R语言将字符值分配为数据集名称

Question

I need to process hundreds of data frames through the same set of code. For illustrative purposes, I create 3 data frames:

ds.1 = as.data.frame(matrix(c(1, 0.15, 0.15, 0.15,
                          0.2, 1, 0.15, 0.15,
                          0.2, .15, 1, 0.15,
                          0.2, 0.15, 0.15, 1), nrow=4, ncol=4))
ds.2 = as.data.frame(matrix(c(1, 0.25, 0.25, 0.25,
                          0.2, 1, 0.25, 0.25,
                          0.2, .25, 1, 0.25,
                          0.2, 0.25, 0.25, 1), nrow=4, ncol=4))
ds.3 = as.data.frame(matrix(c(1, 0.50, 0.50, 0.50,
                          0.2, 1, 0.50, 0.50,
                          0.2, .50, 1, 0.50,
                          0.2, 0.50, 0.50, 1), nrow=4, ncol=4))

I then assign the data frame names to a vector.

ds.vector <- c("ds.1", "ds.2", "ds.3")  #create a vector of data set names

I count the number of data frames in the vector

ds.in.vector <- length(ds.vector) #count the number of elements in vector

I loop through the vector, and seek to assign the data frame name to a data frame called ds. I would then run the code on ds.

for (i in 1:ds.in.vector)
   {
     ds <- ds.vector[i] #copy each of the data sets into ds
     #There would be a bunch of code here. For this example, 
     # I will just try to print
     print(ds)
   }

The line: ds <- ds.vector[i] does not copy the data frame whose name is in the vector, into ds. Rather, it copies the character string of the vector into ds.

Answer 1

These data.frames belong into a list. You should assign them into one when you create them. Having hundreds of data.frames in your global environment is just crazy and really not practical.

ds.list <- list(

ds.1 = as.data.frame(matrix(c(1, 0.15, 0.15, 0.15,
                              0.2, 1, 0.15, 0.15,
                              0.2, .15, 1, 0.15,
                              0.2, 0.15, 0.15, 1), nrow=4, ncol=4)),
ds.2 = as.data.frame(matrix(c(1, 0.25, 0.25, 0.25,
                              0.2, 1, 0.25, 0.25,
                              0.2, .25, 1, 0.25,
                              0.2, 0.25, 0.25, 1), nrow=4, ncol=4)),
ds.3 = as.data.frame(matrix(c(1, 0.50, 0.50, 0.50,
                              0.2, 1, 0.50, 0.50,
                              0.2, .50, 1, 0.50,
                              0.2, 0.50, 0.50, 1), nrow=4, ncol=4))
)

ds.vector <- c("ds.1", "ds.2", "ds.3") 


for (i in seq_along(ds.vector)) {
  ds <- ds.list[ds.vector[i]] #copy each of the data sets into ds
  #There would be a bunch of code here. For this example, 
  # I will just try to print
  print(ds)
}

Of course, if they all have the same structure, you can combine them into one data structure:

library(data.table)
DT <- rbindlist(ds.list, idcol=TRUE)
DT[, print(.SD), by = .id]

Answer 2

one way to do it (although not very efficient) is the following:

ds.cumulated <- get(ds.vector[1])
for (i in 2:ds.in.vector)
{
 ds.i <- get(ds.vector[i])
 ds.cumulated <- rbind(ds.cumulated, ds.i)
}