在函数中使用 data.table i 和 j 参数 [英] Using data.table i and j arguments in functions

问题描述

我正在尝试编写一些包装函数来减少与 data.table 的代码重复.

I am trying to write some wrapper functions to reduce code duplication with data.table.

这是一个使用 mtcars 的示例.首先，设置一些数据:

Here is an example using mtcars. First, set up some data:

library(data.table)
data(mtcars)
mtcars$car <- factor(gsub("(.*?) .*", "\1", rownames(mtcars)), ordered=TRUE)
mtcars <- data.table(mtcars)

现在，这是我通常会写的按组汇总计数的内容.在这种情况下，我按 car 分组:

Now, here is what I would usually write to get a summary of counts by group. In this case I am grouping by car:

mtcars[, list(Total=length(mpg)), by="car"][order(car)]

      car Total
      AMC     1
 Cadillac     1
   Camaro     1
...
   Toyota     2
  Valiant     1
    Volvo     1

复杂之处在于，由于参数 i 和 j 是在 data.table 的框架中计算的，因此必须使用eval(...) 如果你想传入变量:

The complication is that, since the arguments i and j are evaluated in the frame of the data.table, one has to use eval(...) if you want to pass in variables:

这行得通:

group <- "car"
mtcars[, list(Total=length(mpg)), by=eval(group)]

但现在我想按相同的分组变量对结果进行排序.我无法得到以下任何变体来给我正确的结果.注意我总是得到单行结果，而不是有序集.

But now I want to order the results by the same grouping variable. I can't get any variant of the following to give me correct results. Notice how I always get a single row of results, rather than the ordered set.

mtcars[, list(Total=length(mpg)), by=eval(group)][order(group)]
   car Total
 Mazda     2

我知道为什么:这是因为 group 是在 parent.frame 中评估的，而不是 data.table 的框架.

I know why: it's because group is evaluated in the parent.frame, not the frame of the data.table.

如何在 data.table 的上下文中评估 group?

How can I evaluate group in the context of the data.table?

更一般地说，我如何在函数中使用它?我需要以下函数来给我所有的结果，而不仅仅是第一行数据:

More generally, how can I use this inside a function? I need the following function to give me all the results, not just the first row of data:

tableOrder <- function(x, group){
  x[, list(Total=length(mpg)), by=eval(group)][order(group)]
}

tableOrder(mtcars, "car")

推荐答案

Gavin 和 Josh 是对的.这个答案只是为了添加更多背景.这个想法是，您不仅可以将变量列名传递给这样的函数，还可以使用 quote() 将列名的表达式.

Gavin and Josh are right. This answer is only to add more background. The idea is that not only can you pass variable column names into a function like that, but expressions of column names, using quote().

group = quote(car)
mtcars[, list(Total=length(mpg)), by=group][order(group)]
      group Total
        AMC     1
   Cadillac     1
     ...
     Toyota     2
    Valiant     1
      Volvo     1

虽然，诚然，开始更难，但它可以更灵活.反正就是这个想法.在函数内部你需要 substitute()，像这样:

Although, admitedly more difficult to start with, it can be more flexible. That's the idea, anyway. Inside functions you need substitute(), like this :

tableOrder = function(x,.expr) {
    .expr = substitute(.expr)
    ans = x[,list(Total=length(mpg)),by=.expr]
    setkeyv(ans, head(names(ans),-1))    # see below re feature request #1780
    ans
}

tableOrder(mtcars, car)
      .expr Total
        AMC     1
   Cadillac     1
     Camaro     1
      ...
     Toyota     2
    Valiant     1
      Volvo     1

tableOrder(mtcars, substring(car,1,1))  # an expression, not just a column name
      .expr Total
 [1,]     A     1
 [2,]     C     3
 [3,]     D     3
 ...
 [8,]     P     2
 [9,]     T     2
[10,]     V     2

tableOrder(mtcars, list(cyl,gear%%2))   # by two expressions, so head(,-1) above
     cyl gear Total
[1,]   4    0     8
[2,]   4    1     3
[3,]   6    0     4
[4,]   6    1     3
[5,]   8    1    14

在 v1.8.0(2012 年 7 月)中添加了一个新参数 keyby，使其更简单:

A new argument keyby was added in v1.8.0 (July 2012) making it simpler :

tableOrder = function(x,.expr) {
    .expr = substitute(.expr)
    x[,list(Total=length(mpg)),keyby=.expr]
}

欢迎在 i、j 和 by 变量表达式领域提出意见和反馈.您可以做的另一件事是有一个表，其中一列包含表达式，然后查找要放入 i、j 或 by 的表达式从那个表.

Comments and feedback in the area of i,j and by variable expressions are most welcome. The other thing you can do is have a table where a column contains expressions and then look up which expression to put in i, j or by from that table.

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