R：在可行的情况下快速（条件）子集化

Question

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

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

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

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)}

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

Answer 1

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)
}

Usage

> 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

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.

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(.).

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.

Answer 2

An interesting approach could be developed using modified filter function offered in dplyr . In case of conditions not being met the non_empty_filter filter function returns original data set.

Notes

• IMHO, this is fairly non-standard behaviour and should be reported via warning . Of course, this can be removed and has no bearing on the function results.

Function

library(tidyverse)
library(rlang) # enquo
non_empty_filter <- function(df, expr) {
    expr <- enquo(expr)

    res <- df %>% filter(!!expr)

    if (nrow(res) > 0) {
        return(res)
    } else {
        # Indicate that filter is not applied
        warning("No rows meeting conditon")
        return(df)
    }
}

Condition met

Behaviour: Returning one row for which the condition is met.

dat %>%
    non_empty_filter(x > 119 & y > 219)

Results

# id        x        y        z
# 1 55 119.2634 219.0044 315.6556

Condition not met

Behaviour: Returning the full data set as the whole condition is not met due to y > 1e6 .

dat %>%
    non_empty_filter(x > 119 & y > 219 & y > 1e6)

Results

# 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
# 7:   7 112.1879 209.5716 319.6732
# 8:   8 106.1344 202.2453 312.9427
# 9:   9 101.2702 210.5923 309.2864
# 10:  10 106.1071 211.8266 301.0645

Condition met/not met one-by-one

Behaviour: Skipping filter that would return an empty data set.

dat %>%
    non_empty_filter(y > 1e6) %>% 
    non_empty_filter(x > 119) %>% 
    non_empty_filter(y > 219)

Results

# id        x        y        z
# 1 55 119.2634 219.0044 315.6556