how to generate separate array of array in ruby based on one sub array value

Question

I have an array of array as below:

a = [
      ["a", "v", 1], ["b", "w", 2], ["c", "x", 1], ["d", "y", 1],
      ["e", "z", 2], ["f", "one" , 3 ], ["g", "two" , 3 ], ["g", "one" , 4 ],
      ["f", "one" , 1 ], ["h", "one" , 5 ], ["f", "one" , 4 ],
      # ...
    ]

Then , i expect the result as 5 different arrays

a1 = [ ["a", "v", 1],["c", "x", 1], ["d", "y", 1], ["f", "one" , 1 ] ]
a2 = [ ["b", "w", 2], ["e", "z", 2] ] 
a3 = [ ["f", "one", 3], ["g", "two", 3] ]
a4 = [ ["g", "one", 4], ["f", "one", 4] ]
a5 = [ ["h", "one" , 5 ] ]
# ...
an = []

By doing the below code, i was able to sort it.

b = a.sort{|c,d|c[2] <=> d[2]}

How can i generate such a list.

Please help.Thanks in advance.

Stefan · Accepted Answer · 2016-02-29 10:54:14Z

4

You could use partition:

a = [["a", "v", 1], ["b", "w", 2], ["c", "x", 1], ["d", "y", 1], ["e", "z", 2]]

a1, a2 = a.partition { |e| e[2] == 1 }

a1 #=> [["a", "v", 1], ["c", "x", 1], ["d", "y", 1]]
a2 #=> [["b", "w", 2], ["e", "z", 2]]

With more than 2 values, you could use group_by:

a = [
  ["a", "v", 1], ["b", "w", 2], ["c", "x", 1], ["d", "y", 1],
  ["e", "z", 2], ["f", "one" , 3 ], ["g", "two" , 3 ], ["g", "one" , 4 ],
  ["f", "one" , 1 ], ["h", "one" , 5 ], ["f", "one" , 4 ]
]

hash = a.group_by { |e| e[2] }
#=> { 1=>[["a", "v", 1], ["c", "x", 1], ["d", "y", 1], ["f", "one", 1]], 
#     2=>[["b", "w", 2], ["e", "z", 2]],
#     3=>[["f", "one", 3], ["g", "two", 3]],
#     4=>[["g", "one", 4], ["f", "one", 4]],
#     5=>[["h", "one", 5]] }

To access the arrays, you can use hash[1], hash[2], etc. or you can assign them to variables:

a1, a2, a3, a4, a5 = hash.values_at(1, 2, 3, 4, 5)

a1 #=> [["a", "v", 1], ["c", "x", 1], ["d", "y", 1], ["f", "one", 1]]
a2 #=> [["b", "w", 2], ["e", "z", 2]]
a3 #=> [["f", "one", 3], ["g", "two", 3]]
a4 #=> [["g", "one", 4], ["f", "one", 4]]
a5 #=> [["h", "one", 5]]

edited Feb 29, 2016 at 10:54

answered Feb 27, 2016 at 10:45

Stefan

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2 Comments

Cary Swoveland Over a year ago

Fussy readers might be thinking, "given was an index into a, from which 1 was computed".

jissy Over a year ago

Perfect. Thank you @Stefan.

fabriciofreitag · Accepted Answer · 2016-02-27 15:36:34Z

0

Select would be the more readable option:

a.select do |v| v.third == 1 end
 => [["a", "v", 1], ["c", "x", 1], ["d", "y", 1]]

answered Feb 27, 2016 at 15:36

fabriciofreitag

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