如何使用循环形式的r重命名多个数据集中的列子集

Question

I know that this question has been asked before but I can not get it to work for me and I swear I tried many ways do do it from for file in loops to lapply. I have tables in which I want to replace the columns 2 to 7 'S headers which are now in this format: "X1","X2","X3","X4","X5","X6","X7" into "Kingdom", "Phylum", "Class", "Order", "Family", "Genus", "Species".

Each table does not have the same amount of row nor column.

My 31 tables are listed as this:

step4 <- list.files(pattern="*.coldrop.tsv")

Also, and this is a ''sub-problem'', I am doing it from the 2nd column because RAM keeps adding row numbers (1,2,3,4,5,6....n). If anyone can help me here that would be great.. I need to do it on all these ''step4'' list of tables. here are some ''samples'' of what I want to do.

when I fisrt was trying I opted for the for file in loop option:

colnames <- c("Kingdom", "Phylum", "Class", "Order", "Family", "Genus", "Species")

The following works on a single file

names(Omlo_run11_table.tsv.step1.tsv.step2.tsv.step3.tsv.coldrop.tsv)[2:8] <- c("Kingdom", "Phylum", "Class", "Order", "Family", "Genus", "Species")

i = 1
for(i in 1:length(step4)){
  names(step4[i])[2:8] <- c("Kingdom","Phylum","Class","Order","Family","Genus","Species") 

}

I get this: Error in names(step4[i])[2:8] <- c("Kingdom", "Phylum", "Class", "Order", : 'names' attribute [8] must be the same length as the vector [1]

names(get(step4[i]))[names(get(step4[i])) == "X1","X2","X3","X4","X5","X6","X7"] <- c("Kingdom","Phylum","Class","Order","Family","Genus","Species")

I get this: Error in names(get(step4[i]))[names(get(step4[i])) == "X1", "X2", "X3", : incorrect number of subscripts

for(i in 1:length(step4)){
  nm <- paste0("step4[i]")
  tmp <- get(nm)
  colnames(tmp)[2:8] <- c("Kingdom", "Phylum", "Class", "Order", "Family", "Genus", "Species")
  assign(nm, tmp)
}

I get this: Error in get(nm) : object 'step4[i]' not found

lapply (step4, function(df) { colnames(df)[2:length(step4)] <-colnames[1:length(step4)]-1)}

and so on... I am more of a for file in type of person but I am open to lapply options. I encountered solutions with setnames but could not figure it out either.. Can please someone help me...

Answer 1

Simply create a list of dataframes using your step4 character vector as @Gregor comments. Then, rename columns of each df iteratively which can all be handled in one lapply() anonymous function. Also, since you are working with tab separated files, you want the generalized read.table() function (of which read.csv is a special wrapper for comma separated files):

step4 <- list.files(path = tsvfilepath, pattern=".*tsv$", full.names = TRUE)

dfList <- lapply(step4, function(i) {
        df <- read.table(i, sep="\t", quote="", header=TRUE, as.is=FALSE)
        names(df)[2:8] <- c("Kingdom","Phylum","Class","Order","Family","Genus","Species") 
        return(df)
})

---

This list becomes useful for various needs such as for individual dataframes or one master dataframe.

For individual dfs, consider setNames() to name each individually and list2env() to create separate environment objects. Below gives each df the same name as their corresponding file name:

dfList <- setNames(dfList, step4)

list2env(dfList, envir=.GlobalEnv)

For one large master df, where you append all dataframes together, you have the challenge of the incomplete columns. Hence, consider third-party packages to fill in for missing columns across dfs:

library(plyr)
rbind.fill(dfList)

library(dplyr)
bind_rows(dfList)

library(data.table)    
rbindlist(dfList, fill=TRUE)