CODEWITHSUNDEEP
rlistdataframedplyrdata.table
Alexander
Alexander

Reputation: 4635

How to process and combine data.frames in a list with faster way

Finally, I come to an issue that very slow data processing and appending rows of multiple data.frames. I use lapply and dplyr combination for data processing. OTH, the process becomes very slower as I have 20000 rows in each data frame multiplied with 100 files in the directory.

Currently this is a huge bottle neck for me as even after lapply process finishes I don't have enough memory to bind_rows process.

Here is my data processing method,

first make a list of files

files <- list.files("file_directory",pattern = "w.*.csv",recursive=T,full.names = TRUE)

then process this list of files

  library(tidyr)
  library(dplyr)

data<- lapply(files,function(x){
    tmp <- read.table(file=x, sep=',', header = T,fill=F,skip=0, stringsAsFactors = F,row.names=NULL)%>%

      select(A,B, C)%>%
      unite(BC,BC,sep='_')%>%

      mutate(D=C*A)%>%
      group_by(BC)%>%
      mutate(KK=median(C,na.rm=TRUE))%>%
      select(BC,KK,D)
  })

data <- bind_rows(data)

I am getting an error which says,

“Error: cannot allocate vector of size ... Mb” ...

Depends on how much left in my ram. I have 8 Gb ram but seems still struggling;(

I also tried do.call but nothing changed! Who is my friendly function or approach for this issue? I use R version 3.4.2 and dplyr 0.7.4.

Upvotes: 2

Views: 112

Answers (2)

manotheshark
manotheshark

Reputation: 4357

Using ldply from the plyr package would eliminate the need to bind the list after processing as it will output a data.frame

library(tidyr)
library(dplyr)
library(plyr)

files <- list.files("file_directory", pattern = "w.*.csv", recursive = TRUE, full.names = TRUE)

data<- ldply(files, function(x){
  read.table(file=x, sep=',', header = TRUE, fill = FALSE, skip = 0, stringsAsFactors = FALSE, row.names = NULL) %>%
    select(A, B, C) %>%
    unite(BC, BC, sep='_') %>%
    mutate(D = C * A) %>%
    group_by(BC) %>%
    mutate(KK = median(C, na.rm = TRUE)) %>%
    select(BC, KK, D)
})

Upvotes: 2

talat
talat

Reputation: 70266

I can't test this answer since there's no reproducible data but I guess it could be something like the following, using data.table:

library(data.table)

data <- setNames(lapply(files, function(x) {
  fread(x, select = c("A", "B", "C"))
}), basename(files))

data <- rbindlist(data, use.names = TRUE, fill = TRUE, id = "file_id")
data[, BC := paste(B, C, sep = "_")]
data[, D := C * A]
data[, KK := median(C, na.rm = TRUE), by = .(BC, file_id)]
data[, setdiff(names(data), c("BC", "KK", "D")) := NULL]

Upvotes: 4

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