color.legend {plotrix} | R Documentation |
Display a color legend on a plot
color.legend(xl,yb,xr,yt,legend,rect.col,cex=1,align="lt",gradient="x",...)
xl,yb,xr,yt |
The lower left and upper right coordinates of the rectange of colors in user coordinates. |
legend |
The labels that will appear next to some or all of the colors. |
rect.col |
The colors that will fill the rectangle. |
cex |
Character expansion factor for the labels. |
align |
How to align the labels relative to the color rectangle. |
gradient |
Whether to have a horizontal (x) or vertical (y) color gradient. |
... |
Additional arguments passed to text. |
color.legend displays a rectangle defined by the first four arguments filled with smaller rectangles of color defined by the rect.col argument. Labels, defined by the legend argument, are placed next to the color rectangle. The position of the labels is determined by whether the color rectangle is horizontal or vertical and the align argument. The default value of lt places the labels at the left of a vertical rectangle or the top of a horizontal one. rb puts them on the other side. To have the labels in the same color as the rectangles, include a col argument that will be passed to text as in the example.
There can be fewer labels than colors. The labels will be evenly spaced along the rectangle in this case. It is possible to use empty labels to get uneven spacing. The user can pass more labels than colors, but the labels will almost certainly be crowded and it is not obvious that this would be of any use. To have complete control over the labels, see gradient.rect and text or mtext.
nil
Jim Lemon
# get some extra room par(mar=c(7,4,4,6)) testcol<-color.gradient(c(0,1),0,c(1,0),nslices=5) col.labels<-c("Cold","Warm","Hot") color2D.matplot(matrix(rnorm(100),nrow=10),c(1,0),0,c(0,1), main="Test color legends") color.legend(11,6,11.8,9,col.labels,testcol,gradient="y") color.legend(10.2,2,11,5,col.labels,testcol,align="rb",gradient="y") color.legend(0.5,-2,3.5,-1.2,col.labels,testcol) color.legend(7,-1.8,10,-1,col.labels,testcol,align="rb",col=testcol[c(1,3,5)]) par(mar=c(5,4,4,2))