Question

如何在R中按组计算订单统计信息。我想根据列聚合结果，然后每组只返回1行。根据某种顺序，该行应该是该组的第n个元素。理想情况下，我只想使用基本功能。

x <- data.frame(Group=c("A","A", "A", "C", "C"), 
                Name=c("v", "u", "w", "x", "y"), 
                Quantity=c(3,3,4,2,0))
> x
  Group Name Quantity
1     A    v        3
2     A    u        3
3     A    w        4
4     C    x        2
5     C    y        0

我想根据数量和名称的排序取得第n名。对于N = 2，这是

  Group Name Quantity
1     A    u        3
5     C    y        0

For N=1
  Group Name Quantity
3     A    w        4
4     C    x        2

我尝试了以下操作，但收到了无法提供信息的错误消息。

 aggregate.data.frame(x, list(x$Group), function(y){ max(y[,'Quantity'])})
 Error in `[.default`(y, , "Quantity") (from #1) : incorrect number of dimensions"

Answer 1

x <- 
    data.frame(
        Group = c("A","A", "A", "C", "C", "A", "A") , 
        Name = c("v", "u", "w", "x", "y" ,"v", "u") , 
        Quantity = c(3,3,4,2,0,4,1)
    )

# sort your data to start..
# note that Quantity vs. Group and Name
# are sorted in different directions,
# so the -as.numeric() flips them
x <- 
    x[ 
        order( 
            -as.numeric( x$Group ) , 
            x$Quantity , 
            -as.numeric( x$Name ) , 
            decreasing = TRUE 
        ) , 
    ]
# once your data frame is sorted the way you want your Ns to occur, the rest is easy

# rank your data..  
# just create the numerical order, 
# but within each group..
# (or you could add those ranks directly to the data frame if you like)
ranks <- 
    unlist( 
        tapply( 
            order( x$Group ) , 
            as.numeric( x$Group ) , 
            order 
        ) 
    )

# N = 1
x[ ranks == 1 , ]

# N = 2
x[ ranks == 2 , ]

Answer 2

一些聚合魔法：

f <- function(x, N) {
  sel <- function(x) {                                   # Choose the N-th highest value from the set, or lowest element if there < N unique elements.  Is there a built-in for this? 
    z <- unique(x)                                       # This assums that you wan the N-th highest unique value.  Simply don't filter by unique if not.
    z[order(z, decreasing=TRUE)][min(N, length(z))]
  }

  xNq <- aggregate(Quantity ~ Group, data=x,   sel)      # Choose the N-th highest quantity within each "Group"
  xNm <- merge(x, xNq)                                   # Add the matching "Name" values
  x <- aggregate(Name ~ Quantity + Group, data=xNm, sel) # Choose the N-th highest Name in each group
  x[c('Group', 'Name', 'Quantity')]                      # Put into original order
}


> f(x, 2)
##   Group Name Quantity
## 1     A    u        3
## 2     C    y        0

> f(x, 1)
##   Group Name Quantity
## 1     A    w        4
## 2     C    x        2

Answer 3

# define ordering function, increasing on Quantity, decreasing on Name
in.order <- function(group) with(group, group[order(Quantity, -rank(Name)), ])

# set desired rank for each Group
N <- 2

# get Nth row by Group, according to in.order
group.rows <- by(x, x$Group, function(group) head(tail(in.order(group), N), 1))

# collapse rows into data.frame
do.call(rbind, group.rows)

#   Group Name Quantity
# A     A    u        3
# C     C    y        0

您在aggregate.data.frame看到错误的原因是因为此函数根据FUN参数在每列上应用by，而不是每个列的data.frame （这就是by函数的用途，如上所示）。使用aggregate时，您提供给FUN的任何内容都应该是接受列，而不是data.frame。在您的示例中，您尝试将向量y索引为data.frame，因此维度错误。

Answer 4

我去了

do.call(rbind, by(x, x$Group, function(x)
      x[order(-x$Quantity, x$Name),][1,]))

根据别人的建议。我发现它比我的思考过程更适合其他发布的解决方案（我很欣赏）。

如何在R中按组创建订单统计信息？

4 个答案: