Accessing elements of nested hashes in ruby

Hash and Array have a method called dig that solves this problem entirely.

value = structure.dig(:a, :b)

It returns nil if the key is missing at any level.

If you are using a version of Ruby older than 2.3, you can use the ruby_dig gem.


The way I usually do this these days is:

h = Hash.new { |h,k| h[k] = {} }

This will give you a hash that creates a new hash as the entry for a missing key, but returns nil for the second level of key:

h['foo'] -> {}
h['foo']['bar'] -> nil

You can nest this to add multiple layers that can be addressed this way:

h = Hash.new { |h, k| h[k] = Hash.new { |hh, kk| hh[kk] = {} } }

h['bar'] -> {}
h['tar']['zar'] -> {}
h['scar']['far']['mar'] -> nil

You can also chain indefinitely by using the default_proc method:

h = Hash.new { |h, k| h[k] = Hash.new(&h.default_proc) }

h['bar'] -> {}
h['tar']['star']['par'] -> {}

The above code creates a hash whose default proc creates a new Hash with the same default proc. So, a hash created as a default value when a lookup for an unseen key occurs will have the same default behavior.

EDIT: More details

Ruby hashes allow you to control how default values are created when a lookup occurs for a new key. When specified, this behavior is encapsulated as a Proc object and is reachable via the default_proc and default_proc= methods. The default proc can also be specified by passing a block to Hash.new.

Let's break this code down a little. This is not idiomatic ruby, but it's easier to break it out into multiple lines:

1. recursive_hash = Hash.new do |h, k|
2.   h[k] = Hash.new(&h.default_proc)
3. end

Line 1 declares a variable recursive_hash to be a new Hash and begins a block to be recursive_hash's default_proc. The block is passed two objects: h, which is the Hash instance the key lookup is being performed on, and k, the key being looked up.

Line 2 sets the default value in the hash to a new Hash instance. The default behavior for this hash is supplied by passing a Proc created from the default_proc of the hash the lookup is occurring in; ie, the default proc the block itself is defining.

Here's an example from an IRB session:

irb(main):011:0> recursive_hash = Hash.new do |h,k|
irb(main):012:1* h[k] = Hash.new(&h.default_proc)
irb(main):013:1> end
=> {}
irb(main):014:0> recursive_hash[:foo]
=> {}
irb(main):015:0> recursive_hash
=> {:foo=>{}}

When the hash at recursive_hash[:foo] was created, its default_proc was supplied by recursive_hash's default_proc. This has two effects:

  1. The default behavior for recursive_hash[:foo] is the same as recursive_hash.
  2. The default behavior for hashes created by recursive_hash[:foo]'s default_proc will be the same as recursive_hash.

So, continuing in IRB, we get the following:

irb(main):016:0> recursive_hash[:foo][:bar]
=> {}
irb(main):017:0> recursive_hash
=> {:foo=>{:bar=>{}}}
irb(main):018:0> recursive_hash[:foo][:bar][:zap]
=> {}
irb(main):019:0> recursive_hash
=> {:foo=>{:bar=>{:zap=>{}}}}

Traditionally, you really had to do something like this:

structure[:a] && structure[:a][:b]

However, Ruby 2.3 added a method Hash#dig that makes this way more graceful:

structure.dig :a, :b # nil if it misses anywhere along the way

There is a gem called ruby_dig that will back-patch this for you.