Generate a sequence of Data frame from function - r

Question

I searched but I couldn't find a similar question, so Apologies in advance if this is a duplicate question. I am trying to Generate a data frame from within a for loop in R.
what I want to do:
Define each columns of each data frame by a function,
Generate n data frames (length of my sequence of data frame) using loop,
As example I will use n=100 :
n<-100
k<-8
d1 <- data.frame()
for(i in 1:(k)) {d1 <- rbind(d1,c(a="i+1",b="i-1",c="i/1"))}
d2 <- data.frame()
for(i in 1:(k+2)) {d2 <- rbind(d2,c(a="i+2",b="i-2",c="i/2"))}
...
d100 <- data.frame()
for(i in 1:(k+100)) {d100 <- rbind(d100,c(i+100, i-100, i/100))}
It is clear that It'll be difficult to construct one by one each data.frame. I tried this:
d<-list()
for(j in 1:100) {
d[j] <- data.frame()
for(i in 1:(k+j)) {d[j] <- rbind(d[j] ,c(i+j, i-j, i/j))}
But I cant really do anything with it, I run into an error :
Error in d[j] <- data.frame() : replacement has length zero
In addition: Warning messages:
1: In d[j] <- rbind(d[j], c(i + j, i - j, i/j)) :
number of items to replace is not a multiple of replacement length
And a few more remarks about your example:
the number of rows in each data frame are not the same : d1 has 8 rows, d2 has 10 rows, and d100 has 8+100 rows,
the algorithm should give us : D=(d1,d2, ... ,d100).
It would be great to get an answer using the same approach (rbind) and a more base like approach. Both will aid in my understanding. Of course, please point out where I'm going wrong if it's obvious.

Answer 1

Here's how to create an empty data.frame (and it's not what you are trying):
Create an empty data.frame
And you should not be creating 100 separate dataframes but rather a list of dataframes. I would not do it with rbind, since that would be very slow. Instead I would create them with a function that returns a dataframe of the required structure:
make_df <- function(n,var) {data.frame( a=(1:n)+var,b=(1:n)-var,c=(1:n)/var) }
mylist <- setNames(
lapply(1:100, function(n) make_df(n,n)) , # the dataframes
paste0("d_", 1:100)) # the names for access
head(mylist,3)
#---------------
$d_1
a b c
1 2 0 1
$d_2
a b c
1 3 -1 0.5
2 4 0 1.0
$d_3
a b c
1 4 -2 0.3333333
2 5 -1 0.6666667
3 6 0 1.0000000
Then if you want the "d_40" dataframe it's just:
mylist[[ "d_40" ]]
Or
mylist$d_40
If you want to perform the same operation or get a result from all of them at nce; just use lapply:
lapply(mylist, nrow) # will be a list
Or:
sapply(mylist, nrow) #will be a vector because each value is the same length.