I have two hashes which have a structure something similar to this:
hash_a = { :a => { :b => { :c => \"d\" } } }
hash_b = { :a => { :b => { :x
Ruby's existing Hash#merge
allows a block form for resolving duplicates, making this rather simple. I've added functionality for merging multiple conflicting values at the 'leaves' of your tree into an array; you could choose to pick one or the other instead.
hash_a = { :a => { :b => { :c => "d", :z => 'foo' } } }
hash_b = { :a => { :b => { :x => "y", :z => 'bar' } } }
def recurse_merge(a,b)
a.merge(b) do |_,x,y|
(x.is_a?(Hash) && y.is_a?(Hash)) ? recurse_merge(x,y) : [*x,*y]
end
end
p recurse_merge( hash_a, hash_b )
#=> {:a=>{:b=>{:c=>"d", :z=>["foo", "bar"], :x=>"y"}}}
Or, as a clean monkey-patch:
class Hash
def merge_recursive(o)
merge(o) do |_,x,y|
if x.respond_to?(:merge_recursive) && y.is_a?(Hash)
x.merge_recursive(y)
else
[*x,*y]
end
end
end
end
p hash_a.merge_recursive hash_b
#=> {:a=>{:b=>{:c=>"d", :z=>["foo", "bar"], :x=>"y"}}}
If you change the first line of recursive_merge to
merged_hash = merge_to.clone
it works as expected:
recursive_merge(hash_a, hash_b)
-> {:a=>{:b=>{:c=>"d", :x=>"y"}}}
Changing the hash as you move through it is troublesome, you need a "work area" to accumulate your results.
Try this monkey-patching solution:
class Hash
def recursive_merge(hash = nil)
return self unless hash.is_a?(Hash)
base = self
hash.each do |key, v|
if base[key].is_a?(Hash) && hash[key].is_a?(Hash)
base[key].recursive_merge(hash[key])
else
base[key]= hash[key]
end
end
base
end
end
You can do it in one line :
merged_hash = hash_a.merge(hash_b){|k,hha,hhb| hha.merge(hhb){|l,hhha,hhhb| hhha.merge(hhhb)}}
If you want to imediatly merge
the result into hash_a, just replace the method merge by the method merge!
If you are using rails 3 or rails 4 framework, it is even easier :
merged_hash = hash_a.deep_merge(hash_b)
or
hash_a.deep_merge!(hash_b)
In order to merge one into the other as the ticket suggested, you could modify @Phrogz function
def recurse_merge( merge_from, merge_to )
merge_from.merge(merge_to) do |_,x,y|
(x.is_a?(Hash) && y.is_a?(Hash)) ? recurse_merge(x,y) : x
end
end
In case there is duplicate key, it will only use the content of merge_from
hash
Here is even better solution for recursive merging that uses refinements and has bang method alongside with block support. This code does work on pure Ruby.
module HashRecursive
refine Hash do
def merge(other_hash, recursive=false, &block)
if recursive
block_actual = Proc.new {|key, oldval, newval|
newval = block.call(key, oldval, newval) if block_given?
[oldval, newval].all? {|v| v.is_a?(Hash)} ? oldval.merge(newval, &block_actual) : newval
}
self.merge(other_hash, &block_actual)
else
super(other_hash, &block)
end
end
def merge!(other_hash, recursive=false, &block)
if recursive
self.replace(self.merge(other_hash, recursive, &block))
else
super(other_hash, &block)
end
end
end
end
using HashRecursive
After using HashRecursive
was executed you can use default Hash::merge
and Hash::merge!
as if they haven't been modified. You can use blocks with these methods as before.
The new thing is that you can pass boolean recursive
(second argument) to these modified methods and they will merge hashes recursively.
Example usage for answering the question. It's extremely easy:
hash_a = { :a => { :b => { :c => "d" } } }
hash_b = { :a => { :b => { :x => "y" } } }
puts hash_a.merge(hash_b) # Won't override hash_a
# output: { :a => { :b => { :x => "y" } } }
puts hash_a # hash_a is unchanged
# output: { :a => { :b => { :c => "d" } } }
hash_a.merge!(hash_b, recursive=true) # Will override hash_a
puts hash_a # hash_a was changed
# output: { :a => { :b => { :c => "d", :x => "y" } } }
For advanced example take a look at this answer.
Also take a look at my recursive version of Hash::each
(Hash::each_pair
) here.