i have below code which returns hash value at the end of the method . what i want to do is to return same response without declaring response = {} in the beginning of the method. how to do this in ruby ?
def test
response = {}
objects.each do |object|
response = object.fetch(55)
break unless response.nil?
end
response
end
You want to return the first result of fetch(55)? I'd write it down like this
def test
objects.lazy.map{|o| o.fetch(55)}.reject(&:nil?).first
end
o.fetch(55, nil) or o[55].
If obtaining a value for each element of the given array (presumably a hash) is cheap, you could write the following.
def test1(arr)
arr.find { |h| h[55] }&.[](55)
end
else I'd suggest
def test2(arr)
v = nil
arr.find { |h| v = h[55] } && v
end
even though it doesn't meet your requirement. & in the first is the safe navigation operator, which made its debut in Ruby v2.3. By "cheap" I mean that no extensive calculation is required to determine the value of a key (computing pi to one million digits, for example). See Hash#[].
Suppose
a1 = [{ 9=>'cat', 14=>'dog', 12=>'pig' }, { 9=>'cat', 55=>'dog', 12=>'pig' }]
a2 = [{ 9=>'cat', 14=>'dog', 12=>'pig' }, { 9=>'cat', 12=>'dog', 12=>'pig' }]
then
test1(a1)
#=> 'dog'
test1(a2)
#=> nil
test2(a1)
#=> 'dog'
test2(a2)
#=> nil
As pointed out in a comment, the following is a variant of test1.
def test1(arr)
arr.find(->{{}}) { |h| h[55] }[55]
end
See Enumerable#find, specifically, if no match is found and find has an argument which responds to call (namely, a proc or method), that argument is called and the result is returned by find. (To use a method rather than a proc: def m() {} end; arr.find(method(:m)) { |h| h[55] }[55].)
arr.find (->{{}}) { |h| h[55] }[55]?(arr.find ->{{}} { |h| h[55] })[55], which works, but I don't understand it. Please explain.arr.find(->{{}}) { |h| h[55] }[55]. Enumerable#find accepts an ifnone callable object to be called in the event no matching element is found, so in this case we are just passing a lambda that returns an empty Hash. Then we use Hash#[] to act on the empty hash or the hash returned by the find block.Why not return from the block?
def test
objects.find do |object|
response = object.fetch(55)
return response if response
end
end
Note that this only works in a method. You get a LocalJumpError if you try this in irb, for example.
If you just want to get the value out of the block and not actually return from the method, use break:
def test
response = objects.find do |object|
response = object.fetch(55)
break response if response
end
end
Ruby automatically returns the value that is placed in the last line so you can use that logicc
objects.each {...} returns objects (each always returns its receiver), so removing the line response would cause objects to be returned.If you're just looking for the first instance in the collection that matches a conditional, ruby has an idiomatic solution: #detect
object = [{1 => true}, {2 => true}, {55 => 'foo'}, {55 => 'bar'}]
object.detect { |o| o.fetch(55) { false } } #=> {55=>"foo"}
