Subset by multiple ranges [duplicate]

问题:

This question already has an answer here:

• Efficient way to filter one data frame by ranges in another 3 answers

I want to get a list of values that fall in between multiple ranges.

library(data.table)
values <- data.table(value = c(1:100))
range <-  data.table(start = c(6, 29, 87), end = c(10, 35, 92)) 

I need the results to include only the values that fall in between those ranges:

 results <- c(6, 7, 8, 9, 10, 29, 30, 31, 32, 33, 34, 35, 87, 88, 89, 90, 91, 92)

I am currently doing this with a for loop,

results <- data.table(NULL)
for (i in 1:NROW(range){ 
          results <- rbind(results, 
              data.table(result = values[value >= range[i, start] & 
                 value <= range[i, end], value]))}

however the actual dataset is quite large and I am looking for a more efficient way.

Any suggestions are appreciated! Thank you!

回答1:

Using the non-equi join possibility of data.table:

values[range, on = .(value >= start, value <= end), .(results = x.value)]

which gives:

    results
 1:       6
 2:       7
 3:       8
 4:       9
 5:      10
 6:      29
 7:      30
 8:      31
 9:      32
10:      33
11:      34
12:      35
13:      87
14:      88
15:      89
16:      90
17:      91
18:      92

---

Or as per the suggestion of @Henrik: values[value %inrange% range]. This works also very well on data.table's with multiple columns:

# create new data
set.seed(26042017)
values2 <- data.table(value = c(1:100), let = sample(letters, 100, TRUE), num = sample(100))

> values2[value %inrange% range]
    value let num
 1:     6   v  70
 2:     7   f  77
 3:     8   u  21
 4:     9   x  66
 5:    10   g  58
 6:    29   f   7
 7:    30   w  48
 8:    31   c  50
 9:    32   e   5
10:    33   c   8
11:    34   y  19
12:    35   s  97
13:    87   j  80
14:    88   o   4
15:    89   h  65
16:    90   c  94
17:    91   k  22
18:    92   g  46

回答2:

If you have the latest CRAN version of data.table you can use non-equi joins. For example, you can create an index which you can then use to subset your original data:

idx <- values[range, on = .(value >= start, value <= end), which = TRUE]
# [1]  6  7  8  9 10 29 30 31 32 33 34 35 87 88 89 90 91 92
values[idx]

回答3:

Here is one method using lapply and %between%

rbindlist(lapply(seq_len(nrow(range)), function(i) values[value %between% range[i]]))

This method loops through the ranges data.table and subsets values in each iteration according to the variable in ranges. lapply returns a list, which rbindlist constructs into a data.table. If you want a vector, replace rbindlist with unlist.

---

benchmarks

Just to check the speeds of each suggestion on the given data, I ran a quick comparison

microbenchmark(
  lmo=rbindlist(lapply(seq_len(nrow(range)), function(i) values[value %between% range[i]])),
  dd={idx <- values[range, on = .(value >= start, value <= end), which = TRUE]; values[idx]},
  jaap=values[range, on = .(value >= start, value <= end), .(results = x.value)],
  inrange=values[value %inrange% range])

This returned

Unit: microseconds
    expr      min        lq      mean    median       uq      max neval cld
     lmo 1238.472 1460.5645 1593.6632 1520.8630 1613.520 3101.311   100   c
      dd  688.230  766.7750  885.1826  792.8615  825.220 3609.644   100  b 
    jaap  798.279  897.6355  935.9474  921.7265  970.906 1347.380   100  b 
 inrange  463.002  518.3110  563.9724  545.5375  575.758 1944.948   100 a 

As might be expected, my looping solution is quite a bit slower than the others. However, the clear winner is %inrange%, which is essentially a vectorized extension of %between%.