zqe - 8 months ago 50

Ruby Question

I have a set of students who have each selected a certain number of courses that they want to take next semester, represented as an array of hashes:

`[`

{"student"=>"1", "English"=>true, "Algebra"=>true, "History"=>false},

{"student"=>"2", "English"=>false, "Algebra"=>false, "History"=>true},

{"student"=>"3", "English"=>false, "Algebra"=>true, "History"=>false},

{"student"=>"4", "English"=>true, "Algebra"=>false, "History"=>true}

]

I want to build a matrix showing how many conflicts there are between each course, end result something like this:

`English Algebra History`

English 2 1 1

Algebra 1 2 -

History 1 - 2

Where the number at the intersection is the number of students who have selected

I tried looking at the ruby docs for the Matrix class and it seemed like that addresses more mathematical matrices- I'm not sure how to retool it for this purpose, or if it's an appropriate class for this problem.

What kind of approach could I try researching/googling to efficiently construct a matrix like this?

Answer Source

The following works for any number of subjects.

```
arr = [
{"student"=>"1", "English"=>true, "Algebra"=>true, "History"=>false},
{"student"=>"2", "English"=>false, "Algebra"=>false, "History"=>true},
{"student"=>"3", "English"=>false, "Algebra"=>true, "History"=>false},
{"student"=>"4", "English"=>true, "Algebra"=>false, "History"=>true}
]
pairs = (arr.first.keys - ["student"]).repeated_combination(2).to_a
#=> [["English", "English"], ["English", "Algebra"], ["English", "History"],
# ["Algebra", "Algebra"], ["Algebra", "History"], ["History", "History"]]
h = pairs.product([0]).to_h
#=> {["English", "English"]=>0, ["English", "Algebra"]=>0, ["English", "History"]=>0,
# ["Algebra", "Algebra"]=>0, ["Algebra", "History"]=>0, ["History", "History"]=>0}
arr.each_with_object(h) { |g,h|
pairs.each { |sub1, sub2| h[[sub1, sub2]] += 1 if (g[sub1] && g[sub2]) } }
#=> {["English", "English"]=>2, ["English", "Algebra"]=>1, ["English", "History"]=>1,
# ["Algebra", "Algebra"]=>2, ["Algebra", "History"]=>0, ["History", "History"]=>2}
```

See Array#repeated_combination.

The steps are as follows.

```
a = arr.first.keys - ["student"]
#=> ["English", "Algebra", "History"]
b = a.repeated_combination(2)
#=> #<Enumerator: ["English", "Algebra", "History"]:repeated_combination(2)>
pairs = b.to_a
#=> [["English", "English"], ["English", "Algebra"], ["English", "History"],
# ["Algebra", "Algebra"], ["Algebra", "History"], ["History", "History"]]
enum0 = arr.each_with_object(Hash.new(0))
#=> #<Enumerator: [
# {"student"=>"1", "English"=>true, "Algebra"=>true, "History"=>false},
# {"student"=>"2", "English"=>false, "Algebra"=>false, "History"=>true},
# {"student"=>"3", "English"=>false, "Algebra"=>true, "History"=>false},
# {"student"=>"4", "English"=>true, "Algebra"=>false, "History"=>true}
# ]:each_with_object({})>
```

The first element of `enum1`

is generated and pass to the block and the block variables are assigned to that object, using *parallel assignment*.

```
g,h = enum0.next
#=> [{"student"=>"1", "English"=>true, "Algebra"=>true, "History"=>false}, {}]
g #=> {"student"=>"1", "English"=>true, "Algebra"=>true, "History"=>false}
h #=> {}
```

We now encounter a second enumerator.

```
enum1 = pairs.each
#=> #<Enumerator: [
# ["English", "English"], ["English", "Algebra"], ["English", "History"],
# ["Algebra", "Algebra"], ["Algebra", "History"], ["History", "History"]
# ]:each>
```

The first element of that enumerator is passed to the block and the values of the block variables are computed.

```
sub1, sub2 = enum1.next
#=> ["English", "English"]
sub1
#=> "English"
sub2
#=> "English"
```

We next perform the block calculation.

```
c = g[sub1] && g[sub2]
#=> g["English"] && g["English"]
#=> true && true
#=> true
```

Because `c #=> true`

we execute

```
h[[sub1, sub2]] += 1
#=> h[["English", "English"]] += 1
#=> 1
```

so now

```
h #=> {["English", "English"]=>1}
```

The remaining calculations are similar.