R:在可行的情况下快速(条件)子集 [英] R: fast (conditional) subsetting where feasible

问题描述

我想对我的数据行进行子集化

I would like to subset rows of my data

library(data.table); set.seed(333); n <- 100
dat <- data.table(id=1:n, x=runif(n,100,120), y=runif(n,200,220), z=runif(n,300,320))

> head(dat)
   id        x        y        z
1:  1 109.3400 208.6732 308.7595
2:  2 101.6920 201.0989 310.1080
3:  3 119.4697 217.8550 313.9384
4:  4 111.4261 205.2945 317.3651
5:  5 100.4024 212.2826 305.1375
6:  6 114.4711 203.6988 319.4913

分几个阶段.我知道我可以按顺序应用 subset(.) 来实现这一点.

in several stages. I am aware that I could apply subset(.) sequentially to achieve this.

> s <- subset(dat, x>119)
> s <- subset(s, y>219)
> subset(s, z>315)
   id        x        y        z
1: 55 119.2634 219.0044 315.6556

我的问题是我需要自动执行此操作，并且子集可能为空.在这种情况下，我想跳过导致空集的步骤.例如，如果我的数据是

My problem is that I need to automate this and it might happen that the subset is empty. In this case, I would want to skip the step(s) that result in an empty set. For example, if my data was

dat2 <- dat[1:50]
> s <-subset(dat2,x>119)
> s
   id        x        y        z
1:  3 119.4697 217.8550 313.9384
2: 50 119.2519 214.2517 318.8567

第二步 subset(s, y>219) 会空出来，但我仍然想应用第三步 subset(s,z>315).只有当它导致非空集时，有没有办法应用子集命令?我想像 subset(s, y>219, nonzero=TRUE) 这样的东西.我想避免像

the second step subset(s, y>219) would come up empty but I would still want to apply the third step subset(s,z>315). Is there a way to apply a subset-command only if it results in a non-empty set? I imagine something like subset(s, y>219, nonzero=TRUE). I would want to avoid constructions like

s <- dat
if(nrow(subset(s, x>119))>0){s <- subset(s, x>119)}
if(nrow(subset(s, y>219))>0){s <- subset(s, y>219)}
if(nrow(subset(s, z>318))>0){s <- subset(s, z>319)}

因为我担心 if-then 丛林会相当慢，特别是因为我需要使用 lapply(.) 将所有这些应用到列表中的不同 data.tables.这就是为什么我希望找到一个针对速度优化的解决方案.

because I fear the if-then jungle would be rather slow, especially since I need to apply all of this to different data.tables within a list using lapply(.). That's why I am hoping to find a solution optimized for speed.

PS.为了清楚起见，我只选择了 subset(.)，解决方案例如如果不是更多的话，data.table 也会受到欢迎.

PS. I only chose subset(.) for clarity, solutions with e.g. data.table would be just as welcome if not more so.

推荐答案

我同意 Konrad 的回答，即这应该引发警告或至少报告以某种方式发生的情况.这是一种利用索引的 data.table 方式(有关详细信息，请参阅包小插曲):

I agree with Konrad's answer that this should throw a warning or at least report what happens somehow. Here's a data.table way that will take advantage of indices (see package vignettes for details):

f = function(x, ..., verbose=FALSE){
  L   = substitute(list(...))[-1]
  mon = data.table(cond = as.character(L))[, skip := FALSE]

  for (i in seq_along(L)){
    d = eval( substitute(x[cond, verbose=v], list(cond = L[[i]], v = verbose)) )
    if (nrow(d)){
      x = d
    } else {
      mon[i, skip := TRUE]
    }    
  }
  print(mon)
  return(x)
}

用法

> f(dat, x > 119, y > 219, y > 1e6)
        cond  skip
1:   x > 119 FALSE
2:   y > 219 FALSE
3: y > 1e+06  TRUE
   id        x        y        z
1: 55 119.2634 219.0044 315.6556

verbose 选项将打印 data.table 包提供的额外信息，因此您可以查看何时使用索引.例如，使用 f(dat, x == 119, verbose=TRUE)，我明白了.

The verbose option will print extra info provided by data.table package, so you can see when indices are being used. For example, with f(dat, x == 119, verbose=TRUE), I see it.

因为我担心 if-then 丛林会相当缓慢，特别是因为我需要使用 lapply(.) 将所有这些应用到列表中的不同 data.tables.

because I fear the if-then jungle would be rather slow, especially since I need to apply all of this to different data.tables within a list using lapply(.).

如果它用于非交互式使用，最好让函数返回 list(mon = mon, x = x) 以便更轻松地跟踪查询内容和发生的情况.此外，可以捕获并返回详细的控制台输出.

If it's for non-interactive use, maybe better to have the function return list(mon = mon, x = x) to more easily keep track of what the query was and what happened. Also, the verbose console output could be captured and returned.

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