input variable stored in list into loop in R - r

Question

I'm sure there are much better ways of doing this, I'm open to suggestions.
I have these vectors:
vkt1 <- c("df1", "df2", "df3")
vector2 <- paste("sample", wSheatx, sep="_")
The first vector contains a list of the names of dataframes stored in the environment. These are stored as strings, but I'd like to call them as variable names.
The second vector is just the first one adding "sample" at the beggining, equivalent to:
vector2 <- c('sample_df1', 'sample_df2', 'sample_df3')
These strings from vector2 would serve as the names of new data frames to be created.
Alrighty, so now I want to do something like this:
for (i in 1:length(vector){ # meaning for i in 1,2,3
vector2[i] = data.frame(which(eval(parse(text = vkt1[i])) == "Some_String", arr.ind=TRUE))
addStyle(wb, vkt1[i], cols = 1:ncol(eval(parse(text = vkt1[i]))), rows = vector2[[i]][,1]+1, style = duppedStyle, gridExpand = TRUE)
}
It may look complicated, but the idea is to make a data frames named as the strings contained in vector2, being a subset of the data frames from vkt1 when "Some_String" is found.
Then, use that created data frame and add a style to the entire row when said string is present.
vector2[[i]][,1]+1 is intended to deploy as sample_df1[,1]+1 (in the first iteration)
Note that I'm using eval(parse(text = vkt1[i])) to get the variables from the strings of vkt1. So, say, eval(parse(text = vkt1[1])) is equal do df1 (the data frame, not the string)
Like this, the code gives the following error:
In file(filename, "r") :
cannot open file 'noCoinColor_Concat': No such file or directory
Been trying to get it working like so, but I'm beginning to feel this approach might be very wrong.

Answer 1

It is easier to manage code and data when you keep them in a list instead of separate dataframes.
You can use mget to get all the dataframes in vkt1 in a string and let's say you want to search for 'Some_String' in the first column of each dataframe, so you can do :
new_data <- lapply(mget(vkt1), function(df) df[df[[1]] == 'Some_String', ])
I haven't included the addStyle code here because I don't know from which package it is and what it does but you can easily include it in lapply's anonymous function.

Answer 2

Is it not easier to combine your data frames into a list and then use apply or map family functions to adjust your data frames?
data(mtcars)
df1 <- mtcars %>% filter(cyl == 4)
df2 <- mtcars %>% filter(cyl == 6)
df3 <- mtcars %>% filter(cyl == 8)
df_old_names <- c("df1", "df2", "df3")
df_new_names <- c("df_cyl_4", "df_cyl_6", "df_cyl_8")
df_list <- lapply(df_old_names, get)
names(df_list) <- df_new_names