13

I successfully create a plot using the following:

# suppose I have a p <- ggplot(data=df, ...) then the following works 
# I get those two segments plotted correctly
p <- p + geom_segment(aes(x=1,y=103,xend=1,yend=107))
p <- p + geom_segment(aes(x=5,y=103,xend=5,yend=107))

However if I do:

values <- c(1, 5)
for (i in values) {
   p <- p + geom_segment(aes(x=i,y=103,xend=i,yend=107))
}

It doesn't work, only the last segment is created. Can anyone advice what's wrong here?

2 Answers 2

21

It has to do with the lazy evaluation of the aes() values. You are binding to the variable i but not actually doing anything with it in the loop. The mappings aren't resolved till you actually print(p). Essentially this means they are all being bound to i and after the loop exits, i will have the value it had during the final loop.

So the problem really is you shounld't be using aes() here as you don't really want active binding. Just set the x and xend values outside the aes(). (And since the y's are constant they should be outside the aes() as well).

values <- c(1, 5)
for (i in values) {
   p <- p + geom_segment(x=i, y=103, xend=i, yend=107)
}
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3 Comments

What i posted will fix it. Don't use aes() in the geom_segment() in the loop.
That's what I thought. Do you know why adding a print statement or using force in the for loop doesn't fix the problem?
@csgillespie Because in this case you're still binding to the variable i, not the value. You can't force aes to resolve the name till it actually draws the plot. force is sometimes useful in situations like this, but with the non-standard evaluation of aes it won't help in this case.
21

An alternative approach would be to avoid using a loop at all. You can pack your segment data up in a separate data.frame from your main data and use aes() to plot everything at once like so:

segment_data = data.frame(
    x = c(1, 5),
    xend = c(1, 5), 
    y = c(103, 103),
    yend = c(107, 107)
)

p = ggplot(df, ...) +
geom_segment(data = segment_data, aes(x = x, y = y, xend = xend, yend = yend))

2 Comments

This does seem better. I can't even get the other answer to work!
@SkyWalker can you make this the answer?

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