R，如何向量化这个操作

Question

在 R 中，我怎样才能最好地矢量化这个操作？

我有一个参考值表，有一个下限 (A) 和上限 (B)。

我还有一个值表 (X) 可以对照上表进行查找。

对于 X 的每个值，我需要确定它是否位于参考表中 A 和 B 的任何值之间。

为了演示以上内容，这里有一个使用循环的解决方案：

#For Reproduceability,
set.seed(1); 

#Set up the Reference and Lookup Tables
nref = 5; nlook = 10
referenceTable <- data.frame(A=runif(nref,min=0.25,max=0.5),
                             B=runif(nref,min=0.50,max=0.75));
lookupTable    <- data.frame(X=runif(nlook),IsIn=0)

#Process for each row in the lookup table
#search for at least one match in the reference table where A <= X < B
for(x in 1:nrow(lookupTable)){
  v   <- lookupTable$X[x]
  tmp <- subset(referenceTable,v >= A & v < B)
  lookupTable[x,'IsIn'] = as.integer(nrow(tmp) > 0)
}

我正在尝试删除 for(x in .... ) 组件，因为我现实生活中的问题表有成千上万条记录。

Answer 1

我找不到确切的骗子，所以这里有一个使用data.table::foverlaps 的可能解决方案。首先，我们需要向lookupTable 添加一个额外的列，以便在两侧创建边界。然后 key referenceTable （foverlaps 工作所必需的）然后只运行一个简单的重叠连接，同时只选择第一个连接，因为你想要 any 连接（我使用了@987654326 @ 以便转换为二进制，因为您不想要实际位置）

library(data.table)
setDT(lookupTable)[, Y := X] # Add an additional boundary column
setkey(setDT(referenceTable)) # Key the referenceTable data set
lookupTable[, IsIn := 0 ^ !foverlaps(lookupTable, 
                                     referenceTable, 
                                     by.x = c("X", "Y"),
                                     mult = "first", 
                                     nomatch = 0L, 
                                     which = TRUE)]
#             X IsIn         Y
#  1: 0.2059746    0 0.2059746
#  2: 0.1765568    0 0.1765568
#  3: 0.6870228    1 0.6870228
#  4: 0.3841037    1 0.3841037
#  5: 0.7698414    0 0.7698414
#  6: 0.4976992    1 0.4976992
#  7: 0.7176185    1 0.7176185
#  8: 0.9919061    0 0.9919061
#  9: 0.3800352    1 0.3800352
# 10: 0.7774452    0 0.7774452

Answer 2

使用findInterval

referenceTable2 = cbind(-Inf, referenceTable)

for(x in 1:nrow(referenceTable2)){
  tmp <- findInterval(lookupTable$X, referenceTable2[x,])
  lookupTable[,'IsIn'] = lookupTable[,'IsIn'] + (tmp == 2)
}
lookupTable[,'IsIn'] = sign(lookupTable[,'IsIn'])

如您所见，您的参考表仍然存在循环，因此如果您的参考表仍然很小，此解决方案尤其适用。一些基准：

1) 样本集：

> microbenchmark(nicholas = {set.seed(1); nref = 5; nlook = 10; referenceTable <- data.frame(A=runif(nref,min=0.25,max=0.5), B=runif(nref,min=0.50,max=0.75)); lookupTable <- data.frame(X=runif(nlook),IsIn=0); for(x in 1:nrow(lookupTable)){v   <- lookupTable$X[x]; tmp <- subset(referenceTable,v >= A & v < B); lookupTable[x,'IsIn'] = as.integer(nrow(tmp) > 0)}}, 
+                mts = {set.seed(1); nref = 5; nlook = 10; referenceTable <- data.frame(A=runif(nref,min=0.25,max=0.5), B=runif(nref,min=0.50,max=0.75)); lookupTable <- data.frame(X=runif(nlook),IsIn=0); referenceTable2 = cbind(-Inf, referenceTable); for(x in 1:nrow(referenceTable2)){tmp <- findInterval(lookupTable$X, referenceTable2[x,]); lookupTable[,'IsIn'] = lookupTable[,'IsIn'] + (tmp == 2);}; lookupTable[,'IsIn'] = sign(lookupTable[,'IsIn'])},
+                david = {set.seed(1); nref = 5; nlook = 10; referenceTable <- data.frame(A=runif(nref,min=0.25,max=0.5), B=runif(nref,min=0.50,max=0.75)); lookupTable <- data.frame(X=runif(nlook),IsIn=0); setDT(lookupTable)[, Y := X]; setkey(setDT(referenceTable)); lookupTable[, IsIn := 0 ^ !foverlaps(lookupTable, referenceTable, by.x = c("X", "Y"), mult = "first", nomatch = 0L, which = TRUE)]},
+                times = 100)
Unit: milliseconds
     expr      min       lq     mean   median       uq       max neval
 nicholas 1.948096 2.091311 2.190386 2.150790 2.254352  4.092121   100
      mts 2.520489 2.711986 2.883299 2.803421 2.885990  5.165999   100
    david 6.466129 7.013798 7.344596 7.197132 7.422916 12.274029   100

2)nref = 10; nlook = 1000

Unit: milliseconds
     expr        min         lq       mean     median         uq        max neval
 nicholas 152.804680 160.681265 164.601443 163.304849 165.387296 243.250708   100
      mts   4.434997   4.720027   5.025555   4.819624   4.991995  11.325172   100
    david   6.505314   6.920032   7.181116   7.111529   7.331950   9.318765   100

3)nref = 200; nlook = 1000

Unit: milliseconds
     expr        min         lq       mean     median         uq       max neval
 nicholas 172.939666 179.672397 183.337253 181.191081 183.694077 264.59672   100
      mts  77.836588  81.071752  83.860281  81.991919  83.484246 168.22290   100
    david   6.870116   7.404256   7.736445   7.587591   7.836234  11.54349   100

我认为大卫的解决方案显然是赢家。 当参考区间非常少时，此解决方案具有优势。请注意，在您的示例中，其中许多是重叠的，事先将它们组合起来可能会进一步改善结果。