Question

我有两个data.tables有34列，其中列完全相同。

Month SpId1 SpId2 ... SpId33

编辑：以下是使用Reproducible Example

中的重现功能的示例数据

AltSuitSp1 <- data.table(structure(list(Month = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L,12L, 12L, 12L), .Label = c("1", "10", "11", "12", "2", "3", "4","5", "6", "7", "8", "9"), class = "factor"), SpdSpSuit = c(0,0, 0, 0, 0, 0, 0, 0, 0, 0), SpdIncSuit = c(0, 0, 0,0, 0, 0, 0, 0, 0, 0), SpdGrowSuit = c(0.4625, 0.4625, 0.4625, 0.4625, 0.4625, 0.4625, 0.4625,0.4625, 0.4625, 0.4625), RzbSpSuit = c(0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333,0.283333333, 0.283333333, 0.283333333, 0.283333333), RzbIncSuit = c(0.34,0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34), RzbGrowSuit = c(0.283333333,0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333), FMSSpSuit = c(0.34,0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34), FMSIncSuit = c(0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425), FMSGrowSuit = c(0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333), BhsSpSuit = c(0.283333333,0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.233333333, 0.233333333, 0.233333333), BhsIncSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), BhsGrowSuit = c(0.283333333,0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.283333333, 0.233333333, 0.233333333, 0.233333333), BrtSpSuit = c(0.866666667,0.866666667, 0.866666667, 0.866666667, 0.866666667, 0.866666667,0.866666667, 0.54, 0.54, 0.54), BrtIncSuit = c(0.8, 0.8, 0.8, 0.8, 0.8,0.8, 0.8, 0.43, 0.43, 0.43), BrtGrSuit = c(0.8, 0.8, 0.8, 0.8, 0.8, 0.8,0.8, 0.86, 0.86, 0.86), CcfSpSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), CcfIncSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0),CcfGrSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), GsfSpSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), GsfIncSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), GsfGrSuit = c(0, 0, 0, 0, 0,0, 0, 0, 0, 0), RbtSpSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), RbtIncSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0,0), RbtGrSuit = c(0.95, 0.95, 0.95, 0.95, 0.95, 0.95, 0.95, 0.925, 0.925, 0.925), SmbSpSuit = c(0.7, 0.7, 0.7, 0.7, 0.7, 0.7, 0.7, 0.675, 0.675,0.675), SmbIncSuit = c(0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 0.766666667,0.766666667, 0.766666667), SmbGrSuit = c(0.9, 0.9, 0.9, 0.9, 0.9, 0.9, 0.9, 0.0875, 0.0875, 0.0875), StbSpSuit = c(0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425, 0.425), StbIncSuit = c(0, 0,0, 0, 0, 0, 0, 0, 0, 0), StbGrSuit = c(0, 0, 0, 0, 0,0, 0, 0, 0, 0), HbcSpSuit = c(0, 0, 0, 0, 0, 0, 0,0, 0, 0), HbcIncSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0,0), HbcGrSuit = c(0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.425, 0.425, 0.425)), .Names = c("Month", "SpdSpSuit", "SpdIncSuit", "SpdGrowSuit","RzbSpSuit", "RzbIncSuit", "RzbGrowSuit", "FMSSpSuit", "FMSIncSuit","FMSGrowSuit", "BhsSpSuit", "BhsIncSuit", "BhsGrowSuit", "BrtSpSuit","BrtIncSuit", "BrtGrSuit", "CcfSpSuit", "CcfIncSuit", "CcfGrSuit","GsfSpSuit", "GsfIncSuit", "GsfGrSuit", "RbtSpSuit", "RbtIncSuit","RbtGrSuit", "SmbSpSuit", "SmbIncSuit", "SmbGrSuit", "StbSpSuit","StbIncSuit", "StbGrSuit", "HbcSpSuit", "HbcIncSuit", "HbcGrSuit"), class = c("data.table", "data.frame"), row.names = c(NA, -10L))) 

AltSuitDates <- data.table(structure(list(Month = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 10L, 11L,12L), SpdSpT = c(NA, NA, NA, NA, NA, 1L, 1L, NA, NA, NA), SpdIncT = c(NA,NA, NA, NA, NA, 1L, 1L, NA, NA, NA), SpdGrT = c(1L, 1L, 1L, 1L,1L, 1L, 1L, 1L, 1L, 1L), RzbSpT = c(NA, NA, NA, 1L, 1L, 1L, NA,NA, NA, NA), RzbIncT = c(NA, NA, NA, 1L, 1L, 1L, NA, NA, NA,NA), RzbGrT = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), FmsSpT = c(NA,NA, 1L, 1L, NA, NA, NA, NA, NA, NA), FmsIncT = c(NA, NA, 1L,1L, 1L, NA, NA, NA, NA, NA), FMSGrT = c(1L, 1L, 1L, 1L, 1L, 1L,1L, 1L, 1L, 1L), BhsSpT = c(NA, NA, NA, 1L, 1L, 1L, NA, NA, NA,NA), BhsIncT = c(NA, NA, NA, 1L, 1L, 1L, NA, NA, NA, NA), BhsGrT = c(1L,1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), BRTsp = c(1L, 1L, 1L, NA,NA, NA, NA, 1L, 1L, 1L), BRTinc = c(1L, 1L, 1L, 1L, NA, NA, NA,1L, 1L, 1L), BRTgr = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), CCFsp = c(NA, NA, NA, NA, 1L, 1L, 1L, NA, NA, NA), CCFinc = c(NA,NA, NA, NA, 1L, 1L, 1L, NA, NA, NA), CCFgr = c(1L, 1L, 1L, 1L,1L, 1L, 1L, 1L, 1L, 1L), GSFsp = c(NA, NA, NA, NA, 1L, 1L, 1L,NA, NA, NA), GSFinc = c(NA, NA, NA, NA, 1L, 1L, 1L, NA, NA, NA), GSFgr = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), RBTsp = c(1L,1L, 1L, 1L, 1L, 1L, 1L, NA, NA, NA), RBTinc = c(1L, 1L, 1L, 1L,1L, 1L, 1L, NA, NA, NA), RBTgr = c(1L, 1L, 1L, 1L, 1L, 1L, 1L,1L, 1L, 1L), SMBsp = c(NA, NA, NA, 1L, 1L, 1L, 1L, NA, NA, NA), SMBinc = c(NA, NA, NA, 1L, 1L, 1L, 1L, NA, NA, NA), SMBgr = c(1L,1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), STBsp = c(NA, NA, NA, NA,NA, 1L, 1L, NA, NA, NA), STBinc = c(NA, NA, NA, NA, NA, 1L, 1L,NA, NA, NA), STBgr = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), HBCsp = c(NA, NA, NA, 1L, 1L, 1L, NA, NA, NA, NA), HBCinc = c(NA,NA, NA, 1L, 1L, 1L, NA, NA, NA, NA), HBCgr = c(1L, 1L, 1L, 1L,1L, 1L, 1L, 1L, 1L, 1L)), .Names = c("Month", "SpdSpT", "SpdIncT","SpdGrT", "RzbSpT", "RzbIncT", "RzbGrT", "FmsSpT", "FmsIncT","FMSGrT", "BhsSpT", "BhsIncT", "BhsGrT", "BRTsp", "BRTinc", "BRTgr","CCFsp", "CCFinc", "CCFgr", "GSFsp", "GSFinc", "GSFgr", "RBTsp","RBTinc", "RBTgr", "SMBsp", "SMBinc", "SMBgr", "STBsp", "STBinc","STBgr", "HBCsp", "HBCinc", "HBCgr"), class = c("data.table","data.frame"), row.names = c(NA, -10L)))

其中SpId是物种标识符。一个DT是500多万行（AltSuitSp1），另一个是12（AltSuitDates）。我正在使用12行长（相当于12个月）的DT来更新更大的DT。目前我在for循环中使用嵌套的if，else if结构来检查条件并根据较小的DT更新较大的DT（参见下面的代码）

h <- 1
n <- length(AltSuitSp1[,Month])
stm <- AltSuitSp1[,Month]  #  AltSuitSp1 is the 5+ million row DT

cond1 <- which(stm == 1)  #  list of all rows of AltSuitSp1 where the Month is = 1
cond2 <- which(stm == 2)  #  list of all rows of AltSuitSp1 where the Month is = 2
...
cond12 <- which(stm == 12)

for (h in seq(n)){
    if (any(cond1 == h)){
        set(AltSuitSp1,h,2:34,(AltSuitSp1[h,2:34,with=F] * AltSuitDates[1,2:34,with=F]))
    }else if (any(cond2 == h)){
        set(AltSuitSp1,h,2:34,(AltSuitSp1[h,2:34,with=F] * AltSuitDates[2,2:34,with=F]))
    }else if ...
    }else if (any(cond12)){
        set(AltSuitSp1,h,2:34,(AltSuitSp1[h,2:34,with=F] * AltSuitDates[12,2:34,with=F]))
    }else
        break
}

现在，我已经运行了这段代码1分钟，并检查了它已经推进了多远。目前我看到每秒大约29-30个循环，并且h已经发展到大约1800次迭代。但是，即使每秒30个循环（速度相当慢Using Set in DT），此代码也需要大约2天才能完成。但是，正如下面的输出所示，它正在做我想要/期望它做的事情。

AltSuitSp1Results <- data.table(structure(list(Month = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L,12L, 12L, 12L), .Label = c("1", "10", "11", "12", "2", "3", "4","5", "6", "7", "8", "9"), class = "factor"), SpdSpSuit = c(NA,NA, NA, NA, NA, NA, NA, 0, 0, 0), SpdIncSuit = c(NA, NA, NA,NA, NA, NA, NA, 0, 0, 0), SpdGrowSuit = c(0, 0, 0, 0, 0, 0, 0,0.4625, 0.4625, 0.4625), RzbSpSuit = c(NA, NA, NA, NA, NA, NA,NA, 0.283333333, 0.283333333, 0.283333333), RzbIncSuit = c(NA,NA, NA, NA, NA, NA, NA, 0.34, 0.34, 0.34), RzbGrowSuit = c(0,0, 0, 0, 0, 0, 0, 0.283333333, 0.283333333, 0.283333333), FMSSpSuit = c(NA,NA, NA, NA, NA, NA, NA, 0.34, 0.34, 0.34), FMSIncSuit = c(NA,NA, NA, NA, NA, NA, NA, 0.425, 0.425, 0.425), FMSGrowSuit = c(0,0, 0, 0, 0, 0, 0, 0.283333333, 0.283333333, 0.283333333), BhsSpSuit = c(NA,NA, NA, NA, NA, NA, NA, 0.233333333, 0.233333333, 0.233333333), BhsIncSuit = c(NA, NA, NA, NA, NA, NA, NA, 0, 0, 0), BhsGrowSuit = c(0,0, 0, 0, 0, 0, 0, 0.233333333, 0.233333333, 0.233333333), BrtSpSuit = c(0.866666667,0.866666667, 0.866666667, 0.866666667, 0.866666667, 0.866666667,0.866666667, 0, 0, 0), BrtIncSuit = c(0.8, 0.8, 0.8, 0.8, 0.8,0.8, 0.8, 0.43, 0.43, 0.43), BrtGrSuit = c(0, 0, 0, 0, 0, 0,0, 0.86, 0.86, 0.86), CcfSpSuit = c(NA, NA, NA, NA, NA, NA, NA,0, 0, 0), CcfIncSuit = c(NA, NA, NA, NA, NA, NA, NA, 0, 0, 0),CcfGrSuit = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), GsfSpSuit = c(NA,NA, NA, NA, NA, NA, NA, 0, 0, 0), GsfIncSuit = c(NA, NA,NA, NA, NA, NA, NA, 0, 0, 0), GsfGrSuit = c(0, 0, 0, 0, 0,0, 0, 0, 0, 0), RbtSpSuit = c(NA, NA, NA, NA, NA, NA, NA,0, 0, 0), RbtIncSuit = c(NA, NA, NA, NA, NA, NA, NA, 0, 0,0), RbtGrSuit = c(0, 0, 0, 0, 0, 0, 0, 0.925, 0.925, 0.925), SmbSpSuit = c(NA, NA, NA, NA, NA, NA, NA, 0.675, 0.675,0.675), SmbIncSuit = c(NA, NA, NA, NA, NA, NA, NA, 0.766666667,0.766666667, 0.766666667), SmbGrSuit = c(0, 0, 0, 0, 0, 0,0, 0.0875, 0.0875, 0.0875), StbSpSuit = c(NA, NA, NA, NA,NA, NA, NA, 0.425, 0.425, 0.425), StbIncSuit = c(NA, NA,NA, NA, NA, NA, NA, 0, 0, 0), StbGrSuit = c(0, 0, 0, 0, 0,0, 0, 0, 0, 0), HbcSpSuit = c(NA, NA, NA, NA, NA, NA, NA,0, 0, 0), HbcIncSuit = c(NA, NA, NA, NA, NA, NA, NA, 0, 0,0), HbcGrSuit = c(0, 0, 0, 0, 0, 0, 0, 0.425, 0.425, 0.425)), .Names = c("Month", "SpdSpSuit", "SpdIncSuit", "SpdGrowSuit","RzbSpSuit", "RzbIncSuit", "RzbGrowSuit", "FMSSpSuit", "FMSIncSuit","FMSGrowSuit", "BhsSpSuit", "BhsIncSuit", "BhsGrowSuit", "BrtSpSuit","BrtIncSuit", "BrtGrSuit", "CcfSpSuit", "CcfIncSuit", "CcfGrSuit","GsfSpSuit", "GsfIncSuit", "GsfGrSuit", "RbtSpSuit", "RbtIncSuit","RbtGrSuit", "SmbSpSuit", "SmbIncSuit", "SmbGrSuit", "StbSpSuit","StbIncSuit", "StbGrSuit", "HbcSpSuit", "HbcIncSuit", "HbcGrSuit"), class = c("data.table", "data.frame"), row.names = c(NA, -10L)))

显然，我并没有以有效的方式解决这个问题，而是在做一些草率的编程。但是，我正在努力弄清楚我可以在哪里优化我的代码。我是否试图重新发明内置的DT函数DT？我在其中一个圈子里;我错过了一个可以矢量化的地方：R Inferno？

基本上，我需要根据AltSuitDates DT更新AltSuitSp1 DT中的第2:34列，使用Month列作为条件，知道从AltSuitDates DT使用哪一行来更新AltSuitSp1。任何帮助表示赞赏。

Answer 1

编辑，更新为与发布的数据一起运行。

这应该有效：

AltSuitSp1$Month <- as.integer(levels(AltSuitSp1$Month))[AltSuitSp1$Month]
setkey(AltSuitDates, Month)
d.cols <- ncol(AltSuitDates) - 1L
AltSuitDates[AltSuitSp1, ][, 
  c(list(Month=Month), 
   mapply(
     `*`, 
     .SD[, 2:(d.cols + 1), with=F],
     .SD[, (d.cols + 2):(2 * d.cols + 1), with=F],
     SIMPLIFY=FALSE
) ) ]

基本上，您首先通过Month（第三行）加入两个表，然后使用.SD，这是一个引用数据表本身的特殊对象，以传递第一个集合来自AltSuitSp1的行和来自AltSuitDates的第二个集合（现在都在同一个表中）到mapply，以便它们可以将它们相乘。这是我使用的数据：

library(data.table)
set.seed(1)
AltSuitSp1 <- do.call(rbind, replicate(3, data.table(Month=1:12, a=runif(12), b=runif(12), c=runif(12)), s=F))
AltSuitDates <- data.table(Month=1:12, a=runif(12, 5, 10), b=runif(12, 5, 10), c=runif(12, 5, 10))

Answer 2

根据@ BrodieG的评论，这已被重新设计，以容纳SP2中的34列。在您的大小（5e6行，34列）的数据集上，它在大约4分钟内运行。

基本方法是将引用列（此处称为Dates）附加到SP1，一次一个。然后更新SP1的相应列，然后重复下一列。这在内存方面相当高效（在任何点只有1个额外的列），并且仍然利用data.table的引用设置。

library(data.table)
set.seed(1)
ncol <- 34
nrow <- 5e6
m    <- matrix(sample(10000:99999,nrow*ncol,replace=T),ncol=ncol)
SP1  <- data.table(Month=sample(1:12,nrow(m),replace=T),m)
m    <- matrix(sample(1:12,12*ncol,replace=T),ncol=ncol)
SP2  <- data.table(Month=sample(1:12,12),m)
cols <- paste0("SpId",(1:ncol(m)))
setnames(SP1,2:(ncol(m)+1),cols)
setnames(SP2,2:(ncol(m)+1),cols)

system.time({
  setkey(SP1,"Month")
  setkey(SP2,"Month")
  lapply(1:ncol,function(i){
    setnames(SP2,cols[i],"Dates")  # don't want colname collision in merge
    SP1[SP2[,c("Month","Dates"),with=F],Dates:=Dates]
    SP1[,cols[i]:=.SD[,cols[i],with=F]*Dates,with=F]
    setnames(SP2,"Dates",cols[i])  # set it back so next iteration works
  })
})
#    user  system elapsed 
#  219.54   22.45  242.59

Answer 3

试试这个。我认为它可能与BrodieG类似，但是当我复制粘贴BrodieG的答案时，它对我不起作用，我不知道如何阅读mapply部分来弄清楚我做错了什么...

comb<-merge(AltSuitSp1,AltSuitDates,by="Month")
sp<-colnames(AltSuitSp1)[2:NCOL(AltSuitSp1)]
dat<-colnames(AltSuitDates)[2:NCOL(AltSuitDates)]
comb[,eval(parse(text=paste0("list(Month,",paste0(sp,"=",sp,"*",dat,collapse=","),")")))]

我知道ol {eval(parse(text=通常不被认为是一种好习惯，但是当你拥有的只是一把锤子时，它就会让螺丝钉进木头里。

Data.table循环效率

3 个答案: