在 R 中以编程方式分配的效率 [英] Efficiency in assigning programmatically in R

问题描述

总而言之，我有一个脚本可以导入存储在多个 txt 文件中的大量数据.在一个单一文件中不是所有的行都被放在同一个表中(DF现在切换到DT)，所以对于每个文件我选择属于同一个DF的所有行，get DF 和 assign 给它行.

In summary, I have a script for importing lots of data stored in several txt files. In a sigle file not all the rows are to be put in the same table (DF now switching to DT), so for each file I select all the rows belonging to the same DF, get DF and assign to it the rows.

我第一次创建一个名为 table1 的 DF 时:

The first time I create a DF named ,say, table1 I do:

name <- "table1" # in my code the value of name will depend on different factors
                 # and **not** known in advance
assign(name, someRows)

然后，在执行过程中，我的代码可能会(在其他文件中)找到要放入 table1 数据框中的其他行，因此:

Then, during the execution my code may find (in other files) other lines to be put in the table1 data frame, so:

name <- "table"
assign(name, rbindfill(get(name), someRows))

我的问题是:assign(get(string), anyObject) 是以编程方式进行赋值的最佳方式吗?谢谢

My question is: is assign(get(string), anyObject) the best way for doing assignment programmatically? Thanks

这是我的代码的简化版本:(dataSource 中的每一项都是 read.table() 的结果，所以是一个文本文件)

here is a simplified version of my code: (each item in dataSource is the result of read.table() so one single text file)

set.seed(1)
#
dataSource <- list(data.frame(fileType = rep(letters[1:2], each=4),
                              id       = rep(LETTERS[1:4], each=2),
                              var1     = as.integer(rnorm(8))),
                   data.frame(fileType = rep(letters[1:2], each=4),
                              id       = rep(LETTERS[1:4], each=2),
                              var1     = as.integer(rnorm(8))))
#                   #                                                                                          #
#                          
library(plyr)
#
tablesnames <- unique(unlist(lapply(dataSource,function(x) as.character(unique(x[,1])))))
for(l in tablesnames){
  temp <- lapply(dataSource, function(x) x[x[,1]==l, -1])
  if(exists(l)) assign(l, rbind.fill(get(l), rbind.fill(temp))) else assign(l, rbind.fill(temp))
}
#
#            
# now two data frames a and b are crated
#
#
# different method using rbindlist in place of rbind.fill (faster and, until now, I don't # have missing column to fill)
#
rm(a,b)
library(data.table)
#
tablesnames <- unique(unlist(lapply(dataSource,function(x) as.character(unique(x[,1])))))
for(l in tablesnames){
  temp <- lapply(dataSource, function(x) x[x[,1]==l, -1])
  if(exists(l)) assign(l, rbindlist(list(get(l), rbindlist(temp)))) else assign(l, rbindlist(temp))
}

推荐答案

我建议使用命名的list，并跳过使用 assign 和 get.许多很酷的 R 特性(例如 lapply)在列表上工作得非常好，并且不能与使用 assign 和 get 一起使用.此外，您可以轻松地将列表传递给函数，而这对于将变量组与 assign 和 get 结合使用可能会有些麻烦.

I would recommend using a named list, and skip using assign and get. Many of the cool R features (lapply for example) work very well on lists, and do not work with using assign and get. In addition, you can easily pass lists in to a function, while this can be somewhat cumbersome with groups of variables combined with assign and get.

如果你想将一组文件读入一个大的 data.frame 我会使用这样的东西(假设 csv 像文本文件):

If you want to read a set of files into one big data.frame I'd use something like this (assuming csv like text files):

library(plyr)
list_of_files = list.files(pattern = "*.csv")
big_dataframe = ldply(list_of_files, read.csv)

或者如果您想将结果保存在列表中:

or if you want to keep the result in a list:

big_list = lapply(list_of_files, read.csv)

并可能使用 rbind.fill:

big_dataframe = do.call("rbind.fill", big_list)

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