如何将数据帧分配给 R 列表中的特定位置

Question

I've created 15 individual dataframes via a for loop, each one representing scores for each of the survey items comprising 15 different scales/topic areas. They're named df1 through df15 based on the original order of my topic areas, but I'd like to put them into a single list based on a different sort order. I have a separate dataframe, called topic.area, listing each of the 15 topic areas, their original order, and it is arranged in the new sort order:

Topic.Area...... orig.order... sort.order
Leadership........3................1
Engagement..... 1...............2
Innovation.........2................3
etc...

I essentially want to do the following, but the code isn't working for me. I'm very new to R, so I appreciate any solutions that aren't too advanced so I understand the logic behind them.

myList<- list()

for(m in 1:15) {
 myList[[m]]<- paste("df", topic.area$orig.order[m], sep = "") 
}

I think the issue is you can't use the paste function in an assignment command, but I don't know to work around that.

Answer 1

Get all the dataframe in a list. As you have mentioned this is already ordered by orig.order .

list_df <- mget(paste0('df', 1:15))

You can arrange them based on sort.order directly by doing:

sorted_list_df <- list_df[topic.area$sort.order]

Answer 2

Here is a completely reproducible example based on Pokémon Stats data. Instead of just writing the input files to a list() , I use assign() to simulate a set of data frames in the global environment.

download.file("https://raw.githubusercontent.com/lgreski/pokemonData/master/pokemonData.zip",
              "pokemonData.zip",
              method="curl",mode="wb")
unzip("pokemonData.zip")

thePokemonFiles <- list.files("./pokemonData",
                              full.names=TRUE)
filenames <- substr(list.files("./pokemonData",full.names = FALSE),1,5) 
results <- lapply(1:length(thePokemonFiles),function(x) {
     # read data and assign to a name in global environment 
     data <- read.csv(thePokemonFiles[x],stringsAsFactor = FALSE)
     assign(filenames[x],data,globalenv())
     return(TRUE)
})

At this point we have 7 data frames named gen01 to gen07 in the global environment.

Next, we generate the desired sort order as a vector. We'll use this along with the names from the filenames vector to get the data and build the output list in the desired sort order of data frames.

desiredSortOrder <- c(1,3,5,7,2,4,6)
sortedDataFrames <- lapply(desiredSortOrder,function(x){
     get(filenames[x])
})
# verify generation numbers by printing first row of each data frame 
lapply(sortedDataFrames,function(x){
     x[1,c(1,2,12)]
})

Since the Generation column in each data frame matches its associated file name, we can verify the sort order by printing the first row of each data frame.

>lapply(sortedDataFrames,function(x){
+      x[1,c(1,2,12)]
+ })
[[1]]
  Number      Name Generation
1      1 Bulbasaur          1

[[2]]
  Number    Name Generation
1    252 Treecko          3

[[3]]
  Number    Name Generation
1    494 Victini          5

[[4]]
  Number   Name Generation
1    722 Rowlet          7

[[5]]
  Number      Name Generation
1    152 Chikorita          2

[[6]]
  Number    Name Generation
1    387 Turtwig          4

[[7]]
  Number    Name Generation
1    650 Chespin          6

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