How to dynamically modify multiple dataframes in a loop?

Question

I am creating multiple data frames inside a loop using assign, but once they have been created, how do I modify them inside the same loop?

For example the following code...

for(i in 1:3) {
assign(paste0("df", i),data.frame(A=c('a','b','c'),B=c("","","")))
}

create three dataframe, namely df1,df2, and df3 each looking like this...

  A B
1 a  
2 b  
3 c

With the B column blank.

Desired output is to have the value of 'i' inside the B column while creating df[i]. So df2 would be...

  A B
1 a 2
2 b 2
3 c 2

Important: Note that while I could do that inside the assign command itself in this case, in the larger problem that I am working on, I need to do the assignment outside the assign command as the dataframe I am creating is actually a subset of a larger dataframe not a new dataframe itself.

I have tried...

for(i in 1:3) {
assign(paste0("df", i),data.frame(A=c('a','b','c'),B=c("","","")))
paste0("df", i)$B <- i
}

... which doesn't work. What can work in place of paste0("df", i)$B <- i?

why clutter your global environment? Why not have all your dataframes in a list? And if the dataframe is a subset of the larger dataframe, why not use any of the packages often used for grouped operations? — Heroka
– Heroka, Commented Dec 14, 2015 at 11:03
@Heroka, would having the dataframe in a list make this dynamic modification easier? — Gaurav
– Gaurav, Commented Dec 14, 2015 at 11:05
@Heroka, I have to use dynamic subsetting, where even the number of subsets vary based on input. This is the easiet way I could figure out. — Gaurav
– Gaurav, Commented Dec 14, 2015 at 11:06
can you elaborate a bit on the larger problem you're trying to solve? How are subsets determined/made? Do they overlap? — Heroka
– Heroka, Commented Dec 14, 2015 at 11:07
@Heroka... I have a larger dataframe to be split into multiple subsets based where my input is the 'number of subsets' and '% of observations'. So I can have a single subset containing 100% values of larger set of 3 subsets containing 50%, 30% and 20% values respectively. And I have to give labels to each subset which specify the percentage. — Gaurav
– Gaurav, Commented Dec 14, 2015 at 11:11

Heroka · Accepted Answer · 2015-12-14 11:38:46Z

I always try to keep things I will doing the same things to inside something. that can be a dataframe or a list, but it won't clutter my global environment and object manipulation is easier.

###create some data

set.seed(123)
nobs=100
dat <- data.frame(id=1:nobs,x=rnorm(nobs),y=runif(nobs))

n_subsets = 3
percentages = c(50,30,20)

#create sample_flags: list of 'labels', with
#each label being x percent of total

subset_labels <- sprintf("%.f%%",percentages)
subset_flags <- sample(rep(subset_labels , times=percentages*nrow(dat)/100))

#or, depending on larger problem
#subset_flags <- sample(subset_labels, size=nrow(dat), prob=percentages/100, replace=T)

#(might not work nicely with all numbers of obs, but I'm guessing you've solved that)

#random part
dat$mysubset <- subset_flags

#do stuff for each subset, like mean of y orcount
library(data.table)

setDT(dat)[,.(.N, meany=mean(y)),mysubset]

> setDT(dat)[,.(.N, meany=mean(y)),mysubset]
   mysubset  N     meany
1:      30% 30 0.5632690
2:      50% 50 0.4717880
3:      20% 20 0.405884

or if you really want a list

mylist  <- lapply(1:3,function(x){
  data.frame(A=c('a','b','c'),B=c("","",""))
}
)
mylist <- lapply(1:3, function(i){
  r <- mylist[[i]]
  r$i <- i
  r
})

MikeRSpencer · Accepted Answer · 2015-12-14 11:31:45Z

2

As comments say, this is a good opportunity to use a list (based on the limited information given).

I would:

l = lapply(1:3, function(i){
   data.frame(A=c("a", "b", "c"), B=i)
})

If you desperately want to assign column B separately:

l = lapply(1:3, function(i){
   x = data.frame(A=c("a", "b", "c"))
   x$B = i
   x
})

As a side note, I'm not sure what benefit you get from assign so I've not included it in my example.

answered Dec 14, 2015 at 11:31

MikeRSpencer

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