This is my code:
def mainFunction
#notes=Hash.new(Array.new)
#notes["First"].push(true)
#notes["First"].push(false)
#notes["First"].push(true)
#notes["Second"].push(true)
output #notes.size.to_s
end
Why is the output 0? It should be 2 since notes has two keys, "First" and "Second".
When initializing Hash values when accessed for the first time (ie. pushing onto key'd values that don't yet exist), you need to set the key on the Hash when it is requested.
#notes = Hash.new {|h, k| h[k] = []}
For reference, see the following result in the ruby repl initializing the Hash as you have
irb(main):063:0> #notes = Hash.new(Array.new)
=> {}
irb(main):064:0> #notes[:foo].push(true)
=> [true]
irb(main):065:0> #notes.has_key?(:foo)
=> false
irb(main):066:0> puts #notes.size
=> 0
and now the proper way.
irb(main):067:0> #notes = Hash.new {|h, k| h[k] = []}
=> {}
irb(main):068:0> #notes[:foo].push(true)
=> [true]
irb(main):069:0> #notes.has_key?(:foo)
=> true
irb(main):070:0> #notes.size
=> 1
The following statement:
#notes=Hash.new(Array.new)
Creates a hash with a default value: Array.new. When you access the hash with an unknown key, you will receive this default value.
Your other statements therefor change that default array:
#notes["First"].push(true)
This adds true to the default (empty) array.
You need to initialize the "First" array first:
#notes["First"] = []
And then you can add values:
#notes["First"].push(true)
Or more "rubyish":
#notes["First"] << true
The size now is:
#notes.size # => 1
#notes = Hash.new(Array.new) sets the default value for all elements to the same array. Since #notes contains no key "First", it returns that default value, and the .push adds to the end of it. But the key has not been added to the hash. The other pushes add to the end of the same default array.
Read this Hash.new(obj). It states:
If obj is specified, this single object will be used for all default
values.
Hence, when you do: #notes=Hash.new(Array.new). The default object will be an array.
#notes.default # => []
#notes['First'].push(true)
#notes.default # => [true]
#notes['First'].push(false)
#notes.default # => [true, false]
But, there won't be any entry as a key in #notes, of course you use access those values by giving any key(which may or may not exist), like this:
#notes['unknown_key'] #=> [true, false]
Hash.new(object) method call form just return object as a default value but does not update the hash.
You are looking for a block form where you can do the update:
#notes=Hash.new {|h, k| h[k] = []}
Related
Suppose I have an hash = {1=>1,2=>2} to clear the values I can do either hash = {} or hash.clear . What is the difference between them?
According to the docs, if you see the source code for clear method, it iterates through each element in the hash and removes it.
When you do hash = {}, it will just create a new empty hash object but the old object will still be in memory ready for garbage collection if there are no other references for that object.
They are completely different.
The first one will simply bind the local variable hash to a different object. In particular, it will not "clear the values", as you claim:
hash = {i: 'am', still: 'there'}
another = hash
hash = {}
another
# => {i: 'am', still: 'there'}
The second one will actually clear the values:
hash = {i: 'am', still: 'there'}
another = hash
hash.clear
another
# => {}
Atri's answer is right, but beyond the Hash still existing in memory for GC when you use reassignment -- is if the array is still actually referenced by another variable.
A Hash is an object, a variable is just a pointer or reference to it. #clear clears the object itself, reassigning the variable just points the variable to a new Hash.
hash1 = {1=>1,2=>2}
hash2 = hash1
puts hash1 # => {1=>1,2=>2}
puts hash2 # => {1=>1,2=>2}
hash1 = {}
puts hash1 # => {}
puts hash2 # => {1=>1,2=>2}
hash1 = {1=>1,2=>2}
hash2 = hash1
hash1.clear
puts hash1 # => {}
puts hash2 # => {}
This question already has answers here:
Strange, unexpected behavior (disappearing/changing values) when using Hash default value, e.g. Hash.new([])
(4 answers)
Closed 3 years ago.
I want to initialize a Hash with an empty Array and for every new key push a certain value to that array.
Here's what I tried to do:
a = Hash.new([])
# => {}
a[1] << "asd"
# => ["asd"]
a
# => {}
The expected output for a was {1 => ["asd"]} but that didn't happen. What am I missing here?
Ruby version:
ruby 2.0.0p598 (2014-11-13 revision 48408) [x86_64-linux]
Just do
a = Hash.new { |h, k| h[k] = [] }
a[1] << "asd"
a # => {1=>["asd"]}
Read the below lines from the Hash::new documentation. It really explains why you didn't get the desired result.
new(obj) → new_hash
If obj is specified, this single object will be used for all default values.
new {|hash, key| block } → new_hash
If a block is specified, it will be called with the hash object and the key, and should return the default value. It is the block’s responsibility to store the value in the hash if required.
You can test by hand :
a = Hash.new([])
a[1].object_id # => 2160424560
a[2].object_id # => 2160424560
Now with the above style of Hash object creation, you can see every access to an unknown key, returning back the same default object. Now the other way, I meant block way :
b = Hash.new { |h, k| [] }
b[2].object_id # => 2168989980
b[1].object_id # => 2168933180
So, with the block form, every unknown key access, returning a new Array object.
What's the best way to add a key-value pair to an hash object, from within the hash object itself?
The common way I know to add a new key to a hash is as follows:
hash = Hash.new
hash[:key] = 'value'
hash[:key] # => 'value'
What if I wan't to create a new hash which already has this key after its creation?
hash = Hash.new
hash[:key] # => 'value'
Is this possible? Thanks!
To create a Hash with an already initialized set of values you can do:
hash = { :key => 'value' }
hash[:key] # ===> This evaluates to 'value'
Just remember, the idiomatic way to create an empty hash in Ruby is:
hash = {}
Not hash = Hash.new like you exemplified.
Do you mean set the default value? is so, you could do with:
hash = Hash.new('value')
hash[:key] # => 'value'
Not sure what you mean i the other answers aren't what you want, you can create a hash with some keys and values allready filled like this
hash = {:key => 'value'} #{:key=>"value"}
and like the others said, the default value for key's not allready present is given by passing a block tot the hash at creation time like
hash = Hash.new('value') #{}
hash[:test] #"value"
or
h = Hash.new { |hash, key| hash[key] = "Go Fish: #{key}" } #{}
h[:test] #"Go Fish: test"
Thje last sample teken from http://www.ruby-doc.org/core-1.9.3/Hash.html
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Strange ruby behavior when using Hash.new([])
This is a simple one, as I'm lost for words.
Why is this happening:
1.9.3-p194 :001 > h = Hash.new([])
=> {}
1.9.3-p194 :002 > h[:key1] << "Ruby"
=> ["Ruby"]
1.9.3-p194 :003 > h
=> {}
1.9.3-p194 :004 > h.keys
=> []
1.9.3-p194 :005 > h[:key1]
=> ["Ruby"]
When you create a hash like this:
h = Hash.new([])
it means, whenever the hash is accessed with a key that has not been defined yet, its going to return:
[]
Now when you do :
h[:key1] << "Ruby"
h[:key1] has returned [] , to which "Ruby" got pushed, resulting in ["Ruby"], as output, as that is the last object returned. That has also got set as the default value to return when 'h' is accessed with an undefined key.
Hence, when you do :
h[:key1] or h[:key2] or h[:whatever]
You will get
"Ruby"
as output.
Hope this helps.
Look at the documentation of Hash.new
new → new_hash
new(obj) → new_hash
new {|hash, key| block } → new_hash
If this hash is subsequently accessed by a key that doesn’t correspond to a hash entry, the value returned depends on the style of new used to create the hash.
In the first form, the access returns nil.
If obj is specified, this single object will be used for all default values.
If a block is specified, it will be called with the hash object and the key, and should return the default value. It is the block’s responsibility to store the value in the hash if required.
irb(main):015:0> h[:abc] # ["Ruby"]
So ["Ruby"] is returned as default value instead of nil if key is not found.
This construction Hash.new([]) returns default value but this value is not initialized value of hash. You're trying to work with hash assuming that the default value is a part of hash.
What you need is construction which will initialize the hash at some key:
hash = Hash.new { |h,k| h[k] = [] }
hash[:key1] << "Ruby"
hash #=> {:key1=>["Ruby"]}
You actually did not set the value with h[:keys] << "Ruby". You just add a value for the returned default array of a not found key. So no key is created.
If you write this, it will be okay:
h = Hash.new([])
h[:keys1] = []
h[:keys1] << "Ruby"
I have to admit this tripped me out too when I read your question. I had a look at the docs and it became clear though.
If obj is specified, this single object will be used for all default values.
So what you actually doing is modifying this one single array object that is used for the default values, without ever assigning to the key!
Check it out:
h = Hash.new([])
h[:x] << 'x'
# => ['x']
h
# => {}
h[:y]
# => ['x'] # CRAZY TIMES
So you need to do assignment somehow - h[:x] += ['x'] might be the way to go.
This question already has answers here:
Strange, unexpected behavior (disappearing/changing values) when using Hash default value, e.g. Hash.new([])
(4 answers)
Closed 7 years ago.
hash = Hash.new(Hash.new([]))
hash[1][2] << 3
hash[1][2] # => [3]
hash # => {}
hash.keys # => []
hash.values # => []
What's going on? Ruby's hiding data (1.9.3p125)
What's going on? Ruby's hiding data (1.9.3p125)
Ruby hides neither data nor its docs.
Default value you pass into the Hash constructor is returned whenever the key is not found in the hash. But this default value is never actually stored into the hash on its own.
To get what you want you should use Hash constructor with block and store default value into the hash yourself (on both levels of your nested hash):
hash = Hash.new { |hash, key| hash[key] = Hash.new { |h, k| h[k] = [] } }
hash[1][2] << 3
p hash[1][2] #=> [3]
p hash #=> {1=>{2=>[3]}}
p hash.keys #=> [1]
p hash.values #=> [{2=>[3]}]
It's simple. If you pass an object to a Hash constructor, it'll become a default value for all missing keys in that hash. What's interesting is that this value is mutable. Observe:
hash = Hash.new(Hash.new([]))
# mutate default value for nested hash
hash[1][2] << 3
# default value in action
hash[1][2] # => [3]
# and again
hash[1][3] # => [3]
# and again
hash[1][4] # => [3]
# set a plain hash (without default value)
hash[1] = {}
# what? Where did my [3] go?
hash[1][2] # => nil
# ah, here it is!
hash[2][3] # => [3]
I get a try with this in irb.
Seams Ruby does not tag element as "visible" except affecting value over default explicitly via = for instance.
hash = Hash.new(Hash.new([]))
hash[1] = Hash.new([])
hash[1][2] = [3]
hash
#=> {1=>{2=>[3]}}
May be some setters are missing this "undefaulting" behavior ...