Passing column names into graph function in ggplot2

Question

I have a large dataset with multiple dependent variables but with only two independent variables (that I will be using over and over again to sort the many dependent variables). Each of the dependent variables was measured twice, once before and once after treatment. I would like to write a function that allows me to obtain a graph for each of these multiple dependent variables, with the arguments of the function as the two column names for whichever of the dependent variables i wish to graph.

I have generated a toy dataset to illustrate my problem. 't1DV1' and 't1DV2' are the pre- and post- treatment scores for dependent variable 1. 't1DV2' and 't2DV2' are pre- and post- treatment scores for dependent variable 2. 'group' is the independent variable.

group <- factor(rep(c("A", "B"), 10))
t1DV1 <- runif(20, min = 0, max = 10)
t2DV1 <- runif(20, min = 0, max = 10)
t1DV2 <- runif(20, min = 0, max = 10)
t2DV2 <- runif(20, min = 0, max = 10)

df <- data.frame(group, t1DV1, t2DV1, t1DV2, t2DV2)

df

I tried writing the following function

DVGraph <- function (DV1, DV2) { 

require(tidyr)

dfLong <- gather(df, prePost, Score, DV1:DV1)

require(ggplot2)

barGraph <- ggplot(dfLong, aes(group, Score, fill = prePost)) + 
  geom_bar(stat = "identity", position = "dodge", size = 0.5) +
  scale_fill_manual(values = c("#999999", "#666666")) +
  xlab("") +
  ylab("Scores") +
  theme_bw()

return(barGraph)

}

And then tried calling it using the first of the repeated measures variables (I could equally have used the second, i.e. t1DV2 and t2DV2)

DVGraph(t1DV1, t2DV1)

But I get an error.

I tried using inverted commas like so

DVGraph("t1DV1", "t2DV1")

But i got another (different) error.

Does anyone know how I might go about this?

Sam Dickson · Accepted Answer · 2016-01-11 14:28:08Z

1

Alter your gather call to the following:

dfLong <- gather(df, prePost, Score, DV1, DV2)

Then when you call your function, use the column numbers instead of the column names:

DVGraph(2, 3)

Alternatively, you can replace gather() with melt() from reshape2 with substitute() in order to be able to call the function with the unquoted variables:

DVGraph <- function (DV1, DV2) { 

  require(tidyr)
  require(reshape2)

  dfLong <- melt(df,measure.vars = c(substitute(DV1),substitute(DV2)),
                 var="prePost",value.name ="Score")

  require(ggplot2)

  barGraph <- ggplot(dfLong, aes(group, Score, fill = prePost)) + 
    geom_bar(stat = "identity", position = "dodge", size = 0.5) +
    scale_fill_manual(values = c("#999999", "#666666")) +
    xlab("") +
    ylab("Scores") +
    theme_bw()

  return(barGraph)

}

DVGraph(t1DV2, t2DV1)

Update:

If you want to do what you asked about in your comment, one quick fix is to recognize that using substitute() forces your vector to be a list, but you can force it to be a character by using as.character(substitute()) as follows:

createFrame <- function (DV1, DV2) { 
  extractCols <- c("group", as.character(substitute(DV1)), as.character(substitute(DV2)))
  newFrame <- df[,extractCols]
  return(newFrame) 
}

createFrame(t1DV1, t2DV1)

edited Jan 11, 2016 at 14:28

answered Jan 8, 2016 at 21:36

Sam Dickson

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2 Comments

llewmills Over a year ago

Thanks Sam! Worked a treat.

llewmills Over a year ago

Sam, I have a lot of variables in my dataset. A LOT. So in addition to the above I really need a way of creating a reduced version of my dataframe within the function above. Using the principles of the 'substitute' function you showed me above i tried this: createFrame <- function (DV1, DV2) { extractCols <- c("group", substitute(DV1), substitute(DV2)) newFrame <- df[,extractCols] return(newFrame) } createFrame(t1DV1, t2DV1) But this does not work. Any idea how I might do this? Or is this perhaps a separate question for the boards?

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