=TRUE)
class_val <- factor(class_val, levels=c('one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight'), ordered=TRUE)
datatouse <- cbind.data.frame(total_data, class_val, colour_vals_pl)
stripchart(total_data~class_val, main='Power law epidemic lengths', ylab="Epidemic duration (days)", xlab="Class", pch=20, vertical=TRUE, col=colour_vals_pl, data=datatouse)
中提供的定义,我希望每一行(一,二,三等)都是不同的颜色。我怎样才能做到这一点?