Array#find!

Ruby
1

How to remove the first occurrence of something?

a=[1,2,3,4]
a.find!{ |e| a==2 } #=> 2

But there is no #find!, how to define it? My solutions seem unduly
complex.

Thanks,
T.

2

On 2/29/08, 7stud – [email protected] wrote:

T.
end
end

Why not just…
arr = [1, 2, 2, 2, 3]
arr.delete_at(arr.index(2))

arr => [1, 2, 2, 3]

Christopher

3

7stud – wrote:

arr.each_with_index do |elmt, i|
if elmt == 2
arr.delete_at(i)
break
end
end

If you’re going that far, why not just

arr.delete_at(arr.index(2))

(Of course that’s limited to #== comparisons. Maybe Array#index should
take a block…)

4

Trans wrote:

How to remove the first occurrence of something?

a=[1,2,3,4]
a.find!{ |e| a==2 } #=> 2

But there is no #find!, how to define it? My solutions seem unduly
complex.

Thanks,
T.

More complex than the following?

arr = [1, 2, 3, 2, 2, 2]

arr.each_with_index do |elmt, i|
if elmt == 2
arr.delete_at(i)
break
end
end

p arr

[1, 3, 2, 2, 2]

5

On Sat, 2008-03-01 at 10:15 +0900, 7stud – wrote:

arr.each_with_index do |elmt, i|
if elmt == 2
arr.delete_at(i)
break
end
end

This won’t work in 1.9: “RuntimeError: can’t modify array during
iteration”.

Best,
Andre

6

From: Joel VanderWerf [mailto:[email protected]]

arr.delete_at(arr.index(2))

(Of course that’s limited to #== comparisons. Maybe

Array#index should

take a block…)

of course, we agree, that is why matz put it in 1.9 :wink:

irb(main):011:0> RUBY_VERSION
=> “1.9.0”
irb(main):012:0> a
=> [“this”, “is”, “a”, “test”]
irb(main):013:0> a.index{|e| e=~ /^is/}
=> 1
irb(main):014:0> a
=> [“this”, “is”, “a”, “test”]
irb(main):015:0> a.delete_at(a.index{|e| e=~ /^is/})
=> “is”
irb(main):016:0> a
=> [“this”, “a”, “test”]

and 1.9 is fast :slight_smile:

kind regards -botp