Related
Given I have this hash:
h = { a: 'a', b: 'b', c: { d: 'd', e: 'e'} }
And I convert to OpenStruct:
o = OpenStruct.new(h)
=> #<OpenStruct a="a", b="b", c={:d=>"d", :e=>"e"}>
o.a
=> "a"
o.b
=> "b"
o.c
=> {:d=>"d", :e=>"e"}
2.1.2 :006 > o.c.d
NoMethodError: undefined method `d' for {:d=>"d", :e=>"e"}:Hash
I want all the nested keys to be methods as well. So I can access d as such:
o.c.d
=> "d"
How can I achieve this?
You can monkey-patch the Hash class
class Hash
def to_o
JSON.parse to_json, object_class: OpenStruct
end
end
then you can say
h = { a: 'a', b: 'b', c: { d: 'd', e: 'e'} }
o = h.to_o
o.c.d # => 'd'
See Convert a complex nested hash to an object.
I came up with this solution:
h = { a: 'a', b: 'b', c: { d: 'd', e: 'e'} }
json = h.to_json
=> "{\"a\":\"a\",\"b\":\"b\",\"c\":{\"d\":\"d\",\"e\":\"e\"}}"
object = JSON.parse(json, object_class:OpenStruct)
object.c.d
=> "d"
So for this to work, I had to do an extra step: convert it to json.
personally I use the recursive-open-struct gem - it's then as simple as RecursiveOpenStruct.new(<nested_hash>)
But for the sake of recursion practice, I'll show you a fresh solution:
require 'ostruct'
def to_recursive_ostruct(hash)
result = hash.each_with_object({}) do |(key, val), memo|
memo[key] = val.is_a?(Hash) ? to_recursive_ostruct(val) : val
end
OpenStruct.new(result)
end
puts to_recursive_ostruct(a: { b: 1}).a.b
# => 1
edit
Weihang Jian showed a slight improvement to this here https://stackoverflow.com/a/69311716/2981429
def to_recursive_ostruct(hash)
hash.each_with_object(OpenStruct.new) do |(key, val), memo|
memo[key] = val.is_a?(Hash) ? to_recursive_ostruct(val) : val
end
end
Also see https://stackoverflow.com/a/63264908/2981429 which shows how to handle arrays
note
the reason this is better than the JSON-based solutions is because you can lose some data when you convert to JSON. For example if you convert a Time object to JSON and then parse it, it will be a string. There are many other examples of this:
class Foo; end
JSON.parse({obj: Foo.new}.to_json)["obj"]
# => "#<Foo:0x00007fc8720198b0>"
yeah ... not super useful. You've completely lost your reference to the actual instance.
Here's a recursive solution that avoids converting the hash to json:
def to_o(obj)
if obj.is_a?(Hash)
return OpenStruct.new(obj.map{ |key, val| [ key, to_o(val) ] }.to_h)
elsif obj.is_a?(Array)
return obj.map{ |o| to_o(o) }
else # Assumed to be a primitive value
return obj
end
end
My solution is cleaner and faster than #max-pleaner's.
I don't actually know why but I don't instance extra Hash objects:
def dot_access(hash)
hash.each_with_object(OpenStruct.new) do |(key, value), struct|
struct[key] = value.is_a?(Hash) ? dot_access(value) : value
end
end
Here is the benchmark for you reference:
require 'ostruct'
def dot_access(hash)
hash.each_with_object(OpenStruct.new) do |(key, value), struct|
struct[key] = value.is_a?(Hash) ? dot_access(value) : value
end
end
def to_recursive_ostruct(hash)
result = hash.each_with_object({}) do |(key, val), memo|
memo[key] = val.is_a?(Hash) ? to_recursive_ostruct(val) : val
end
OpenStruct.new(result)
end
require 'benchmark/ips'
Benchmark.ips do |x|
hash = { a: 1, b: 2, c: { d: 3 } }
x.report('dot_access') { dot_access(hash) }
x.report('to_recursive_ostruct') { to_recursive_ostruct(hash) }
end
Warming up --------------------------------------
dot_access 4.843k i/100ms
to_recursive_ostruct 5.218k i/100ms
Calculating -------------------------------------
dot_access 51.976k (± 5.0%) i/s - 261.522k in 5.044482s
to_recursive_ostruct 50.122k (± 4.6%) i/s - 250.464k in 5.008116s
My solution, based on max pleaner's answer and similar to Xavi's answer:
require 'ostruct'
def initialize_open_struct_deeply(value)
case value
when Hash
OpenStruct.new(value.transform_values { |hash_value| send __method__, hash_value })
when Array
value.map { |element| send __method__, element }
else
value
end
end
Here is one way to override the initializer so you can do OpenStruct.new({ a: "b", c: { d: "e", f: ["g", "h", "i"] }}).
Further, this class is included when you require 'json', so be sure to do this patch after the require.
class OpenStruct
def initialize(hash = nil)
#table = {}
if hash
hash.each_pair do |k, v|
self[k] = v.is_a?(Hash) ? OpenStruct.new(v) : v
end
end
end
def keys
#table.keys.map{|k| k.to_s}
end
end
Basing a conversion on OpenStruct works fine until it doesn't. For instance, none of the other answers here properly handle these simple hashes:
people = { person1: { display: { first: 'John' } } }
creds = { oauth: { trust: true }, basic: { trust: false } }
The method below works with those hashes, modifying the input hash rather than returning a new object.
def add_indifferent_access!(hash)
hash.each_pair do |k, v|
hash.instance_variable_set("##{k}", v.tap { |v| send(__method__, v) if v.is_a?(Hash) } )
hash.define_singleton_method(k, proc { hash.instance_variable_get("##{k}") } )
end
end
then
add_indifferent_access!(people)
people.person1.display.first # => 'John'
Or if your context calls for a more inline call structure:
creds.yield_self(&method(:add_indifferent_access!)).oauth.trust # => true
Alternatively, you could mix it in:
module HashExtension
def very_indifferent_access!
each_pair do |k, v|
instance_variable_set("##{k}", v.tap { |v| v.extend(HashExtension) && v.send(__method__) if v.is_a?(Hash) } )
define_singleton_method(k, proc { self.instance_variable_get("##{k}") } )
end
end
end
and apply to individual hashes:
favs = { song1: { title: 'John and Marsha', author: 'Stan Freberg' } }
favs.extend(HashExtension).very_indifferent_access!
favs.song1.title
Here is a variation for monkey-patching Hash, should you opt to do so:
class Hash
def with_very_indifferent_access!
each_pair do |k, v|
instance_variable_set("##{k}", v.tap { |v| v.send(__method__) if v.is_a?(Hash) } )
define_singleton_method(k, proc { instance_variable_get("##{k}") } )
end
end
end
# Note the omission of "v.extend(HashExtension)" vs. the mix-in variation.
Comments to other answers expressed a desire to retain class types. This solution accommodates that.
people = { person1: { created_at: Time.now } }
people.with_very_indifferent_access!
people.person1.created_at.class # => Time
Whatever solution you choose, I recommend testing with this hash:
people = { person1: { display: { first: 'John' } }, person2: { display: { last: 'Jingleheimer' } } }
If you are ok with monkey-patching the Hash class, you can do:
require 'ostruct'
module Structurizable
def each_pair(&block)
each do |k, v|
v = OpenStruct.new(v) if v.is_a? Hash
yield k, v
end
end
end
Hash.prepend Structurizable
people = { person1: { display: { first: 'John' } }, person2: { display: { last: 'Jingleheimer' } } }
puts OpenStruct.new(people).person1.display.first
Ideally, instead of pretending this, we should be able to use a Refinement, but for some reason I can't understand it didn't worked for the each_pair method (also, unfortunately Refinements are still pretty limited)
h = {
data: {
user: {
value: "John Doe"
}
}
}
To assign value to the nested hash, we can use
h[:data][:user][:value] = "Bob"
However if any part in the middle is missing, it will cause error.
Something like
h.dig(:data, :user, :value) = "Bob"
won't work, since there's no Hash#dig= available yet.
To safely assign value, we can do
h.dig(:data, :user)&.[]=(:value, "Bob") # or equivalently
h.dig(:data, :user)&.store(:value, "Bob")
But is there better way to do that?
It's not without its caveats (and doesn't work if you're receiving the hash from elsewhere), but a common solution is this:
hash = Hash.new {|h,k| h[k] = h.class.new(&h.default_proc) }
hash[:data][:user][:value] = "Bob"
p hash
# => { :data => { :user => { :value => "Bob" } } }
And building on #rellampec's answer, ones that does not throw errors:
def dig_set(obj, keys, value)
key = keys.first
if keys.length == 1
obj[key] = value
else
obj[key] = {} unless obj[key]
dig_set(obj[key], keys.slice(1..-1), value)
end
end
obj = {d: 'hey'}
dig_set(obj, [:a, :b, :c], 'val')
obj #=> {d: 'hey', a: {b: {c: 'val'}}}
interesting one:
def dig_set(obj, keys, value)
if keys.length == 1
obj[keys.first] = value
else
dig_set(obj[keys.first], keys.slice(1..-1), value)
end
end
will raise an exception anyways if there's no [] or []= methods.
I found a simple solution to set the value of a nested hash, even if a parent key is missing, even if the hash already exists. Given:
x = { gojira: { guitar: { joe: 'charvel' } } }
Suppose you wanted to include mario's drum to result in:
x = { gojira: { guitar: { joe: 'charvel' }, drum: { mario: 'tama' } } }
I ended up monkey-patching Hash:
class Hash
# ensures nested hash from keys, and sets final key to value
# keys: Array of Symbol|String
# value: any
def nested_set(keys, value)
raise "DEBUG: nested_set keys must be an Array" unless keys.is_a?(Array)
final_key = keys.pop
return unless valid_key?(final_key)
position = self
for key in keys
return unless valid_key?(key)
position[key] = {} unless position[key].is_a?(Hash)
position = position[key]
end
position[final_key] = value
end
private
# returns true if key is valid
def valid_key?(key)
return true if key.is_a?(Symbol) || key.is_a?(String)
raise "DEBUG: nested_set invalid key: #{key} (#{key.class})"
end
end
usage:
x.nested_set([:instrument, :drum, :mario], 'tama')
usage for your example:
h.nested_set([:data, :user, :value], 'Bob')
any caveats i missed? any better way to write the code without sacrificing readability?
Searching for an answer to a similar question I developmentally stumbled upon an interface similar to #niels-kristian's answer, but wanted to also support a namespace definition parameter, like an xpath.
def deep_merge(memo, source)
# From: http://www.ruby-forum.com/topic/142809
# Author: Stefan Rusterholz
merger = proc { |key, v1, v2| Hash === v1 && Hash === v2 ? v1.merge(v2, &merger) : v2 }
memo.merge!(source, &merger)
end
# Like Hash#dig, but for setting a value at an xpath
def bury(memo, xpath, value, delimiter=%r{\.})
xpath = xpath.split(delimiter) if xpath.respond_to?(:split)
xpath.map!{|x|x.to_s.to_sym}.push(value)
deep_merge(memo, xpath.reverse.inject { |memo, field| {field.to_sym => memo} })
end
Nested hashes are sort of like xpaths, and the opposite of dig is bury.
irb(main):014:0> memo = {:test=>"value"}
=> {:test=>"value"}
irb(main):015:0> bury(memo, 'test.this.long.path', 'value')
=> {:test=>{:this=>{:long=>{:path=>"value"}}}}
irb(main):016:0> bury(memo, [:test, 'this', 2, 4.0], 'value')
=> {:test=>{:this=>{:long=>{:path=>"value"}, :"2"=>{:"4.0"=>"value"}}}}
irb(main):017:0> bury(memo, 'test.this.long.path.even.longer', 'value')
=> {:test=>{:this=>{:long=>{:path=>{:even=>{:longer=>"value"}}}, :"2"=>{:"4.0"=>"value"}}}}
irb(main):018:0> bury(memo, 'test.this.long.other.even.longer', 'other')
=> {:test=>{:this=>{:long=>{:path=>{:even=>{:longer=>"value"}}, :other=>{:even=>{:longer=>"other"}}}, :"2"=>{:"4.0"=>"value"}}}}
A more ruby-helper-like version of #niels-kristian answer
You can use it like:
a = {}
a.bury!([:a, :b], "foo")
a # => {:a => { :b => "foo" }}
class Hash
def bury!(keys, value)
key = keys.first
if keys.length == 1
self[key] = value
else
self[key] = {} unless self[key]
self[key].bury!(keys.slice(1..-1), value)
end
self
end
end
Consider the Following nested Hash:
data1 = {
"3"=>{"passenger_type"=>"ADT", "the_order"=>"3", "last"=>"JONES", "first"=>"ALENA", "middle"=>nil},
"2"=>{"passenger_type"=>"ADT", "the_order"=>"2", "last"=>"JONES", "first"=>"MAXIM", "middle"=>nil},
"1"=>{"passenger_type"=>"ADTT", "the_order"=>"1", "last"=>"JONES", "first"=>"TODD", "middle"=>nil}}
data2 = {
"3"=>{"first"=>"ALENA", "the_order"=>"3", "middle"=>"", "passenger_type"=>"ADTT", "last"=>"JONES"},
"2"=>{"first"=>"MAXIM", "the_order"=>"2", "middle"=>"", "passenger_type"=>"ADT", "last"=>"JONES"},
"1"=>{"first"=>"TODD", "the_order"=>"1", "middle"=>"", "passenger_type"=>"ADT", "last"=>"JONESS"}}
The Output Should be like this(difference between both hash listed values):
{"3" => {"passenger_type" => ["ADT", "ADTT"]},
"1" => {"passenger_type" => ["ADTT", "ADT"], "last" => ["JONES", "JONESS"]}
Anyone your suggestion is appreciated, thanks in advance.
You can use the form of Hash#merge that takes a block to produce the desired result in a compact manner:
data1.merge(data2) { |_,ho,hn|
ho.merge(hn) { |_,o,n| (o==n||o==''||n=='') ? nil : [o,n] }
.delete_if { |_,v| v==nil } }
.delete_if { |_,v| v.empty? }
#=> {"3"=>{"passenger_type"=>["ADT", "ADTT"]},
# "1"=>{"passenger_type"=>["ADTT", "ADT"], "last"=>["JONES", "JONESS"]}}
Here's some ugly code:
data3 = {}
data1.each do |k, v|
v2 = data2[k]
v.each do |item, val|
if v2.has_key?(item) then
if (val == nil or val == '') and (v2[item] == nil or v2[item] == '') then
next
end
if val != v2[item] then
data3[k] ||= {}
data3[k][item] = [val, v2[item]]
end
end
end
end
puts data3
prints
{"3"=>{"passenger_type"=>["ADT", "ADTT"]}, "1"=>{"passenger_type"=>["ADTT", "ADT"], "last"=>["JONES", "JONESS"]}}
What's a convenient way to get a list of all Hash keys (with nesting) separated by dots?
Given I have a hash:
{ level1: { level21: { level31: 'val1',
level32: 'val2' },
level22: 'val3' }
}
Desired output (array of strings) which represents all key paths in a hash:
level1.level21.level31
level1.level21.level32
level1.level22
My current solution:
class HashKeysDumper
def self.dump(hash)
hash.map do |k, v|
if v.is_a? Hash
keys = dump(v)
keys.map { |k1| [k, k1].join('.') }
else
k.to_s
end
end.flatten
end
end
It also available as gist (with specs).
Well, it depends on what you mean by cleaner, but here's a smaller version that…
Will work on subclasses Hashes or Hash-alikes
Extends Hash, making it look cleaner in your code.
class Hash
def keydump
map{|k,v|v.keydump.map{|a|"#{k}.#{a}"} rescue k.to_s}.flatten
end
end
results:
{ level1: { level21: { level31: 'val1',
level32: 'val2' },
level22: 'val3' }
}.keydump
=> ["level1.level21.level31", "level1.level21.level32", "level1.level22"]
Here is my vision of this:
h = { 'level1' => { 'level2' => { 'level31' => 'val1', 'level32' => 'val2' } } }
class Hash
def nested_keys
self.inject([]) { |f, (k,v)| f += [k, v.is_a?(Hash) ? v.nested_keys : []] }.flatten
end
end
keys = h.nested_keys
p keys
#=> ["level1", "level2", "level31", "level32"]
k1, k2 = keys.shift, keys.shift
puts [k1, k2, keys.shift].join('.')
#=> level1.level2.level31
puts [k1, k2, keys.shift].join('.')
#=> level1.level2.level32
Here is a Working Demo
I just committed some code to RubyTree that adds from_hash() which would allow you to do this:
require 'rubytree'
Tree::TreeNode.from_hash(hash).each_leaf.map{|n| "#{n.name}.#{n.parentage.map(&:name).reverse.join('.')}" }
=> ["level1.level21.level31", "level1.level21.level32", "level1.level22"]
Aside from the gem require, it's a one-liner :)
I want to programmatically convert this:
{
"a"=>
{"1"=>
{"A"=>
{"Standard"=>"true"}
}
},
"b"=>
{"1"=>
{"A"=>
{"Standard"=>"true"}
}
}
}
to an array like this:
['a/1/A/Standard', 'b/1/A/Standard']
def extract_keys(hash)
return [] unless hash.is_a?(Hash)
hash.each_pair.map {|key, value| [key, extract_keys(value)].join('/') }
end
extract_keys(hash)
=> ["a/1/A/Standard", "b/1/A/Standard"]
From one of my other answers - adapted for your situation. See the link for a more verbose solution to flat_hash
def flat_hash(hash, k = "")
return {k => hash} unless hash.is_a?(Hash)
hash.inject({}){ |h, v| h.merge! flat_hash(v[-1], k + '/' + v[0]) }
end
example = {...} # your example hash
foo = flat_hash(example).keys
=> ["/a/1/A/Standard", "/b/1/A/Standard"]
Found this flatten lambda definition.
h = {
"a"=>
{"1"=>
{"A"=>
{"Standard"=>"true"}
}
},
"b"=>
{"1"=>
{"A"=>
{"Standard"=>"true"}
}
}
}
a = []
flatten =
lambda {|r|
(recurse = lambda {|v|
if v.is_a?(Hash)
v.to_a.map{|v| recurse.call(v)}.flatten
elsif v.is_a?(Array)
v.flatten.map{|v| recurse.call(v)}
else
v.to_s
end
}).call(r)
}
h.each do |k,v|
a << k + "/" + flatten.call(v).join("/")
end
Output:
["a/1/A/Standard/true", "b/1/A/Standard/true"]