I have an array filled with 0's, 1's, 3's and 4's and want to check inside the array, so that once a certain amount of the 4's have been found, a statement turns false.
5 Answers
Use the count method.
alist.count(4) >= n
# => true or false
n would be the certain number you want to check for.
answered Dec 18, 2014 at 12:12
Bharadwaj Srigiriraju
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Comments
The documentation is the best place to start answering questions like this, ruby has a very rich set of functions on Array and Enumerable
def enough_4s?(array, how_many)
array.count(4) >= how_many
end
array = [1,1,1,2,3,4,4,4]
enough_4s?(array, 3)
#=> true
enough_4s?(array, 4)
#=> false
[0,1,3,4,3].count(4) > 2 => false
(e.g 2 is the expected count)
Comments
If the goal is to immediately stop searching upon locating enough 4s, use each or find or equivalent, and return or break, rather than count. For instance:
n = 2
([0,1,3,4]*5).find do |i|
n -= 1 if i == 4
break false if n == 0
end
# false
answered Dec 18, 2014 at 13:36
Denis de Bernardy
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5 Comments
Uri Agassi
return is no good here (LocalJumpError: unexpected return)
Denis de Bernardy
@UriAgassi: The example I wrote is not using
return. It's using break. One with each could do the same. One with map would be similar too (don't forget to return i in that case, else you'll get an array of nil.)Uri Agassi
You say in your explanation "... and
return or break" - return won't work...Denis de Bernardy
@UriAgassi: well, yeah, it won't work outside of a function. But I would assume OP isn't using IRB.
def foo; n = 2; ([0,1,3,4]*5).each { |i| n -= 1 if i == 4; return false if n == 0 }; end; fooUri Agassi
You did not enclose it in a functions, so not using IRB may cause more confusion, as the whole function this snippet will appear in will return unexpectedly.
Another solution, which stops as soon as it can is based on using any?:
def has_at_least?(arr, val, min)
arr.any? do |i|
i == val && (min -= 1).zero?
end
end
has_at_least?([0,1,3,4]*5, 4, 2)
# => true
has_at_least?([0,1,3,4]*5, 4, 7)
# => false