使用data.table功能将长结构化的data.table重塑为广泛的结构吗？ [英] Reshape long structured data.table into a wide structure using data.table functionality?

问题描述

> library(data.table)
> A <- data.table(x = c(1,1,2,2), y = c(1,2,1,2), v = c(0.1,0.2,0.3,0.4))
> A
   x y   v
1: 1 1 0.1
2: 1 2 0.2
3: 2 1 0.3
4: 2 2 0.4
> B <- dcast(A, x~y)
Using v as value column: use value.var to override.
> B
  x   1   2
1 1 0.1 0.2
2 2 0.3 0.4

很明显，我可以使用fx将data.table从长整形改成宽整形dcast软件包reshape2。但是data.table附带了一个重载的括号运算符，提供了 by和 group之类的参数，这让我想知道是否可以使用它来实现（针对data.table的特定功能）？

Apparently I can reshape a data.table from long to wide using f.x. dcast of package reshape2. But data.table comes along with an overloaded bracket-operator offering parameters like 'by' and 'group', which make me wonder if it is possible to achieve it using this (to data.table specific functionality)?

这是手册中的一个随机示例：

Just one random example from the manual:

DT[,lapply(.SD,sum),by=x]

这看起来很棒-但我不喜欢

That looks awesome - but I don't fully understand the usage yet.

我既没有找到方法也没有例子，所以也许是不可能的，甚至是不应该的-因此，当然，一个肯定的不，因为……是不可能的也是一个有效的答案。

I neither found a way nor an example for this so maybe it is just not possible maybe it isn't even supposed to be - so, a definite "no, is not possible because ..." is then of course also a valid answer.

推荐答案

我会选择一个具有不相等组的示例，以便在一般情况下更易于说明：

I'll pick an example with unequal groups so that it's easier to illustrate for the general case:

A <- data.table(x=c(1,1,1,2,2), y=c(1,2,3,1,2), v=(1:5)/5)
> A
   x y   v
1: 1 1 0.2
2: 1 2 0.4
3: 1 3 0.6
4: 2 1 0.8
5: 2 2 1.0

第一步是获取每个 x组的元素/条目数相同。在这里，对于x = 1，存在3个y值，但是对于x = 2，只有2个值。因此，我们必须首先用NA来解决x = 2，y = 3。

The first step is to get the number of elements/entries for each group of "x" to be the same. Here, for x=1 there are 3 values of y, but only 2 for x=2. So, we'll have to fix that first with NA for x=2, y=3.

setkey(A, x, y)
A[CJ(unique(x), unique(y))]

现在，要使其具有较宽的格式，我们应该按 x分组，并在 v 上使用 as.list 如下：

Now, to get it to wide format, we should group by "x" and use as.list on v as follows:

out <- A[CJ(unique(x), unique(y))][, as.list(v), by=x]
   x  V1  V2  V3
1: 1 0.2 0.4 0.6
2: 2 0.8 1.0  NA

现在，您可以使用带有 setnames 的引用来设置重塑列的名称，如下所示：

Now, you can set the names of the reshaped columns using reference with setnames as follows:

setnames(out, c("x", as.character(unique(A$y)))

   x   1   2   3
1: 1 0.2 0.4 0.6
2: 2 0.8 1.0  NA

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