Question

我在R中有一个data.table：

library(data.table)
set.seed(1)
DT = data.table(
  group=sample(letters[1:2],100,replace=TRUE), 
  year=sample(2010:2012,100,replace=TRUE),
  v=runif(100))

按组和年份将这些数据汇总到汇总表中简单而优雅：

table <- DT[,mean(v),by='group, year']

但是，将这些数据汇总到汇总表（包括小计和总计）中会稍微困难一点，并且不那么优雅：

library(plyr)
yearTot <- DT[,list(mean(v),year='Total'),by='group']
groupTot <- DT[,list(mean(v),group='Total'),by='year']
Tot <- DT[,list(mean(v), year='Total', group='Total')]
table <- rbind.fill(table,yearTot,groupTot,Tot)
table$group[table$group==1] <- 'Total'
table$year[table$year==1] <- 'Total'

这会产生：

table[order(table$group, table$year), ]

是否有一种简单的方法可以使用data.table指定小计和总计，例如plyr的margins=TRUE命令？我更喜欢在我的数据集上使用data.table over plyr，因为它是一个非常大的数据集，我已经拥有data.table格式。

Answer 1

在最近的devel data.table中，您可以使用名为“分组集”的新功能来生成子总计：

$ pip install git+http://github.com/networkx/networkx

Answer 2

我不知道一个简单的方法。这是对实现的第一次尝试。我不知道plyr中的margins=TRUE，这是做什么的？

crossby = function(DT, j, by) {
    j = substitute(j)
    ans = rbind(
        DT[,eval(j),by],
        DT[,list("Total",eval(j)),by=by[1]],
        cbind("Total",DT[,eval(j),by=by[2]]),
        list("Total","Total",DT[,eval(j)]),
        use.names=FALSE
        # 'use.names' argument added in data.table v1.8.0
    )
    setkeyv(ans,by)
    ans
}

crossby(DT, mean(v), c("group","year"))

      group  year        V1
 [1,]     a  2010 0.2926945
 [2,]     a  2011 0.4176346
 [3,]     a  2012 0.4227796
 [4,]     a Total 0.3901875
 [5,]     b  2010 0.5231845
 [6,]     b  2011 0.4997119
 [7,]     b  2012 0.4306871
 [8,]     b Total 0.4835788
 [9,] Total  2010 0.4278093
[10,] Total  2011 0.4463616
[11,] Total  2012 0.4271160
[12,] Total Total 0.4350153

Answer 3

请参阅下面的解决方案 - 类似于上面的@ MattDowle - 可以使用任意数量的组。

crossby2 <- function(data, j, by, grand.total = T, total.label = "(all)", value.label = "value") {
  j = substitute(j)

  # Calculate by each group
  lst <- lapply(1:length(by), function(i) {
    x <- data[, list(..VALUE.. = eval(j)), by = eval(by[1:i])]
    if (i != length(by)) x[, (by[-(1:i)]) := total.label]
    return(x)
  })

  # Grand total
  if (grand.total) lst <- c(lst, list(data[, list(..VALUE.. = eval(j))][, (by) := total.label]))

  # Combine all tables
  res <- rbindlist(lst, use.names = T, fill = F)

  # Change value column name
  setnames(res, "..VALUE..", value.label)

  # Set proper column order
  setcolorder(res, c(by, value.label))

  # Sort values
  setkeyv(res, by)

  return(res)
}

Answer 4

使用当前答案我已经添加了对多个度量和聚合函数的支持，并且可以添加聚合级别指示符。

#' @title SQL's ROLLUP function
#' @description Returns data.table of aggregates value for each level of hierarchy provided in `by`.
#' @param x data.table input data.
#' @param j expression to evaluate in `j`, support multiple measures.
#' @param by character a hierarchy level for aggregations.
#' @param level logical, use `TRUE` to add `level` column of sub-aggregation.
#' @seealso [postgres: GROUPING SETS, CUBE, and ROLLUP](http://www.postgresql.org/docs/9.5/static/queries-table-expressions.html#QUERIES-GROUPING-SETS), [SO: Aggregating sub totals and grand totals with data.table](http://stackoverflow.com/a/24828162/2490497)
#' @return data.table
#' @examples 
#' set.seed(1)
#' x = data.table(group=sample(letters[1:2],100,replace=TRUE),
#'                year=sample(2010:2012,100,replace=TRUE),
#'                v=runif(100))
#' rollup(x, .(vmean=mean(v), vsum=sum(v)), by = c("group","year"))
library(data.table)
rollup = function(x, j, by, level=FALSE){
    stopifnot(is.data.table(x), is.character(by), length(by) >= 2L, is.logical(level))
    j = substitute(j)
    aggrs = rbindlist(c(
        lapply(1:(length(by)-1L), function(i) x[, eval(j), c(by[1:i])][, (by[-(1:i)]) := NA]), # subtotals
        list(x[, eval(j), c(by)]), # leafs aggregations
        list(x[, eval(j)][, c(by) := NA]) # grand total
    ), use.names = TRUE, fill = FALSE)
    if(level) aggrs[, c("level") := sum(sapply(.SD, is.na)), 1:nrow(aggrs), .SDcols = by]
    setcolorder(aggrs, neworder = c(by, names(aggrs)[!names(aggrs) %in% by]))
    setorderv(aggrs, cols = by, order=1L, na.last=TRUE)
    return(aggrs[])
}
set.seed(1)
x = data.table(group=sample(letters[1:2],100,replace=TRUE),
               year=sample(2010:2012,100,replace=TRUE),
               month=sample(1:12,100,replace=TRUE),
               v=runif(100))
rollup(x, .(vmean=mean(v), vsum=sum(v)), by = c("group","year","month"), level=TRUE)

Answer 5

借用此答案（https://stackoverflow.com/a/39536828/4241780），下面提供了一个全子集摘要（与crossby2不同，而rollup似乎错过了OP＆＃39的第9行到第11行; s期望的输出）。此函数可扩展为任意数量的by或aggregate变量，但在其当前状态下仅允许一种类型的聚合函数。非常适合通过组交互计算行子站点（我用它来做）。

add_col_sums.data.table <- function(data, aggvars, byvars, FUN = sum, level = "level") {

  # Find all possible subsets of your data
  subsets <- lapply(0:length(byvars), combn, x = byvars, simplify = FALSE)
  subsets <- do.call(c, subsets)

  # Calculate summary value by each subset
  agg_values <- lapply(subsets, function(x) 
    data[,lapply(.SD, FUN), by = x, .SDcols = aggvars])

  # Pull them all into one dataframe
  dat_out <- rbindlist(agg_values, fill = TRUE)

  # Order columns and rows
  setorderv(dat_out, byvars, na.last = TRUE)
  setcolorder(dat_out, c(byvars, aggvars))

  # Add level indication
  dat_out[, c(level) := Reduce("+", lapply(.SD, is.na))]

  # Return data.table
  dat_out[]

}

add_col_sums.data.table(DT, "v", c("group", "year"), FUN = mean)

使用data.table聚合子总计和总计

5 个答案: