(Ruby) How to Check Whether a Range Contains a Subset of Another Range

How do you check whether a range contains a subset of another range?

Be careful using this with large ranges but this is an elegant way to do it:

(x.to_a & y.to_a).empty?

Check if an array is subset of another array in Ruby

Use sets. Then you can use set.subset?. Example:

require 'set'

a1 = Set[3,6,4]
a2 = Set[1,2,3,4,5,6,7,8,9]

puts a1.subset?(a2)

Output:

true

See it working online: ideone

Check if two ranges overlap in ruby

Two ranges overlap for a given range A when:

1. range B starts within range A,
2. range B ends within range A or
3. range B starts before range A and ends after range A

Examples:

Range A    |-----|
             |-----|  Case 1
         |-----|      Case 2
             |-|      Case 1 + 2
         |---------|  Case 3

Looking closer the rule is: Two ranges overlap when Range B starts before the range A ends and range B ends after the range A starts.

def ranges_overlap?(range_a, range_b)
  range_b.begin <= range_a.end && range_a.begin <= range_b.end 
end 

How to determine if one array contains all elements of another array

a = [5, 1, 6, 14, 2, 8]
b = [2, 6, 15]

a - b
# => [5, 1, 14, 8]

b - a
# => [15]

(b - a).empty?
# => false

How to check whether a string contains a substring in Ruby

You can use the include? method:

my_string = "abcdefg"
if my_string.include? "cde"
   puts "String includes 'cde'"
end

How to select array elements in a given range in Ruby?

You can use ranges in the array subscript:

arr[100..200]

Subset a table by whether two columns fall within any reference range in a lookup table in R

If I understand you correctly, you want to find variants which (for at least one range in table2, r) meet at least one of the following conditions:

1. table1$end is within r
2. table1$pos is within r
3. table1$pos and table2$end are on opposite sides of r

Below are schematic depictions of all possibilities with r delimited by brackets and a variant represented by hyphens together with the numbers of conditions met.

-[-  ]   1.
 [  -]-  2.
 [ - ]   1. and 2.
-[---]-  3.

Please note that the code below assumes that (unlike table1) table2[, 2] <= table2[, 3].

# simplified table1
table1 <- data.frame(pos=c(2431, 98041, 8743), end=c(2100, 99100, 9000))
table2 <- data.frame(exon='Exon', from=c(2001, 8700), to=c(2500, 8750))

is.within <- apply(table1[, c('pos', 'end')], 1, function(x) {
  
  x <- sort(x)  # make sure x[1] <= x[2], i.e. pos <= end
  any((x[1] >= table2[, 2] & x[1] <= table2[, 3]) |      # 1.
        (x[2] >= table2[, 2] & x[2] <= table2[, 3]) |    # 2.
        (x[1] <= table2[, 2] & x[2] >= table2[, 3]))     # 3.
        
})
table1[is.within, ]

(Ruby) If the array intersection operator ( & ) is inefficient, why is it available?

& is not a particularly inefficient method. I think you misunderstood the criticism of the accepted answer.

Your preferred solution is inefficient because it converts the ranges to arrays.

A range such as 1..10000 has a relatively small memory footprint - it only stores the start and end points. But if you convert it to an array, you allocate memory for all 10,000 entries.

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