Showing correct label in ggplot using a different variable

Question

I have a data with 4 variables that looks like:

id|name|year|value|
1   A    16   500
1   A    15   400
1   A    14   430
2   B    16   200
2   B    15   180
2   B    14   170
3   A    16   620
3   A    15   510
3   A    14   200

and then, I have to create in ggplot a temporal line chart for each id but showing up its label instead of its id. What I did was:

ggplot(db, aes(x=year, y= value)) + geom_line() + facet_wrap(~db$id)

but it shows the charts written their ids instead of their names, so I tried:

ggplot(db, aes(x=year, y= value)) + geom_line() + facet_wrap(~db$name)

It has created the line charts with their correct labels, however id 1 and id 3 have both the same name, so, at the end, it created only 2 charts instead of 3 with one of them with 6 observations instead of 3.

Is there a way to concatenate name with id? and then do by name corrected by the id concatenation.

It's more likely that we will be able to help you if you provide a [complete minimal reproducible example](http://stackoverflow.com/help/mcve) to go along with your question. Something we can work from and use to show you how it might be possible to answer your question. — Eric Fail, Oct 18 '17 at 17:33
well, what I just showed is minimal, complete and verifiable. — gustavomoty, Oct 18 '17 at 17:37
See example in `?labeller`: "_# Or use character vectors as lookup tables:_". E.g. `lab <- c("1" = "A", "2" = "B", "3" = "A")`; `facet_wrap(~id, labeller = labeller(id = lab))` — Henrik, Oct 18 '17 at 17:49

Mako212 · Answer 1 · 2017-10-18T17:54:35.723

0

Update:

Create a combined name and id column called name2 with id on a new line, then graph.

db$name2 <- paste(db$name,db$id,sep="\n") 

ggplot(db, aes(x=year, y= value)) + geom_line() + facet_wrap(~name2)

You can do it like this:

ggplot(db, aes(x=year, y= value)) + 
  geom_line() + 
  facet_wrap(~interaction(id, name))

Which creates a separate chart for each combination of name and id. If you want to exclude the 1. and 2. we'll have to adjust the solution.

Data:

db <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), name = structure(c(1L, 
1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L), .Label = c("A", "B"), class = "factor"), 
    year = c(16L, 15L, 14L, 16L, 15L, 14L, 16L, 15L, 14L), value = c(500L, 
    400L, 430L, 200L, 180L, 170L, 620L, 510L, 200L)), .Names = c("id", 
"name", "year", "value"), class = "data.frame", row.names = c(NA, 
-9L))

edited Oct 18 '17 at 17:54

answered Oct 18 '17 at 17:37

Mako212

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it worked perfect! However it would be nice to drop .1 and .2. – gustavomoty Oct 18 '17 at 17:41
or if we make the label in two lines works, because I do not have actually those ids and names, but ids and names much biggers in some cases. So if we put name in first line, skip and then id in the second line, would be the perfect scenario. – gustavomoty Oct 18 '17 at 17:49
@gustavomoty see my update! – Mako212 Oct 18 '17 at 17:54
@gustavomoty also, Henrik's solution in the comments works if you really want to exclude `id` completely – Mako212 Oct 18 '17 at 18:03
Thanks, worked as I wanted!! @mako212 – gustavomoty Oct 18 '17 at 18:11

Eric Fail · Answer 2 · 2017-10-20T16:34:23.367

Marko212's answer is definitely better. However, in case you need to concatenate name with id at some future point.

dta <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L),
         name = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L), 
         .Label = c("A", "B"), class = "factor"), 
         year = c(16L, 15L, 14L, 16L, 15L, 14L, 16L, 15L, 14L),
         value = c(500L, 400L, 430L, 200L, 180L, 170L, 620L, 510L, 200L)),
         .Names = c("id", "name", "year", "value"), 
         class = "data.frame", row.names = c(NA,  -9L))

dta$name_id <- with(dta, paste0(id, name))
dta
#>   id name year value name_id
#> 1  1    A   16   500      1A
#> 2  1    A   15   400      1A
#> 3  1    A   14   430      1A
#> 4  2    B   16   200      2B
#> 5  2    B   15   180      2B
#> 6  2    B   14   170      2B
#> 7  3    A   16   620      3A
#> 8  3    A   15   510      3A
#> 9  3    A   14   200      3A

Showing correct label in ggplot using a different variable

2 Answers2