CODEWITHSUNDEEP
rubyarraysindexingelementminimum
kyrylo
kyrylo

Reputation: 1738

Ruby: How to find the index of the minimum array element?

Is there any way to rewrite this more elegant? I think, that it's a bad piece of code and should be refactored.

>> a = [2, 4, 10, 1, 13]
=> [2, 4, 10, 1, 13]
>> index_of_minimal_value_in_array = a.index(a.min)
=> 3

Upvotes: 27

Views: 23561

Answers (4)

andersonvom
andersonvom

Reputation: 11851

I believe this will traverse the array only once and is still easy to read:

numbers = [20, 30, 40, 50, 10]           # => [20, 30, 40, 50, 10]
elem, idx = numbers.each_with_index.min  # => [10, 4]

Upvotes: 53

Leo Liang
Leo Liang

Reputation: 171

I actually like @andersonvom 's answer, it only need to loop the array once and still get the index.

And in case you don't want to use ary.each_with_index.min, here is what you can do:

ary = [2,3,4,5,1]                                             # => [2,3,4,5,1]
_, index_of_minimal_value_in_array = ary.each_with_index.min  # => [1, 4]
index_of_minimal_value_in_array                               # => 4

Upvotes: 4

x17y19
x17y19

Reputation: 91

This traverses the array only once whereas ary.index(ary.min) would traverse it twice:

ary.each_with_index.inject(0){ |minidx, (v,i)| v < a[minidx] ? i : minidx }

Upvotes: 9

the Tin Man
the Tin Man

Reputation: 160551

It would be interesting to read about other situations (finding all and only last minimal element). 

ary = [1, 2, 1]

# find all matching elements' indexes
ary.each.with_index.find_all{ |a,i| a == ary.min }.map{ |a,b| b } # => [0, 2]
ary.each.with_index.map{ |a, i| (a == ary.min) ? i : nil }.compact # => [0, 2]

# find last matching element's index
ary.rindex(ary.min) # => 2

Upvotes: 8

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