How to sort array of hashes by value? - ruby

Question

I'm trying to sort an array of hashes based on the given key-value, and will return that value on top of the array first, then preceded by the remaining data.
Example would be:
students = [{name: "John Doe", age: 16, adviser: "Mrs. Robinson"},
{name: "John Smith", age: 18, adviser: "Mrs. Williams"},
{name: "Michael Rodriguez", age: 17, adviser: "Mr. Lee"}]
def sort_by_adviser(data, name)
...
end
> sort_by_adviser(students, "Mr. Lee")
=> [{name: "Michael Rodriguez", age: 17, adviser: "Mr. Lee"},
{name: "John Doe", age: 16, adviser: "Mrs. Robinson"},
{name: "John Smith", age: 18, adviser: "Mrs. Williams"}]
> sort_by_adviser(students, "Mrs. Williams")
=> [{name: "John Smith", age: 18, adviser: "Mrs. Williams"},
{name: "Michael Rodriguez", age: 17, adviser: "Mr. Lee"},
{name: "John Doe", age: 16, adviser: "Mrs. Robinson"}]
Here the output brings the adviser's name on top of the list, then preceded by other hashes in the array.
> sort_by_keyvalue(data, "Z")
=> [{letter: 'Z'},
{letter: 'A'},
.
.
.
{letter: 'Y'}]
> sort_by_keyvalue(data, 5)
=> [{number: 5, value: 'value1'},
{number: 5, value: 'value2'},
{number: 5, value: 'value3'},
{number: 9, value: 'value1'},
{number: 9, value: 'value2'},
{number: 8, value: 'value1'},
{number: 8, value: 'value2'},
{number: 7, value: 'value1'},
{number: 6, value: 'value1'},
{number: 4, value: 'value1'},
{number: 3, value: 'value1'},
{number: 2, value: 'value1'},
{number: 1, value: 'value1'},
{number: 1, value: 'value2'},
{number: 0, value: 'value1'}]
Anyone knows how to do it?

Answer 1

Another implemention :)
def sort_by_adviser(data, name)
data.each_with_index do |hash,index|
if hash[:adviser]==name
data.delete_at index #delete from array
data.unshift hash
break
end
end
data
end
> sort_by_adviser(students, "Mr. Lee")
#=> [{:name=>"Michael Rodriguez", :age=>17, :adviser=>"Mr. Lee"}, {:name=>"John Doe", :age=>16, :adviser=>"Mrs. Robinson"}, {:name=>"John Smith", :age=>18, :adviser=>"Mrs. Williams"}]

Answer 2

def creamy_sort(key, value, arr)
top, bottom = arr.partition{|e| e[key] == value }
top.concat(bottom.sort{|a,b| b[key] <=> a[key]})
end
creamy_sort(:adviser, "Mr. Lee", students)

Answer 3

You can do that:
def sort_by_adviser(data, name)
data = data.sort{|x,y|x[:adviser] <=> y[:adviser]}
i = data.index{|h|h[:adviser] = name}
h = data.delete_at i
data.unshift h
end

Answer 4

I have this solution:
students = [{name: "John Doe", age: 16, adviser: "Mrs. Robinson"},
{name: "John Smith", age: 18, adviser: "Mrs. Williams"},
{name: "Michael Rodriguez", age: 17, adviser: "Mr. Lee"}]
def sort_by_adviser(data, *name)
data.sort_by{| entry |
[
name.index(entry[:adviser]) || 999,
entry[:age], entry[:name] #2nd sort criteria
]
}
end
p sort_by_adviser(students, "Mr. Lee")
#[{:name=>"Michael Rodriguez", :age=>17, :adviser=>"Mr. Lee"}, {:name=>"John Doe", :age=>16, :adviser=>"Mrs. Robinson"}, {:name=>"John Smith", :age=>18, :adviser=>"Mrs. Williams"}]
p sort_by_adviser(students, "Mrs. Williams")
# [{:name=>"John Smith", :age=>18, :adviser=>"Mrs. Williams"}, {:name=>"John Doe", :age=>16, :adviser=>"Mrs. Robinson"}, {:name=>"Michael Rodriguez", :age=>17, :adviser=>"Mr. Lee"}]
I didn't understand, what's the sorting of the remaining entry.
You wrote: then preceded by the remaining data. What's tho order criteria of the hash?
I selected age, followed by name. But you may adapt it for your need.

Answer 5

def weird_sort(array, key, value)
return array.sort_by{|d| 2 <=> (d[key] == value).object_id}
end
This is based on the fact that true.object_id equals 2 in ruby.
Kind of a weird solution that's why it's a weird_sort :p It also messes around the other value ordering ... so it only guarantees you that values that are equal go on top!