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我正在寻求帮助编写一个函数，该函数可以识别数据集中给定客户的值的趋势（“正/负/混合”，请参见下面的定义）。

我有以下交易数据；所有客户每人都有 3-13 笔交易。

customer_ID transaction_num sales
Josh         1              $35
Josh         2              $50
Josh         3              $65
Ray          1              $65
Ray          2              $52
Ray          3              $49
Ray          4              $15
Eric         1              $10 
Eric         2              $13
Eric         3              $9

我想在 R 中编写一个函数来填充一个新的数据框，如下所示

Customer_ID     Sales_Slope  
Josh              Positive
Ray               Negative
Eric               Mixed

在哪里：

Josh 的斜率是正数，因为他的所有交易销售成本都会随着每增加一个购物点而继续增加

Ray 的斜率是负数，因为他的所有交易销售成本随着每增加一个购物点而继续降低

Eric 的斜率参差不齐，因为他所有的交易销售成本都在波动……没有明确的趋势……

我自己已经尝试过非常广泛地这样做，但被卡住了..这是一些我能够放在一起的伪代码

counter = max(transaction_num)
while counter >= 0 
 if sales at max transaction_num are greater than sales at max transaction_num - 1) 
   then counter = counter - 1 ; else "not positive slope trend"
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2 回答 2

2

简单的答案是使用diff. 它只是从下一个值中减去当前值，因此如果所有值diff(x)都高于零，则它正在增加，反之亦然。首先，读取数据：

# Read in some data.
data<-read.table(textConnection('customer_ID transaction_num sales
Josh         1              $35
Josh         2              $50
Josh         3              $65
Ray          1              $65
Ray          2              $52
Ray          3              $49
Ray          4              $15
Eric         1              $10 
Eric         2              $13
Eric         3              $9'),header=TRUE,stringsAsFactors=FALSE)

data$sales<-as.numeric(sub('\\$','',data$sales))

现在代码： 

# Diff subtracts next value from current in a diff.
# so diff(c(1,2,3,4)) is c(1,1,1)
direction<-function(x){
  if(all(diff(x)>0)) return('Increasing')
  if(all(diff(x)<0)) return('Decreasing')
  return('Mixed')
}

# If you want a vector.
c(by(data$sales,data$customer_ID,direction))
#   Eric         Josh          Ray 
# "Mixed" "Increasing" "Decreasing"

# If you want to a little data frame.
aggregate(sales~customer_ID,data,direction)
#   customer_ID      sales
# 1        Eric      Mixed
# 2        Josh Increasing
# 3         Ray Decreasing
于 2014-05-12T03:12:30.957 回答
2

我想我会从这样的事情开始。 data.table通常对于更大的数据集非常有效。

#Make fake data
require("data.table")
data <- data.table(customer_ID=c(rep("Josh",3),rep("Ray",4),rep("Eric",3)),
                   sales=c(35,50,65,65,52,49,15,10,13,9))
data[,transaction_num:=seq(1,.N),by=c("customer_ID")]

现在是实际代码。

data <- data.table(data)

#Calculate difference in rolling two time periods
rolled.up <- data[,list(N.Minus.1=.N-1,Change=list(
  sales[transaction_num+1]-sales[transaction_num])),
  by=c("customer_ID")]

#Sum up positive and negative values
rolled.up[,Pos.Values:=as.numeric(lapply(Change,FUN=function(x) {sum(1*(x>0),na.rm=T)}))]
rolled.up[,Neg.Values:=(N.Minus.1-Pos.Values)]

#Make Sales Slope variable
rolled.up[,Sales_Slope:=ifelse(Pos.Values>0 & Neg.Values==0,"Positive",
      ifelse(Pos.Values==0 & Neg.Values>0,"Negative","Mixed"))]

#Make final table
final.table <- rolled.up[,list(customer_ID,Sales_Slope)]
final.table

#      customer_ID Sales_Slope
# 1:        Josh    Positive
# 2:         Ray    Negative
# 3:        Eric       Mixed

#You can always merge this result back onto your main dataset if you want
data <- merge(x=data,y=final.table,by=c("customer_ID"),all.x=T)
于 2014-05-12T03:35:28.583 回答