CODEWITHSUNDEEP
rfunctionmatrix
CapnShanty
CapnShanty

Reputation: 559

Outer returning multiple errors?

Alright so I have a dataframe of some number of rows and 44 columns.

I am trying, in order to make a heatmap, to create a matrix that is 5 x 44. 

x = seq(1,5,1)
y = seq(1,44,1)
f = function(x, y) { r = mean(r01[r01$weekday == x,y]); r}
z = outer(x, y, f)

Is how I've been trying to do it, but I get a couple of errors:

Error in dim(robj) <- c(dX, dY) : 
  dims [product 220] do not match the length of object [1]
In addition: Warning messages:
1: In r01$weekday == p :
  longer object length is not a multiple of shorter object length
2: In mean.default(r01[r01$weekday == p, q]) :
  argument is not numeric or logical: returning NA

So in my attempts to remove these errors, I have tried the following:

x = seq(1,5,1)
y = seq(1,44,1)
f = function(x,y){ r = x+y; r+1 }
z = outer(x, y, f)

Which works. Now this, which doubles as my reproducible example:

hugs<-data.frame(replicate(44,sample(1:5,1000,rep=TRUE)))
x = seq(1,5,1)
y = seq(1,44,1)
f = function(x, y) { r = mean(hugs[hugs$x1 == x,y]); r}
z = outer(x, y, f)

Fails, with largely the same errors:

Error in dim(robj) <- c(dX, dY) : 
  dims [product 220] do not match the length of object [1]
In addition: Warning message:
In mean.default(hugs[hugs$X1 == x, y]) :
  argument is not numeric or logical: returning NA

And this leaves me largely confused, because I assumed this would work if the one above it did. So I have no clue how to reconcile what works from the one that works with the ones that don't.

The error that's missing from the "hugs" version I reproduced by modifying hugs with the following before running the rest:

hugs$X1[hugs$X1 == 1,] <- 3

But I don't actually know how to fix that error, since my original dataframe has values from 1:5 in weekday, and 44 columns just like my example, so I must just be misunderstanding its source. It's not that the means don't exist:

for (i in x){
  for (j in y){
    print(mean(r01[r01$weekday == i,j]))}}

Returns values for every single one. I guess I could reconstruct my upper logic to just use these two loops, but I want to use outer since it's more efficient(?).

Upvotes: 1

Views: 43

Answers (1)

CapnShanty
CapnShanty

Reputation: 559

Thanks to the comments, I decided not to use outer and instead made use of this equally fast and simple one liner:

aggregate(r01[1:44], r01["weekday"], mean)

Upvotes: 1

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