How to plot graphs through two loops

Question

Though this problem has been 'solved' many times, it turns out there's always another problem.

Without the print function it runs with no errors, but with it I get the following:

Error in .subset2(x, i) : recursive indexing failed at level 2

Which I'm taking to mean it doesn't like graphs being created in two layers of iteration? Changing the method to 'qplot(whatever:whatever)' has the exact same problem.

It's designed to print a graph for every pairing of the variables I'm looking at. There's too many for them to fit in a singular picture, such as for the pairs function, and I need to be able to see the actual variable names in the axes.

load("Transport_Survey.RData")

variables <- select(Transport, "InfOfReceievingWeather", "InfOfReceievingTraffic", "InfOfSeeingTraffic", "InfWeather.Ice", "InfWeather.Rain", "InfWeather.Wind", "InfWeather.Storm", "InfWeather.Snow", "InfWeather.Cold", "InfWeather.Warm", "InfWeather.DarkMorn", "InfWeather.DarkEve", "HomeParking", "WorkParking", "Disability", "Age", "CommuteFlexibility", "Gender", "PassionReduceCongest")
varnames <- list("InfOfReceivingWeather", "InfOfReceivingTraffic", "InfOfSeeingTraffic", "InfWeather.Ice", "InfWeather.Rain", "InfWeather.Wind", "InfWeather.Storm", "InfWeather.Snow", "InfWeather.Cold", "InfWeather.Warm", "InfWeather.DarkMorn", "InfWeather.DarkEve", "HomeParking", "WorkParking", "Disability", "Age", "CommuteFlexibility", "Gender", "PassionReduceCongest")

counterx = 1
countery = 1

for (a in variables) {
  for (b in variables) {
    print(ggplot(variables, mapping=aes(x=variables[[a]], y=variables[[b]],
    xlab=varnames[counterx], ylab=varnames[countery]))+
      geom_point())
    countery = countery+1
  counterx = counterx+1
  }
}

#variables2 <- select(Transport, one_of(InfOfReceivingWeather, InfOfReceivingTraffic, InfOfSeeingTraffic, InfWeather.Ice, InfWeather.Rain, InfWeather.Wind, InfWeather.Storm, InfWeather.Snow, InfWeather.Cold, InfWeather.Warm, InfWeather.DarkMorn, InfWeather.DarkEve, HomeParking, WorkParking, Disability, Age, CommuteFlexibility, Gender, PassionReduceCongest))

Here is a mini-data frame for reference, sampled from the columns I'm using:

structure(list(InfOfReceievingWeather = c(1, 1, 1, 1, 4), InfOfReceievingTraffic = c(1, 
1, 1, 1, 4), InfOfSeeingTraffic = c(1, 1, 1, 1, 4), InfWeather.Ice = c(3, 
1, 3, 5, 5), InfWeather.Rain = c(1, 1, 2, 2, 4), InfWeather.Wind = c(1, 
1, 2, 2, 4), InfWeather.Storm = c(1, 1, 1, 2, 5), InfWeather.Snow = c(1, 
1, 2, 5, 5), InfWeather.Cold = c(1, 1, 1, 2, 5), InfWeather.Warm = c(1, 
1, 1, 1, 3), InfWeather.DarkMorn = c(1, 1, 1, 1, 1), InfWeather.DarkEve = c(1, 
1, 1, 1, 1), HomeParking = c(1, 1, 3, 1, 1), WorkParking = c(1, 
4, 4, 5, 4), Disability = c(1, 1, 1, 1, 1), Age = c(19, 45, 35, 
40, 58), CommuteFlexibility = c(2, 1, 5, 1, 2), Gender = c(2, 
2, 2, 2, 1), PassionReduceCongest = c(0, 0, 2, 0, 2)), row.names = c(NA, 
-5L), class = c("tbl_df", "tbl", "data.frame"))

Does your `variables` object is a dataframe or a list ? based on your code, it looks to be a list (you are using `variables[[a]]`). Can you edit your question to add a small [reproducible example of your dataset](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) ? (for example, the output of `dput(df[1:10,1:10])`) — dc37, Nov 30 '19 at 19:35
@dc37 Varnames is a list, but variables is a dataset. As for a minature dataset, I'll see what I can do... — Piomicron, Nov 30 '19 at 20:14

score 1 · Accepted Answer · answered Dec 01 '19 at 19:40

You get an error in the assignment of your a and b. Basically, when defining a and b in variables, they become the vector of values contained in columns of variables. Thus, in your aes mapping, when you are calling variables[[a]], basically, you are writing (for the first iteration of a in variables): variables[[c(1, 1, 1, 1, 4)]] instead of variables[["InfOfReceievingWeather"]]. So, it can't work.

To get over this issue, you have to either choose between:

for (a in variables) {
  for (b in variables) {
    print(ggplot(variables, mapping=aes(x=a, y=b)) ...

or

for (a in 1:ncol(variables)) {
  for (b in 1:ncol(variables)) {
    print(ggplot(variables, mapping=aes(x=variables[[a]], y=variables[[b]])) ...

Despite the first one seems to be simpler, I will rather prefere the second option because it will allow you to recycle a and b as column indicator to extract colnames of variables for xlab and ylab. At the end, writing something like this should work:

for (a in 1:ncol(variables)) {
  for (b in 1:ncol(variables)) {
    print(ggplot(variables, mapping=aes(x=variables[[a]], y=variables[[b]])) +
            xlab(colnames(variables)[a])+
            ylab(colnames(variables)[b])+
            geom_point())
  }
}

Does it answer your question ?

How to plot graphs through two loops

1 Answers1