Question

相对较新的R，正在处理一个包含数百万行的项目，所以我做了这个例子：
我有一个包含三行不同数据的矩阵。
如果[，1] [，2] [Farm]的组合总共少于两个观察值，则该行的[Farm]值将更改为q99999。这样他们就属于同一组，以便以后分析。

    A <- matrix(c(1,1,2,3,4,5,5), ncol = 7)
    B <- matrix(c(T,T,F,T,F,T,T), ncol = 7)
    C <- matrix(c("Req","Req","Req","fd","as","f","bla"), ncol = 7)
    AB <- rbind.fill.matrix(A,B, C)
    AB <-t(AB)
    colnames(AB) <- c("Col1", "Col2", "Farm")
    format(AB)

     Col1  Col2  Farm
    1 "1  " "1  " "Req"
    2 "1  " "1  " "Req"
    3 "2  " "0  " "Req"
    4 "3  " "1  " "fd "
    5 "4  " "0  " "as "
    6 "5  " "1  " "f  "
    7 "5  " "1  " "bla"

所以预期结果如下：

     Col1  Col2  Farm
    1 "1  " "1  " "Req"
    2 "1  " "1  " "Req"
    3 "2  " "0  " "q99999"
    4 "3  " "1  " "q99999"
    5 "4  " "0  " "q99999"
    6 "5  " "1  " "q99999"
    7 "5  " "1  " "q99999"

现在“Farm”，“Req”和“q99999”

列有两组

在尽可能快地保持性能的同时，R的最佳方法是什么？

Answer 1

使用data.table包的可能解决方案：

library(data.table)

as.data.table(AB)[,Farm:=ifelse(.N>1, Farm, "q99999"),.(Col1, Col2, Farm)][]

#   Col1 Col2   Farm
#1:    1    1    Req
#2:    1    1    Req
#3:    2    0 q99999
#4:    3    1 q99999
#5:    4    0 q99999
#6:    5    1 q99999
#7:    5    1 q99999

或以R为基础ave：

AB[,'Farm'] = ave(AB[,'Farm'], do.call(c,apply(AB,2,list)), FUN=function(x) ifelse(length(x)==1, 'q99999',x))

#  Col1 Col2 Farm    
#1 "1"  "1"  "Req"   
#2 "1"  "1"  "Req"   
#3 "2"  "0"  "q99999"
#4 "3"  "1"  "q99999"
#5 "4"  "0"  "q99999"
#6 "5"  "1"  "q99999"
#7 "5"  "1"  "q99999"

当多行/列组合不符合要求时更改值

1 个答案: