使用R data.table填写缺少的行

Question

我在R中有一个data.table,它是从一个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,我想按摩它,以便日期,标识符,描述和位置的每个组合在表中有一行,其中NA为value1和value2.我知道日期的范围和所有可能的位置值.

我是R和data.table的新手,我的思绪在这一点上很难.我想为上面的示例表提出的结果是:

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/位置组合可以具有任何数量的条目或者对于每个日期根本没有条目.我想达到给定的日期范围(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数据.可行的技巧,但我在消隐.

编辑:我通过合并另一个数据表以较小的比例为一个标识符/描述做了这个,但我不知道如何通过增加多个标识符/描述和位置的复杂性来做到这一点.

非常感谢您的回复.

这是原始数据的输出输出,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新行缺少该列，但可以填充：

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