r - 用 R data.table 填充缺失的行

Question

我在 R 中有一个 data.table，它是从如下所示的数据库中获取的:

date,identifier,description,location,value1,value2
2014-03-01,1,foo,1,100,200
2014-03-01,1,foo,2,200,300
2014-04-01,1,foo,1,100,200
2014-04-01,1,foo,2,100,200
2014-05-01,1,foo,1,100,200
2014-05-01,1,foo,2,100,200
2014-03-01,2,bar,1,100,200
2014-04-01,2,bar,1,100,200
2014-05-01,2,bar,1,100,200
2014-03-01,3,baz,1,100,200
2014-03-01,3,baz,2,200,300
2014-04-01,3,baz,1,100,200
2014-04-01,3,baz,2,100,200
2014-05-01,3,baz,1,100,200
2014-05-01,3,baz,2,100,200
2014-05-01,4,quux,2,100,200
<SNIP>

为了对数据进行一些计算，我想对其进行按摩，以便日期、标识符、描述和位置的每个组合在表中都有一行，NA 为 value1 和 value2。我知道日期范围和位置的所有潜在值。

我是 R 和 data.table 的新手，此时我的头脑很困惑。我想为上述示例表得出的结果是:

date,identifier,description,location,value1,value2
2014-03-01,1,foo,1,100,200
2014-03-01,1,foo,2,200,300
2014-04-01,1,foo,1,100,200
2014-04-01,1,foo,2,100,200
2014-05-01,1,foo,1,100,200
2014-05-01,1,foo,2,100,200
2014-03-01,2,bar,1,100,200
2014-03-01,2,bar,2,NA,NA
2014-04-01,2,bar,1,100,200
2014-04-01,2,bar,2,NA,NA
2014-05-01,2,bar,1,100,200
2014-05-01,2,bar,2,NA,NA
2014-03-01,3,baz,1,100,200
2014-03-01,3,baz,2,200,300
2014-04-01,3,baz,1,100,200
2014-04-01,3,baz,2,100,200
2014-05-01,3,baz,1,100,200
2014-05-01,3,baz,2,100,200
2014-03-01,4,quux,1,NA,NA
2014-03-01,4,quux,2,NA,NA
2014-04-01,4,quux,1,NA,NA
2014-04-01,4,quux,2,NA,NA
2014-05-01,4,quux,1,NA,NA
2014-05-01,4,quux,2,100,200

数据库中的数据是稀疏的，因为给定的标识符/描述/位置组合对于每个日期可能有任意数量的条目或根本没有条目。我想在给定的日期范围内(例如，2014-03-01 到 2014-05-01)每个标识符/描述和位置在表中都有一行。

这似乎是一个有趣的 data.table 技巧要做的事情，但我正在空白。

编辑:我通过合并另一个数据表对一个标识符/描述进行了较小规模的此操作，但我不确定如何在增加多个标识符/描述和位置的复杂性的情况下执行此操作。

非常感谢您的回复。

这是原始数据的 dput 输出，可以很容易地复制到 R 中:

structure(list(date = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 1L, 2L, 3L, 1L, 1L, 2L, 2L, 3L, 3L, 3L), 
.Label = c("2014-03-01", "2014-04-01", "2014-05-01"), class = "factor"), 
identifier = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L),     
description = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 4L), 
.Label = c("bar", "baz", "foo", "quux"), class = "factor"), 
location = c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 2L), 
value1 = c(100L, 200L, 100L, 100L, 100L, 100L, 100L, 100L, 100L, 100L, 200L, 100L, 100L, 100L, 100L, 100L), 
value2 = c(200L, 300L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 300L, 200L, 200L, 200L, 200L, 200L)), 
.Names = c("date", "identifier", "description", "location", "value1", "value2"), 
row.names = c(NA, -16L),
class = c("data.table", "data.frame"))

Answer 1

在@akrun 和@eddi 的帮助下，这是惯用的 (?) 方式:

mycols  = c("description","date","location")
setkeyv(DT0,mycols)
DT1 <- DT0[J(do.call(CJ,lapply(mycols,function(x)unique(get(x)))))]
# alternately: DT1 <- DT0[DT0[,do.call(CJ,lapply(.SD,unique)),.SDcols=mycols]]

identifier新行缺少列，但可以填充:

setkey(DT1,description)
DT1[unique(DT0[,c("description","identifier")]),identifier:=i.identifier]