How to change facet labels

Question

I have used the following ggplot command   ggplot survey  aes x   age     stat bin aes n   nrow h3   y     count     n   binwidth   10      scale y continuous formatter    percent   breaks   c 0  0 1  0 2       facet grid hospital          theme panel background   theme blank      to produce    I d like to change the facet labels  however  to something shorter  like Hosp 1  Hosp 2     because they are too long now and look cramped  increasing the height of the graph is not an option  it would take too much space in the document   I looked at the facet grid help page but cannot figure out how

User · Answer

I have another way to achieve the same goal without changing the underlying data:

ggplot(transform(survey, survey = factor(survey,
        labels = c("Hosp 1", "Hosp 2", "Hosp 3", "Hosp 4"))), aes(x = age)) +
  stat_bin(aes(n = nrow(h3),y=..count../n), binwidth = 10) +
  scale_y_continuous(formatter = "percent", breaks = c(0, 0.1, 0.2)) +
  facet_grid(hospital ~ .) +
  opts(panel.background = theme_blank())

What I did above is changing the labels of the factor in the original data frame, and that is the only difference compared with your original code.

User · Answer

Note that this solution will not work nicely in case ggplot will show less factors than your variable actually contains  which could happen if you had been for example subsetting     library ggplot2   labeli  lt - function variable  value     names li  lt - list  versicolor   versi    virginica   virg     return names li value        dat  lt - subset iris Species   setosa    ggplot dat  aes Petal Length     stat bin     facet grid Species      labeller labeli    A simple solution  besides adding all unused factors in names li  which can be tedious  is to drop the unused factors with droplevels    either in the original dataset  or in the labbeler function  see   labeli2  lt - function variable  value     value  lt - droplevels value    names li  lt - list  versicolor   versi    virginica   virg     return names li value      dat  lt - subset iris Species   setosa   ggplot dat  aes Petal Length     stat bin     facet grid Species      labeller labeli2

User · Answer

I feel like I should add my answer to this because it took me quite long to make this work   This answer is for you if    you do not want to edit your original data  if you need expressions  bquote  in your labels and  if you want the flexibility of a separate labelling name-vector   I basically put the labels in a named vector so labels would not get confused or switched  The labeller expression could probably be simpler  but this at least works  improvements are very welcome   Note the    back quotes  to protect the facet-factor   n  lt - 10 x  lt - seq 0  300  length out   n     I have my data in a  long  format my data  lt - data frame    Type   as factor c rep  dl l   n   rep  alpha   n       T   c x  x     Value   c x 0 1  sqrt x        the label names as a named vector type names  lt - c     nonsense     this is just here because it looks good      dl l    Linear Expansion  Delta L L Ref               bquote expression    alpha    Linear Expansion Coefficient  alpha       1 K         ggplot        geom point data   my data  mapping   aes T  Value        facet wrap     Type  scales  free y                 labeller   label bquote   as expression                 eval parse text   paste0  type names         Type                                 labs x  Temperature  K    y     colour           theme legend position    none

User · Answer

Here s another solution that s in the spirit of the one given by  naught101  but simpler and also does not throw a warning on the latest version of ggplot2   Basically  you first create a named character vector  hospital names  lt - c                       Hospital 1     Some Hospital                        Hospital 2     Another Hospital                        Hospital 3     Hospital Number 3                        Hospital 4     The Other Hospital                          And then you use it as a labeller  just by modifying the last line of the code given by  naught101 to        facet grid hospital      labeller   as labeller hospital names     Hope this helps

User · Answer

The EASIEST way to change WITHOUT modifying the underlying data is   Create an object using the as labeller function adding the back tick mark for each of the default values     Necessary to put RH  into the facet labels hum names  lt - as labeller       c  50     quot RH  50 quot    60     quot RH  60 quot   70     quot RH  70 quot            80     quot RH  80 quot   90     quot RH  90 quot    100     quot RH  100 quot      We add into the GGplot       ggplot dataframe  aes x   Temperature C  y   fit              geom line              facet wrap  Humidity RH   nrow   2  labeller   hum names

User · Answer

Simple solution  from here    p  lt - ggplot mtcars  aes disp  drat     geom point     Example  old labels  p   facet wrap  am    to string  lt - as labeller c  0     Zero    1     One      Example  New labels  p   facet wrap  am  labeller   to string

User · Answer

Change the underlying factor level names with something like     Using the Iris data  gt  i  lt - iris  gt  levels i Species   1   setosa       versicolor   virginica    gt  levels i Species   lt - c  S    Ve    Vi    gt  ggplot i  aes Petal Length     stat bin     facet grid Species

User · Answer

Here is a solution that avoids editing your data   Say your plot is facetted by the group part of your dataframe  which has levels control  test1  test2  then create a list named by those values   hospital names  lt - list     Hospital 1   Some Hospital      Hospital 2   Another Hospital      Hospital 3   Hospital Number 3      Hospital 4   The Other Hospital      Then create a  labeller  function  and push it into your facet grid call   hospital labeller  lt - function variable value     return hospital names value      ggplot survey aes x age     stat bin aes n nrow h3  y   count   n   binwidth 10     facet grid hospital      labeller hospital labeller         This uses the levels of the data frame to index the hospital names list  returning the list values  the correct names      Please note that this only works if you only have one faceting variable  If you have two facets  then your labeller function needs to return a different name vector for each facet  You can do this with something like    plot labeller  lt - function variable value     if  variable   facet1         return facet1 names value       else       return facet2 names value           Where facet1 names and facet2 names are pre-defined lists of names indexed by the facet index names   Hostpital 1   etc       Edit  The above method fails if you pass a variable value combination that the labeller doesn t know  You can add a fail-safe for unknown variables like this   plot labeller  lt - function variable value     if  variable   facet1         return facet1 names value       else if  variable   facet2         return facet2 names value       else       return as character value             Answer adapted from how to change strip text labels in ggplot with facet and margin TRUE    edit  WARNING  if you re using this method to facet by a character column  you may be getting incorrect labels  See this bug report  fixed in recent versions of ggplot2

User · Answer

I think all other solutions are really helpful to do this  but there is yet another way    I assume    you have installed the dplyr package  which has the convenient mutate command  and  your dataset is named survey   survey       mutate Hosp1   Hospital1  Hosp2   Hospital2             This command helps you to rename columns  yet all other columns are kept    Then do the same facet wrap  you are fine now

User · Answer

Both facet wrap and facet grid also accept input from ifelse as an argument  So if the variable used for faceting is logical  the solution is very simple   facet wrap  ifelse variable   Label if true    Label if false      If the variable has more categories  the ifelse statement needs to be nested   As a side effect  this also allows the creation of the groups to be faceted within the ggplot call

User · Answer

The labeller function defintion with variable  value as arguments would not work for me  Also if you want to use expression you need to use lapply and can not simply use arr val   as the argument to the function is a data frame   This code did work   libary latex2exp  library ggplot2  arr  lt - list  virginica  TeX  x 1     versicolor  TeX  x 2     setosa  TeX  x 3    mylabel  lt - function val    return lapply val  function x  arr x      ggplot iris  aes x Sepal Length  y Sepal Width     geom line     facet wrap  Species  labeller mylabel

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Since I m not yet allowed to comment on posts  I m posting this separately as an addendum to Vince s answer and son520804 s answer   Credit goes to them      Son520804       using Iris data       I assume    You have installed the dplyr package  which has the convenient mutate command  and   your dataset is named survey    survey   gt   mutate Hosp1   Hospital1  Hosp2   Hospital2             This command helps you to rename columns  yet all other columns are kept    Then do the same facet wrap  you are fine now    Using the iris example of Vince and the partial code of son520804  I did this with the mutate function and achieved an easy solution without touching the original dataset  The trick is to create a stand-in name vector and use mutate   inside the pipe to correct the facet names temporarily     i  lt - iris  levels i Species   1   setosa       versicolor   virginica   new names  lt - c    rep  Bristle-pointed iris   50      rep  Poison flag iris  50      rep  Virginia iris   50    i   gt   mutate Species new names    gt    ggplot aes Petal Length        stat bin        facet grid Species        In this example you can see the levels of i Species is temporarily changed to corresponding common names contained in the new names vector  The line containing   mutate Species new names    gt     can easily be removed to reveal the original naming   Word of caution  This may easily introduce errors in names if the new name vector is not correctly set up  It would probably be much cleaner to use a separate function to replace the variable strings  Keep in mind that the new name vector may need to be repeated in different ways to match the order of your original dataset  Please double - and - triple check that this is correctly achieved

User · Answer

Adding another solution similar to  domi s with parsing mathematical symbols  superscript  subscript  parenthesis bracket   etc     library tidyverse  theme set theme bw base size   18        create separate name vectors   run  demo plotmath   for more examples of mathematical annotation in R am names  lt - c     0     delta  15  N-NO 3  -        1     sqrt x y        use  scriptstyle  to reduce the size of the parentheses  amp     bgroup  to make adding     possible  cyl names  lt - c     4     scriptstyle bgroup     a        T - 5          6     scriptstyle bgroup     b        T   10 degree C      8     scriptstyle bgroup     c        T   30         ggplot mtcars  aes wt  mpg        geom jitter       facet grid am   cyl               labeller   labeller am    as labeller am names   label parsed                                    cyl   as labeller cyl names  label parsed                      geom text x   4  y   25  size   4  nudge y   1              parse   TRUE  check overlap   TRUE              label   as character expression paste  Log   10    bgroup      frac  x    y                     OR create new variables then assign labels directly   reverse facet orders just for fun mtcars  lt - mtcars   gt      mutate am2    factor am   labels   am names            cyl2   factor cyl  labels   rev cyl names   levels   rev attr cyl names   names          ggplot mtcars  aes wt  mpg        geom jitter       facet grid am2   cyl2               labeller   label parsed      annotate  text   x   4  y   30  size   5             parse   TRUE              label   as character expression paste  speed     m   s  -1               Created on 2019-03-30 by the reprex package  v0 2 1 9000

User · Answer

Here s how I did it with facet grid yfacet xfacet  using ggplot2  version 2 2 1   facet grid      yfacet xfacet      labeller   labeller          yfacet   c  0     an y label    1     another y label            xfacet   c  10     an x label    20     another x label             Note that this does not contain a call to as labeller   -- something that I struggled with for a while   This approach is inspired by the last example on the help page Coerce to labeller function

User · Answer

This is working for me   Define a factor   hospitals factor lt - factor  c  H0   H1   H2       and use  in ggplot     facet grid  hospitals factor hospital

User · Answer

Just extending naught101 s answer -- credit goes to him  plot labeller  lt - function variable value  facetVar1   lt name-of-1st-facetting-var gt    var1NamesMapping  lt pass-list-of-name-mappings-here gt   facetVar2     var2NamesMapping list          print  variable     print  value    if  variable  facetVar1               value  lt - as character value        return var1NamesMapping value            else if  variable  facetVar2               value  lt - as character value        return var2NamesMapping value            else              return as character value             What you have to do is create a list with name-to-name mapping  clusteringDistance names  lt - list     100   100      200   200      300   300      400   400      600   500      and redefine plot labeller   with new default arguments   plot labeller  lt - function variable value  facetVar1  clusteringDistance   var1NamesMapping clusteringDistance names  facetVar2     var1NamesMapping list       And then   ggplot        facet grid clusteringDistance       labeller plot labeller     Alternatively you can create a dedicated function for each of the label changes you want to have

User · Answer

This solution is very close to what  domi has  but is designed to shorten the name by fetching first 4 letters and last number   library ggplot2     simulate some data xy  lt - data frame hospital   rep paste  Hospital     1 3  sep        each   30                    value   rnorm 90    shortener  lt - function string      abb  lt - substr string  start   1  stop   4    fetch only first 4 strings   num  lt - gsub        d 1         1   string    using regular expression  fetch last number   out  lt - paste abb  num    put everything together   out    ggplot xy  aes x   value       theme bw       geom histogram       facet grid hospital      labeller   labeller hospital   shortener

User · Answer

After struggling for a while  what I found is that we can use fct relevel   and fct recode   from forcats in conjunction to change the order of the facets as well fix the facet labels  I am not sure if it s supported by design  but it works  Check out the plots below     library tidyverse   before  lt - mpg   gt     ggplot aes displ  hwy        geom point       facet wrap  class  before     after  lt - mpg   gt     ggplot aes displ  hwy        geom point        facet wrap      vars          Change factor level name       fct recode class   motorbike     2seater     gt              Change factor level order         fct relevel  compact             after     Created on 2020-02-16 by the reprex package  v0 3 0

User · Answer

Have you tried changing the specific levels of your Hospital vector   levels survey hospital  levels survey hospital      Hospital  1    lt -  Hosp 1  levels survey hospital  levels survey hospital      Hospital  2    lt -  Hosp 2  levels survey hospital  levels survey hospital      Hospital  3    lt -  Hosp 3

User · Answer

If you have two facets hospital and room but want to rename just one  you can use   facet grid  hospital   room  labeller   labeller hospital   as labeller hospital names      For renaming two facets using the vector-based approach  as in naught101 s answer   you can do   facet grid  hospital   room  labeller   labeller hospital   as labeller hospital names                                                    room   as labeller room names

[r] How to change facet labels?

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