I have a hash like this below
h = {a: 1, b: 2, c: 3, d: 4.....z: 26}
now user input 1 then I will fetch first 5 if user input 2 then 6 to next 5 means 6 to 11
How can I acheieve this by a best way
4 Answers
h = {a: 1, b: 2, c: 3, d: 4.....z: 26}
user_input = 1
Hash[h.to_a[((user_input - 1) * 5 )..( (user_input * 5) - 1)]]
#=> {:a=>1, :b=>2, :c=>3, :d=>4, :e=>5}
Comments
I assume the question concerns an arbitrary hash.
Code
Three choices:
#1
def select_em(h, which_block_5)
select_range = (5*which_block_5 - 4)..(5*which_block_5)
h.select.with_index(1) { |_,i| select_range.cover?(i) }
end
#2
def select_em(h, which_block_5)
select_array = Array.new(5*(which_block_5-1),false) +
Array.new(5,true) +
Array.new(h.size-5*(which_block_5),false)
h.select { select_array.shift }
end
Note
select_array = Array.new(5*(which_block_5-1),false) +
Array.new(5,true) +
Array.new(26-5*(which_block_5),false)
#=> [false, false, false, false, false, true, true, true, true, true,
# false, false, false, false, false, false, false, false, false,
# false, false, false, false, false, false, false]
#3
def select_em(h, which_block_5)
start = 5*which_block_5 - 4
stop = start + 4
h.select.with_index(1) { |_,i| (i==start..i==stop) ? true : false }
end
This method uses Ruby's flip-flop operator.
All of these methods use Hash#select (which returns a hash), not Enumerable#select (which returns an array).
Examples
h = {:a=>1, :b=>2, :c=>3, :d=>4, :e=>5, :f=>6, :g=>7, :cat=>"meow", :dog=>"woof",
:h=>8, :i=>9, :j=>10, :k=>11, :l=>12, :m=>13, :n=>14, :o=>15,
:p=>16, :q=>17, :r=>18, :s=>19, :t=>20, :u=>21, :v=>22, :w=>23,
:x=>24, :y=>25, :z=>26}
select_em(h, 1)
#=> {:a=>1, :b=>2, :c=>3, :d=>4, :e=>5}
select_em(h, 2)
#=> {:f=>6, :g=>7, :cat=>"meow", :dog=>"woof", :h=>8}
select_em(h, 3)
#=> {:i=>9, :j=>10, :k=>11, :l=>12, :m=>13}
select_em(h, 4)
#=> {:n=>14, :o=>15, :p=>16, :q=>17, :r=>18}
Comments
Here is what i would do:
char_ary = ('a'..'z').to_a
start_idx = (input - 1) * 5
subary = char_ary.slice(start_idx, 5)
subary.inject({}) do |h, c|
h[c.to_sym] = char_ary.index(c) + 1
h
end
This doesn't need a hash to be defined with all the alphabets.
1 Comment
Cary Swoveland
I believe the hash given was just an example. If is was that particular hash, you wouldn't have to do any slicing or selecting, just construct the 5-element hash directly.
So you can simply use slice array function to split out things.
alphabets = [*('a'..'z')].collect.with_index{ |key,index| { key => index+1 } }
user_input = 1 # Capture user input here.
part_of_array = alphabets[( (user_input - 1) * 5 )..( (user_input * 5) - 1)]
Converting into simple one hash use the below code.
part_of_array = eval(alphabets[( (user_input - 1) * 5 )..( (user_input * 5) - 1)].to_s.gsub("{","").gsub("}",""))
Let me know if you have any issues.
3 Comments
Pavel Tkackenko
This code will generate array with standalone hashes:
[{"a"=>1}, {"b"=>2}, {"c"=>3}, {"d"=>4}, {"e"=>5}]
Muaaz Rafi
Simple. eval(alphabets[( (user_input - 1) * 5 )..( (user_input * 5) - 1)].to_s.gsub("{","").gsub("}",""))
Cary Swoveland
My comment on @Jagdeep's answer applies to yours as well.
hshould be a valid Ruby hash (no....), so readers can cut-and-paste rather than having to construct the hash. Also,....leaves the impression that you are lazy. 2. If you mean it to be an arbitrary hash (which I assume is the case), you should not have such a regular pattern (:a,:b,:c,... and1,2,3,...) as some readers will interpret your question as referring to that specific hash.