使用：=在data.table中粘贴（） [英] Using := in data.table with paste()

问题描述

对于大型人口模型，我开始使用 data.table 。到目前为止，我已经印象深刻，因为使用data.table结构减少了我的模拟运行时间约30％。我试图进一步优化我的代码，并包括一个简化的例子。我的两个问题是：1）是否可以使用：= 运算符与此代码？和2）将使用：= 操作符更快（虽然，如果我能回答我的第一个问题，我应该能够回答我的问题2） / p>

我在使用 data.table 1.9.4版的运行Windows 7的机器上使用R版本3.1.2。

这是我可重现的示例：

  ）
 
 ##创建示例表并设置初始条件
 nYears = 10 
 exampleTable = data.table（Site = paste（Site，1：3））
 exampleTable [，growthRate：= c（1.1，1.2，1.3），] 
 exampleTable [，c（paste（popYears，0：nYears，sep =））：= 0， 
 $ b exampleTable [，popYears0：= c（10，12，13）]＃设置初始填充大小
 
（yearIndex in 0：（nYears  -  1） 
 exampleTable [[粘贴（popYears，yearIndex，sep =）]] <
 exampleTable [[粘贴（popYears，yearIndex + 1， 
 exampleTable [，growthRate] 
} 
  

  for（yearIndex in 0：（nYears  -  1））{
 exampleTable [，paste（popYears ，yearIndex + 1，sep =）：= 
 paste（popYears，yearIndex，sep =）* growthRate，] 
} 
  / pre> 
 
 
但是，这不工作，因为粘贴不能与 data.table  ：
  exampleTable [，粘贴（popYears，yearIndex + 1，sep =）] 
＃ 1]popYears10
  
我已浏览过 data.table documentation 。 FAQ的第2.9节使用 cat ，但这会产生一个null输出。 
  exampleTable [，cat（paste（popYears，yearIndex + 1，sep =））] 
 ＃[1] popYears10NULL 
  
此外，我尝试搜索Google和rseek.org，但没有找到任何东西。如果缺少一个明显的搜索字词，我会喜欢搜索提示。我总是发现搜索R运算符是困难的，因为搜索引擎不喜欢符号（例如，：= ）和R可能是模糊的。 
 
 
 
最后，这是我在stackoverflow的第一篇文章，所以我道歉，如果我违反了过帐标准。 
解决方案
 
  ##从第一列三列示例数据开始
 dt< ;  -  exampleTable [，1：3，with = FALSE] 
 
 ##运行1年5年
 nYears < -  5 
（ii in seq_len（nYears） 1）{
 y0<  -  as.symbol（paste0（popYears，ii））
 y1<  -  paste0（popYears，ii + 1）
 dt [， y1）：= eval（y0）* growthRate] 
} 
 
 ##检查它是否工作
 dt 
＃网站growthRate popYears0 popYears1 popYears2 popYears3 popYears4 popYears5 
＃1：Site 1 1.1 10 11.0 12.10 13.310 14.6410 16.10510 
＃2：Site 2 1.2 12 14.4 17.28 20.736 24.8832 29.85984 
＃3：Site 3 1.3 13 16.9 21.97 28.561 37.1293 48.26809 
  
 
 
 
 编辑：
 
 
 
使用 set（）加速这个可能性在评论中不断出现，我会把这个额外的选项放在那里。
  nYears < -  5 
 
 ##只需要计算一次的事情可以从循环中取出
r< -  dt [[growthRate]] 
 yy<  -  paste0（popYears，seq_len（nYears + 1）-1）
 
 ##使用set .table的不错的紧凑语法
 for（ii in seq_len（nYears））{
 set（dt，，yy [ii + 1]，r * dt [[yy [ii]]]）
} 
 
 ##检查结果
 dt 
＃站点growthRate popYears0 popYears1 popYears2 popYears3 popYears4 popYears5 
＃1：Site 1 1.1 10 11.0 12.10 13.310 14.6410 16.10510 
＃2：Site 2 1.2 12 14.4 17.28 20.736 24.8832 29.85984 
＃3：Site 3 1.3 13 16.9 21.97 28.561 37.1293 48.26809 
  
 
I have started using data.table for a large population model. So far, I have been impressed because using the data.table structure decreases my simulation run times by about 30%. I am trying to further optimize my code and have included a simplified example. My two questions are: 1) Is is possible to use the := operator with this code? and 2) Would using the := operator be quicker (although, if I am able to answer my first question, I should be able to answer my question 2!)?

I am using R version 3.1.2 on a machine running Windows 7 with data.table version 1.9.4.

Here is my reproducible example:
library(data.table)

## Create  example table and set initial conditions
nYears = 10
exampleTable = data.table(Site = paste("Site", 1:3))
exampleTable[ , growthRate := c(1.1, 1.2, 1.3), ]
exampleTable[ , c(paste("popYears", 0:nYears, sep = "")) := 0, ]

exampleTable[ , "popYears0" := c(10, 12, 13)] # set the initial population size

for(yearIndex in 0:(nYears - 1)){
    exampleTable[[paste("popYears", yearIndex + 1, sep = "")]] <- 
    exampleTable[[paste("popYears", yearIndex, sep = "")]] * 
    exampleTable[, growthRate]
}
I am trying to do something like:
for(yearIndex in 0:(nYears - 1)){
    exampleTable[ , paste("popYears", yearIndex + 1, sep = "") := 
    paste("popYears", yearIndex, sep = "") * growthRate, ] 
}
However, this does not work because the paste does not work with the data.table, for example:
exampleTable[ , paste("popYears", yearIndex + 1, sep = "")]
# [1] "popYears10"
I have looked through the data.table documentation. Section 2.9 of the FAQ uses cat, but this produces a null output. 
exampleTable[ , cat(paste("popYears", yearIndex + 1, sep = ""))]
# [1] popYears10NULL
Also, I tried searching Google and rseek.org, but didn't find anything. If am missing an obvious search term, I would appreciate a search tip. I have always found searching for R operators to be hard because search engines don't like symbols (e.g., ":=") and "R" can be vague. 

Last, this is my first post on stackoverflow so I apologize if I violated a posting standard. 
解决方案
## Start with 1st three columns of example data
dt <- exampleTable[,1:3,with=FALSE]

## Run for 1st five years
nYears <- 5
for(ii in seq_len(nYears)-1) {
    y0 <- as.symbol(paste0("popYears", ii))
    y1 <- paste0("popYears", ii+1)
    dt[, (y1) := eval(y0)*growthRate]
}

## Check that it worked
dt
#     Site growthRate popYears0 popYears1 popYears2 popYears3 popYears4 popYears5
#1: Site 1        1.1        10      11.0     12.10    13.310   14.6410  16.10510
#2: Site 2        1.2        12      14.4     17.28    20.736   24.8832  29.85984
#3: Site 3        1.3        13      16.9     21.97    28.561   37.1293  48.26809
Edit:

Because the possibility of speeding this up using set() keeps coming up in the comments, I'll throw this additional option out there.
nYears <- 5

## Things that only need to be calculated once can be taken out of the loop
r <- dt[["growthRate"]]
yy <- paste0("popYears", seq_len(nYears+1)-1)

## A loop using set() and data.table's nice compact syntax
for(ii in seq_len(nYears)) {
    set(dt, , yy[ii+1], r*dt[[yy[ii]]])
}

## Check results
dt
#     Site growthRate popYears0 popYears1 popYears2 popYears3 popYears4 popYears5
#1: Site 1        1.1        10      11.0     12.10    13.310   14.6410  16.10510
#2: Site 2        1.2        12      14.4     17.28    20.736   24.8832  29.85984
#3: Site 3        1.3        13      16.9     21.97    28.561   37.1293  48.26809


                        
这篇关于使用：=在data.table中粘贴（）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！