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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顺序获得第 n 高。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，但收到一条信息不完整的错误消息。

 aggregate.data.frame(x, list(x$Group), function(y){ max(y[,'Quantity'])})
 Error in `[.default`(y, , "Quantity") (from #1) : incorrect number of dimensions"
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4 回答 4

2 
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 , ]
于 2013-01-20T03:45:25.900 回答
1

一些聚合合并魔法：

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
于 2013-01-20T03:42:30.530 回答
1 
# 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

您看到该错误的原因aggregate.data.frame是因为此函数FUN根据by参数应用于每一列，而不是完整的每个子集data.frame（这就是by函数的用途，如上所示）。使用时aggregate，您提供的任何内容都FUN应该接受列，而不是data.frames。在您的示例中，您尝试y像 a 那样索引向量data.frame，因此出现尺寸错误。

于 2013-01-20T06:16:06.623 回答
0

我和

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

按照别人的建议。我发现它比其他发布的解决方案更适合我的思维过程（我很欣赏）。

于 2013-01-20T22:47:20.527 回答