Creating an R dataframe row-by-row

Question

I would like to construct a dataframe row-by-row in R. I've done some searching, and all I came up with is the suggestion to create an empty list, keep a list index scalar, then each time add to the list a single-row dataframe and advance the list index by one. Finally, do.call(rbind,) on the list.

While this works, it seems very cumbersome. Isn't there an easier way for achieving the same goal?

Obviously I refer to cases where I can't use some apply function and explicitly need to create the dataframe row by row. At least, is there a way to push into the end of a list instead of explicitly keeping track of the last index used?

Answer 1

One can add rows to NULL:

df<-NULL;
while(...){
  #Some code that generates new row
  rbind(df,row)->df
}

for instance

df<-NULL
for(e in 1:10) rbind(df,data.frame(x=e,square=e^2,even=factor(e%%2==0)))->df
print(df)

Answer 2

You can grow them row by row by appending or using rbind().

That does not mean you should. Dynamically growing structures is one of the least efficient ways to code in R.

If you can, just preserver all your data.frame up front:

N <- 1e4  # some magic number, possibly an overestimate

DF <- data.frame(num=rep(NA,10), txt=rep("", 10), # as many cols as you need
                 stringsAsFactors=FALSE)          # you don't know levels

and then during your operations insert row at a time

DF[i, ] <- c(1.4, "foo")

That should work for arbitrary data.frame and be much more efficient. If you overshot N you can always shrink empty rows out at the end.

Answer 3

This is a silly example of how to use do.call(rbind,) on the output of Map() [which is similar to lapply()]

> DF <- do.call(rbind,Map(function(x) data.frame(a=x,b=x+1),x=1:3))
> DF
  x y
1 1 2
2 2 3
3 3 4
> class(DF)
[1] "data.frame"

I use this construct quite often.