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Home/ Questions/Q 8182313
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T00:49:28+00:00 2026-06-07T00:49:28+00:00

so I have a simple example–a fully crossed three treatment three context experiment, where

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so I have a simple example–a fully crossed three treatmentthree context experiment, where a continuous effect was measured for each treatmentcontext pair. I want to order each treatment by effect, separately according for each context, but I’m stuck on ggplot’s faceting.

here’s my data

df <- data.frame(treatment = rep(letters[1:3], times = 3),
                 context = rep(LETTERS[1:3], each = 3),
                 effect = runif(9,0,1))

and I can get something very close if I collapse treatment and context into a single 9 point scale, as such:

df$treat.con <- paste(df$treatment,df$context, sep = ".")
df$treat.con <- reorder(df$treat.con, -df$effect, )

ggplot(df, aes(x = treat.con, y = effect)) +
           geom_point() +
           facet_wrap(~context, 
                      scales="free_x",
                      ncol = 1)

very close to what i want

except to achieve the separate ordering in each facet, the new x variable I created is potentially misleading, since it doesn’t demonstrate that we’ve used the same treatment in all three contexts.

Is this solved via some manipulation of the underlying factor, or is there a ggplot command for this situation?

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  1. Editorial Team
    Editorial Team
    2026-06-07T00:49:30+00:00Added an answer on June 7, 2026 at 12:49 am

    Try:

    ggplot(df, aes(x = treat.con, y = effect)) +
      geom_point() +
      facet_wrap(~context, scales="free_x", ncol = 1) +
      scale_x_discrete(labels=function(x) substr(x,1,1))
    

    The anonymous function provided to the labels argument does the formatting of the labels. In older versions of ggplot2 you used the formatter argument for this. If your treatment names are of differing lengths, then the substr approach might not work too well, but you could use strsplit, eg:

    + scale_x_discrete(labels=function(x) sapply(strsplit(x,"[.]"),"[",1))
    
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