ggplot multiple dataframes in R using lapply whilst specifying column names in function call

Question

I would like to apply the following ggplot function to a list of dataframes, each with the exact same format. This is the format of the data:

All datasets have same datatypes, column names, and row & column lengths. For each list of dataframes, I want to always plot xvar as Time, and yvar as Zscore.

The function I want to use is as follows.

library(ggplot2)
library(gridExtra)

Tzero = 0 
Tminus120 = -120
TEnd = 120

testIDplot <- function(data1,
                          xvar,
                          yvar){ggplot(data = data1)+
    geom_line(aes(x={{xvar}},y={{yvar}}),colour="#fd8d3c", size=0.5) +
    geom_vline(xintercept=Tzero, colour="black", size=0.8)+
    labs(x = "Time (s)", y = "Z-Score")+
    scale_x_continuous(breaks = seq(Tminus120,TEnd,60), limits = c(Tminus120,NA))+ 
    scale_y_continuous(expand = c(0, 0), limits = c(-3, 6))}

To call this function normally, I would specify testIDplot(newdata,Time,Zscore)

However, I don't know how to specify the columns when calling through a list. I tried lapply:

listofDF <- list(newdata1,newdata2,newdata3,newdata4,newdata5)
plotlist <- lapply(listofDF,testIDplot)
plotlist[1]

But this method doesn't work because of the missing aesthetics x and y.

Does anyone have any suggestions?

Try with lapply(listofDF, \(x) testIDplot(x, Time, Zscore)) — stefan
– stefan, Commented Jul 24, 2023 at 16:27

Andy Baxter · Accepted Answer · 2023-07-24 21:13:43Z

2

You can pass these values in the ... space of lapply:

library(tidyverse)


dfs <- map(1:4, \(x) {
  tibble(
    a = letters,
    b = runif(26, 1, 10)
  )
})

plot_df <- function(df, xvar, yvar) {
  
  ggplot(df, aes({{xvar}}, {{yvar}})) +
    geom_col()
  
}

# Testing function call for one dataset:
plot_df(dfs[[1]], a, b)

# Pass column names after the function argument in `lapply`
# (either as positional, or preferably named arguments)
lapply(dfs, plot_df, xvar = a, yvar = b)

Anonymous functions are a super-helpful and clear way as well, as suggested by Stefan above:

lapply(dfs, \(df) plot_df(df, xvar = a, yvar = b))

answered Jul 24, 2023 at 21:13

Andy Baxter

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Comments

Joan · Accepted Answer · 2023-07-24 17:56:20Z

1

You could potentially try the following, change: aes(x={{xvar}},y={{yvar}} -> x=eval(parse(xvar)), y = eval(parse(var))

Then:

lapply(listofDF, testIDplot, xvar = "Time", y = "Score".

see this other question

Or try this:

aes(x= .data[[x_var]], y= .data[[y_var]])

this other solution

Hope this might help.

answered Jul 24, 2023 at 17:56

Joan

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