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Home/ Questions/Q 7518487
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T01:45:25+00:00 2026-05-30T01:45:25+00:00

I’m facing the following problem. I’ve the following two data frames set up x

  • 0

I’m facing the following problem. I’ve the following two data frames set up

x = c(1,2,3,4,1,2,3,4)
f = c(rep('a',4),rep('b',4))
y = c(2,4,5,6,9,4,2,10)
z = data.frame(x=x,f=f,y=y)

and the other data frame is

x1 = c(2,2.5,3.1,2.8,3.4,3.5)
y1 = c(1,1.2,1,2,3,2)
f1 = c(rep('a',3),rep('b',3))
z1 = data.frame(x=x1,f=f1,y=y1)

What I would like to be able to do is plot the first data frame z using faceting based on the column f. This is easy to do. However, my second data frame represents coordinates of two triangles I would like to plot on each of the plots. How do I work with layers to get this done? I tried the following, but it did not help.

p = ggplot() +
 layer(
 data=z,
 mapping = aes(x,y),
 geom="line"
) +
layer(
 data = z1,
 mapping = aes(x,y),
 geom="polygon",
 color="blue"
) +
facet_grid(f ~ .)

Thanks much for the attention & help.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T01:45:26+00:00Added an answer on May 30, 2026 at 1:45 am

    The important thing is that your facets should have the same column name in each data source:

    Set up the data. (I use the same data as you, but note that you don’t need to predefine your vectors – simply do it inside the data.frame function. This reduces clutter in your workspace:

    library(ggplot2)
    z <- data.frame(
      x = c(1,2,3,4,1,2,3,4),
      y = c(2,4,5,6,9,4,2,10),
      f = c(rep('a',4),rep('b',4))
    )
    
    z1 <- data.frame(
      x = c(2,2.5,3.1,2.8,3.4,3.5),
      y = c(1,1.2,1,2,3,2),
      f = c(rep('a',3),rep('b',3))
    )
    

    The call to ggplot is virtually identical to your original code. I just saved a few keystrokes by not using layer, but the geoms directly:

    ggplot() +
        geom_line(data=z, aes(x,y)) +
        geom_polygon(data=z1, aes(x,y), color="blue") +
        facet_grid(f ~ .)
    

    enter image description here

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