R（data.table）按自定义范围（例如，-18，18-25，…，65+；）对数据进行分组

我在R（使用data.table）中找不到按自定义范围（例如，-18，18-25，…，65+）而不是按单个值对数据进行分组的解决方案

我现在使用的是：

DT[,list(M_Savings=mean(Savings), M_Term=mean(Term)), by=Age] [order (Age)]

这给了我以下结果：

    Age     M_Savings   M_Term
1:  18      6500        5.5 
2:  19      7000        6.2 
3:  20      7200        5.8
...
50: 68      4000        4.2 

理想结果：

    Age     M_Savings   M_Term
1:  18-25   7450        5.5 
2:  25-30   8320        6.2 
...
50: 65+     3862        4.3 

我希望我的解释足够清楚。

---

非常感谢您提供的任何帮助。

带有数字变量的示例，它使以下内容：

          test         BucketName
1    615.59148      01. 0 - 5,000
2   1135.42357      01. 0 - 5,000
3   5302.24208 02. 5,000 - 10,000
4   3794.23109      01. 0 - 5,000
5   2773.70667      01. 0 - 5,000
...

代码是什么

generateLabelsForPivot = function(breaksVector)
{

    startValue = min(breaksVector)
    lastValue  = max(breaksVector)

    lengthOfBreaks = length(breaksVector)
    orders         = seq(1, lengthOfBreaks-1, 1)
    startingPoints = c(breaksVector[-length(breaksVector)])
    finishPoints   = c(breaksVector[-1])

    addingZeros = function(X)
    {
        prefix = ""

        if(nchar(X) == 1)
        {
            prefix = "0"
        } else {
            prefix = ""
        }

        return(paste(prefix, X, ". ", sep = ""))
    }

    orderPrefixes = sapply(orders, addingZeros)
    startingPoints.pretty = prettyNum(startingPoints, scientific=FALSE, big.mark=",", preserve.width = "none")
    finishPoints.pretty   = prettyNum(finishPoints, scientific=FALSE, big.mark=",", preserve.width = "none")
    labels = paste(orderPrefixes, startingPoints.pretty, " - ", finishPoints.pretty, sep = "")
    return(labels)
}


dataFrame = data.frame(test = runif(100, 0, 100*100))

GCV_breaks = c(0, 5000, 10000, 20000, 30000, 1000000) 
GCV_labels = generateLabelsForPivot(GCV_breaks) 
GCV_labels 
GCV_buckets = cut(dataFrame$test, breaks = GCV_breaks, labels = GCV_labels) 

dataFrame$BucketName = GCV_buckets

---

@哈里森是对的：

cut（…）

是正确的选择

library(data.table)
# create sample - you have this already
set.seed(1)   # for reproducibility
DT <- data.table(age=sample(15:70,1000,replace=TRUE),
                 value=rpois(1000,10))

# you start here...
breaks <- c(0,18,25,35,45,65,Inf)
DT[,list(mean=mean(value)),by=list(age=cut(age,breaks=breaks))][order(age)]
#         age      mean
# 1:   (0,18] 10.000000
# 2:  (18,25]  9.579365
# 3:  (25,35] 10.158192
# 4:  (35,45]  9.775510
# 5:  (45,65]  9.969697
# 6: (65,Inf] 10.141414

库（data.table）
#创建示例-您已经有了此示例
设定种子（1）#用于再现性
DT您是否尝试过使用
cut
？或
findInterval
（应该很快）我非常感谢您的回答-cut（）完全符合我的搜索要求！谢谢！