Question

我在OLapCube软件包data.cube中遇到一些问题：

install.packages("data.cube", repos = paste0("https://", c(
    "jangorecki.gitlab.io/data.cube",
    "cloud.r-project.org"
)))

一些测试数据：

 library(data.table)
 set.seed(42)

 dt <- CJ(color = c("green","yellow","red"),
            year = 2011:2015,
            month = 1:12,
            status = c("active","inactive","archived","removed")
 )[sample(600)]

 dt[, "value" := sample(4:7/2, nrow(dt), TRUE)]

现在，我想创建一个多维数据集并在时间维度上应用层次结构。像这样：

library(data.cube)
dc <- as.data.cube(dt, id.vars = c("color", "year", "month", "status"), 
                   measure.vars = "value", 
                   hierarchies = list(time <- list("year, month")))

如果我运行此代码，则会收到错误消息：

Error in as.data.cube.data.table(dt, id.vars = c("color", "year", "month",  : 
  identical(names(hierarchies), id.vars) | identical(names(hierarchies),  .... is not TRUE

如果我尝试类似

hierarchies = list(time <- list("year, month"), color <- list("color"), 
                  status <- list("status"))

我遇到同样的错误。

Answer 1

写得很好的问题。
我看到您是根据?as.data.cube个示例制作的示例，因此我也会尝试使用该示例来回答您的问题

# Original example goes as follows
library(data.cube)
library(data.table)
set.seed(1L)
dt = CJ(color = c("green","yellow","red"),
        year = 2011:2015,
        status = c("active","inactive","archived","removed"))[sample(30)]
dt[, "value" := sample(4:7/2, nrow(dt), TRUE)]

dc = as.data.cube(
  x = dt, id.vars = c("color","year","status"),
  measure.vars = "value",
  hierarchies = sapply(c("color","year","status"),
                       function(x) list(setNames(list(character()), x)),
                       simplify=FALSE)
)
str(dc)

检查层次结构的有效性时，似乎会出现您的错误。
不幸的是，这不是非常有意义的错误，我创建了问题#18，因此有一天会得到改善。
因此，让我们比较手动层次结构和示例中创建的层次结构。

sapply(c("color","year","status"),
       function(x) list(setNames(list(character()), x)),
       simplify=FALSE) -> h
str(h)
#List of 3
# $ color :List of 1
#  ..$ :List of 1
#  .. ..$ color: chr(0) 
# $ year  :List of 1
#  ..$ :List of 1
#  .. ..$ year: chr(0) 
# $ status:List of 1
#  ..$ :List of 1
#  .. ..$ status: chr(0)     

hierarchies = list(time <- list("year, month"), color <- list("color"), 
                   status <- list("status"))
str(hierarchies)
#List of 3
# $ :List of 1
#  ..$ : chr "year, month"
# $ :List of 1
#  ..$ : chr "color"
# $ :List of 1
#  ..$ : chr "status"

我们可以看到手册中的层次结构是命名元素的列表，而您的示例是未命名元素的列表。
我相信您误用了<-，其中应该使用=。 <-并不总是等于=运算符。您可以在3.1.3.1 Assignment <- vs =中详细了解这种情况。

所以让我们看看修复是否足够

hierarchies = list(time = list(c("year, month")), color = list("color"), 
                   status = list("status"))

dc <- as.data.cube(dt, id.vars = c("color", "year", "month", "status"), 
                   measure.vars = "value", 
                   hierarchies = hierarchies)

我们仍然遇到相同的错误，因此需要输入名称，而不是问题的根本原因。仔细查看后，我现在看到您要构建没有主键的 time 维度。
重要说明，您不能将多个列名作为单个字符串传递

"year, month"

应写为

c("year","month")

仍然需要 time 维主键为单个字段， year 和 month 只是这些属性。
因此，让我们为 time 维度创建主键，然后，由于我们的时间维度具有年月粒度，因此我们将在该粒度上创建密钥。

library(data.table)
set.seed(42)

dt <- CJ(color = c("green","yellow","red"),
         year = 2011:2015,
         month = 1:12,
         status = c("active","inactive","archived","removed")
)[sample(600)
  ][, yearmonth:=sprintf("%04d%02d", year, month) # this ensure four numbers for year and 2 numbers for month
    ]

dt[, "value" := sample(4:7/2, nrow(dt), TRUE)]

现在让我们进行层次结构设置，注意year已更改为yearmonth。在下面的层次结构中，值c("year","month")的向量表示这些属性依赖于yearmonth。请参见?as.data.cube中的更多示例，以了解更复杂的层次结构情况。

hierarchies = list(
  color = list(color = list(color = character())),
  yearmonth = list(yearmonth = list(yearmonth = c("year","month"))),
  status = list(status = list(status = character()))
)

dc = as.data.cube(
  x = dt, id.vars = c("color","yearmonth","status"),
  measure.vars = "value",
  hierarchies = hierarchies
)
str(dc)

我们的data.cube已成功创建。让我们尝试使用yearmonth

键进行查询

dc[, .(yearmonth=201105L)] -> d
as.data.table(d)
dc[, .(yearmonth=201105L), drop=FALSE] -> d
as.data.table(d)

现在尝试使用维度，年份和月份以及两者的属性来查询它

dc[, .(year=2011L)] -> d
as.data.table(d) # note that dimension is not being dropped because it still have more than 1 value
dc[, .(month=5L)] -> d
as.data.table(d)
dc[, .(year=2011L, month=5L)] -> d
as.data.table(d) # here dimension has been dropped because there was only single element in that dimension, you can of course use `drop=FALSE` if needed.

希望有帮助，祝你好运！

R数据立方体定义层次结构

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