r-矢量化索引for for循环

Question

我正在尝试模拟汇编线。我有一个零件列表以及他们在每个车站花费多少时间。我试图一次通过一个装配线发送零件，并在每个电台记录时间。但是，我的工作方式是嵌套在a内的循环。必须有一种更好的方法来做到这一点。

parts <- data.frame(JobNum = sample(c('a','b','c','d'),400,replace=TRUE)
,DS.CT = sample.int(10,400,replace=TRUE)
,C1.CT = sample.int(10,400,replace=TRUE)
,C2.CT = sample.int(10,400,replace=TRUE)
,C3.CT = sample.int(10,400,replace=TRUE)
,C4.CT = sample.int(10,400,replace=TRUE)
,C5D5.CT = sample.int(10,400,replace=TRUE)
,C6D6.CT = sample.int(10,400,replace=TRUE)
,C5D7.CT = sample.int(10,400,replace=TRUE)
,C6D8.CT = sample.int(10,400,replace=TRUE)
,C7CD.CT = sample.int(10,400,replace=TRUE)
)
LineParts <- parts[sample(nrow(parts),234,replace=FALSE),]
#Initialize Dip collecting variables
DS <- c()
D1 <- c()
D2 <- c()
D3 <- c()
D4 <- c()
D5 <- c()
D6 <- c()
D7 <- c()
D8 <- c()
D9 <- c()
for(i in 1:dim(parts)[1]){
#Create temporary dataframe for use in indexing line
LinePartsTemp <- data.frame(matrix("",nrow=234,ncol=11))
colnames(LinePartsTemp)=names(LineParts)
LinePartsTemp$JobNum <- as.character(LinePartsTemp$JobNum)
LinePartsTemp$DS.CT <- as.integer(LinePartsTemp$DS.CT)
LinePartsTemp$C1.CT <- as.integer(LinePartsTemp$C1.CT)
LinePartsTemp$C2.CT <- as.integer(LinePartsTemp$C2.CT)
LinePartsTemp$C3.CT <- as.integer(LinePartsTemp$C3.CT)
LinePartsTemp$C4.CT <- as.integer(LinePartsTemp$C4.CT)
LinePartsTemp$C5D5.CT <- as.integer(LinePartsTemp$C5D5.CT)
LinePartsTemp$C6D6.CT <- as.integer(LinePartsTemp$C6D6.CT)
LinePartsTemp$C5D7.CT <- as.integer(LinePartsTemp$C5D7.CT)
LinePartsTemp$C6D8.CT <- as.integer(LinePartsTemp$C6D8.CT)
LinePartsTemp$C7CD.CT <- as.integer(LinePartsTemp$C7CD.CT)
#Index line
for(j in 1:dim(LineParts)[1]){
    LinePartsTemp[j+1,] <- LineParts[j,]
}
#put new part into system
LinePartsTemp[1,] <- parts[i,]
#update the list of parts on the line
LineParts <- LinePartsTemp
#Append CT values at stations
DS <- append(DS,LineParts[1,'DS.CT'])
D1 <- append(D1,LineParts[10,'C1.CT'])
D2 <- append(D2,LineParts[26,'C2.CT'])
D3 <- append(D3,LineParts[42,'C3.CT'])
D4 <- append(D4,LineParts[57,'C4.CT'])
D5 <- append(D5,LineParts[85,'C5D5.CT'])
D6 <- append(D6,LineParts[120,'C6D6.CT'])
D7 <- append(D7,LineParts[167,'C5D7.CT'])
D8 <- append(D8,LineParts[210,'C6D8.CT'])
D9 <- append(D9,LineParts[234,'C7CD.CT'])
}

编辑：添加了示例数据

Answer 1

考虑与列表进行交互，以避免初始化空容器并随后使用许多独立的环境对象附加到它们。除了输入外，下方仅使用两个对象。

1. 首先，构建 lineParts dataframes的列表， lineParts_dflist
2. 然后，将所需的数据点提取到向量列表中， stations_veclist 。

您会注意到数据框列表的lapply使用<<-操作员更新全局对象（本地函数的外部范围），因为 lineParts 需要使用更新值重复使用：

LineParts_dfList <- lapply(seq(nrow(parts)), function(i){      
  #Index line
  LinePartsTemp <- parts[1,]
  LinePartsTemp[2:nrow(LineParts),] <- LineParts[1:nrow(LineParts)-1,]
  #put new part into system
  LinePartsTemp[1,] <- parts[i,]
  #update the list of parts on the line
  LineParts <<- LinePartsTemp      
})
# Extract CT values at stations
stations_veclist <- 
  list(
    DS = vapply(LineParts_dfList, function(df) df[1,'DS.CT'], numeric(1)),
    D1 = vapply(LineParts_dfList, function(df) df[10,'C1.CT'], numeric(1)),
    D2 = vapply(LineParts_dfList, function(df) df[26,'C2.CT'], numeric(1)),
    D3 = vapply(LineParts_dfList, function(df) df[42,'C3.CT'], numeric(1)),
    D4 = vapply(LineParts_dfList, function(df) df[57,'C4.CT'], numeric(1)),
    D5 = vapply(LineParts_dfList, function(df) df[85,'C5D5.CT'], numeric(1)),
    D6 = vapply(LineParts_dfList, function(df) df[120,'C6D6.CT'], numeric(1)),
    D7 = vapply(LineParts_dfList, function(df) df[167,'C5D7.CT'], numeric(1)),
    D8 = vapply(LineParts_dfList, function(df) df[210,'C6D8.CT'], numeric(1)),
    D9 = vapply(LineParts_dfList, function(df) df[234,'C7CD.CT'], numeric(1))
  )

并避免了许多vapply调用，请考虑将所有 lineParts dataFrame项目绑定到一个大数据框中， linePartsallsall （n = 93,600 obs for 234 x 400），然后通过行依次提取值：

LinePartsAll <- do.call(rbind, LineParts_dfList)
otherstations_veclist <- 
  list(
    DS = LinePartsAll[seq(1,93600, by=234),'DS.CT'],
    D1 = LinePartsAll[seq(10,93600, by=234),'C1.CT'],
    D2 = LinePartsAll[seq(26,93600, by=234),'C2.CT'],
    D3 = LinePartsAll[seq(42,93600, by=234),'C3.CT'],
    D4 = LinePartsAll[seq(57,93600, by=234),'C4.CT'],
    D5 = LinePartsAll[seq(85,93600, by=234), 'C5D5.CT'],
    D6 = LinePartsAll[seq(120,93600, by=234), 'C6D6.CT'],
    D7 = LinePartsAll[seq(167,93600, by=234),'C5D7.CT'],
    D8 = LinePartsAll[seq(210,93600, by=234), 'C6D8.CT'],
    D9 = LinePartsAll[seq(234,93600, by=234), 'C7CD.CT']
  )

和检查此更新，更快的方法确实与原始for循环过程相同的端值。要使用已发布的示例数据测试，您必须在两个 parts 和 lineParts 分配之前设置样本种子，set.seed(###)。

all.equal(DS, stationsList$DS)
# [1] TRUE
all.equal(D1, stationsList$D1)
# [1] TRUE
all.equal(D9, stationsList$D9)
# [1] TRUE
all.equal(stations_veclist, otherstations_veclist)
# [1] TRUE