I don't know how to add new item to already existing hash. For example, first I construct hash:
hash = {item1: 1}
After that, I want to add item2, so after this I have hash like this:
{item1: 1, item2: 2}
I don't know what method to do on hash. Could someone help me?
Create the hash:
hash = {:item1 => 1}
Add a new item to it:
hash[:item2] = 2
If you want to add new items from another hash - use merge method:
hash = {:item1 => 1}
another_hash = {:item2 => 2, :item3 => 3}
hash.merge(another_hash) # {:item1=>1, :item2=>2, :item3=>3}
In your specific case it could be:
hash = {:item1 => 1}
hash.merge({:item2 => 2}) # {:item1=>1, :item2=>2}
but it's not wise to use it when you should to add just one element more.
Pay attention that merge will replace the values with the existing keys:
hash = {:item1 => 1}
hash.merge({:item1 => 2}) # {:item1=>2}
exactly like hash[:item1] = 2
Also you should pay attention that merge method (of course) doesn't effect the original value of hash variable - it returns a new merged hash. If you want to replace the value of the hash variable then use merge! instead:
hash = {:item1 => 1}
hash.merge!({:item2 => 2})
# now hash == {:item1=>1, :item2=>2}
hash.store(key, value) - Stores a key-value pair in hash.
Example:
hash #=> {"a"=>9, "b"=>200, "c"=>4}
hash.store("d", 42) #=> 42
hash #=> {"a"=>9, "b"=>200, "c"=>4, "d"=>42}
Documentation
It's as simple as:
irb(main):001:0> hash = {:item1 => 1}
=> {:item1=>1}
irb(main):002:0> hash[:item2] = 2
=> 2
irb(main):003:0> hash
=> {:item1=>1, :item2=>2}
hash[key]=value
Associates the value given by value with the key given by key.
hash[:newKey] = "newValue"
From Ruby documentation:
http://www.tutorialspoint.com/ruby/ruby_hashes.htm
hash_items = {:item => 1}
puts hash_items
#hash_items will give you {:item => 1}
hash_items.merge!({:item => 2})
puts hash_items
#hash_items will give you {:item => 1, :item => 2}
hash_items.merge({:item => 2})
puts hash_items
#hash_items will give you {:item => 1, :item => 2}, but the original variable will be the same old one.
Create hash as:
h = Hash.new
=> {}
Now insert into hash as:
h = Hash["one" => 1]
Related
Tell me how to do this, where you can read about it, because I do not understand at all how to implement it. Thanks.
def initialize
#arr = []
end
def items(init)
arrfinish = init - #arr
#arr = (#arr + init).uniq
yield arrfinish
end
def idefine(find_text)
end
The class has a method(items) that connects arrays by removing duplicate elements. I need to make sure that the idefine
method receives the key by which filtering will be performed when adding new elements, I will give an example below.
app_handler.idefine('id')
app_handler.items([{'id' => 1}, {'id' => 1, 'test_key' => 'Some data'}, {'id' => 2}])
From this example, the second element with id = 1 should be ignored.
Ok, putting aside the class definition, what I understand of the question is:
In an array of hashes, remove the hashes that contain a duplicate value of a given key.
The following function filter the hashes and copy the content of the selected ones in a new array:
require 'set'
def no_dup_val key, arr
previous_values = Set[]
arr.each_with_object([]) do |hash,result|
next unless hash.has_key?(key)
next if previous_values.include?(hash[key])
previous_values << hash[key]
result << hash.dup
end
end
Which gives you:
no_dup_val 'id', [{'id' => 1}, {'id' => 1, 'key' => 'data'}, {'id' => 2}, {'stock' => 3}, {'e-stock'=>0}]
#=> [{"id"=>1}, {"id"=>2}]
Note that the hashes that don't contain the key are also removed, that's my choice, which leads to the following questions:
What happens when the key is not present in a hash?
What happens when the item function is called more than once? Do you take into account the hashes already in #arr?
What happens when you call idefine with a new key? Do you filter the existing elements of #arr with the new key?
As you can see, you need to be a little more specific about what you want to do.
Update
If you don't care about copying the contents of the hashes then these may fit your needs.
Hashes without the id key are removed:
def no_dup_val key, arr
arr.filter{ |h| h.has_key?(key) }.uniq{ |h| h[key] }
end
no_dup_val 'id', [{'id' => 1}, {'id' => 1, 'key' => 'data'}, {'id' => 2}, {'stock' => 3}, {'e-stock'=>0}]
#=> [{"id"=>1}, {"id"=>2}]
Hashes without the id key are treated as having "id" => nil (so the first will be kept):
def no_dup_val key, arr
arr.uniq{ |h| h[key] }
end
no_dup_val 'id', [{'id' => 1}, {'id' => 1, 'key' => 'data'}, {'id' => 2}, {'stock' => 3}, {'e-stock'=>0}]
#=> [{"id"=>1}, {"id"=>2}, {"stock"=>3}]
All the hashes without the id key are kept:
Let's say I have the following array of hashes:
h = [{"name" => "bob"}, {"car" => "toyota"}, {"age" => "25"}]
And I have the following key to match:
k = 'car'
How do I match the 'k' to 'h' and have delete every element after the match so that it returns:
h = [{"name" => "bob"}, {"car" => "toyota"}]
Just convert hash to array, do your task and then convert back
h = {"name" => "bob", "car" => "toyota", "age" => "25"}
array = h.to_a.flatten
index = array.index('car') + 1
h = Hash[*array[0..index]]
=> {"name"=>"bob", "car"=>"toyota"}
By the way, the hash is ordered only since Ruby 1.9
ar = [{"name" => "bob"}, {"car" => "toyota"}, {"age" => "25"}]
p ar[0 .. ar.index{|h| h.key?('car')}] #=>[{"name"=>"bob"}, {"car"=>"toyota"}]
I like megas' version, as its short and to the point. Another approach, which would be more explicit, would be iterating over the keys array of each hash. The keys of a hash are maintained in an ordered array (http://ruby-doc.org/core-1.9.3/Hash.html). They are ordered by when they were first entered. As a result, you can try the following:
newArray = Array.new
h.each do |hash| # Iterate through your array of hashes
newArray << hash
if hash.has_key?("car") # check if this hash is the "car" hash.
break # exits the block
end
end
This all depends, of course, on whether the array was created in the proper order. If it was, then you're golden.
A hash is unordered set by definition, so what you request is somewhat undefined. However you can do something like a hack:
h = {"name" => "bob", "car" => "toyota", "age" => "25"}
matched = false
key_given = "car"
h.each do |k,v|
if matched
h.delete(k)
end
if k == key_given
matched = true
next
end
end
I'm pretty late to the party here. I was looking for a solution to this same problem, but I didn't love these answers. So, here's my approach:
class Array
def take_until(&blk)
i = find_index &blk
take(i + 1)
end
end
h = [{"name" => "bob"}, {"car" => "toyota"}, {"age" => "25"}]
k = 'car'
h.take_until { |x| x.has_key?(k) }
=> [{"name"=>"bob"}, {"car"=>"toyota"}]
I have a set of categories and their values stored as a list of hashes:
r = [{:A => :X}, {:A => :Y}, {:B => :X}, {:A => :X}, {:A => :Z}, {:A => :X},
{:A => :X}, {:B => :Z}, {:C => :X}, {:C => :Y}, {:B => :X}, {:C => :Y},
{:C => :Y}]
I'd like to get a count of each value coupled with its category as a hash like this:
{:A => {:X => 4, :Y => 1, :Z => 1},
:B => {:X => 2, :Z => 1},
:C => {:X => 1, :Y => 3}}
How can I do this efficiently?
Here's what I have so far (it returns inconsistent values):
r.reduce(Hash.new(Hash.new(0))) do |memo, x|
memo[x.keys.first][x.values.first] += 1
memo
end
Should I first compute the counts of all instances of specific {:cat => :val}s and then create the hash? Should I give a different base-case to reduce and change the body to check for nil cases (and assign zero when nil) instead of always adding 1?
EDIT:
I ended up changing my code and using the below method to have a cleaner way of achieving a nested hash:
r.map do |x|
[x.keys.first, x.values.last]
end.reduce({}) do |memo, x|
memo[x.first] = Hash.new(0) if memo[x.first].nil?
memo[x.first][x.last] += 1
memo
end
The problem of your code is: memo did not hold the value.
Use a variable outside the loop to hold the value would be ok:
memo = Hash.new {|h,k| h[k] = Hash.new {|hh, kk| hh[kk] = 0 } }
r.each do |x|
memo[x.keys.first][x.values.first] += 1
end
p memo
And what's more, it won't work to init a hash nested inside a hash directly like this:
# NOT RIGHT
memo = Hash.new(Hash.new(0))
memo = Hash.new({})
Here is a link for more about the set default value issue:
http://www.themomorohoax.com/2008/12/31/why-setting-the-default-value-of-a-hash-to-be-a-hash-is-wrong
Not sure what "inconsistent values" means, but your problem is the hash you're injecting into is not remembering its results
r.each_with_object(Hash.new { |h, k| h[k] = Hash.new 0 }) do |individual, consolidated|
individual.each do |key, value|
consolidated[key][value] += 1
end
end
But honestly, it would probably be better to just go to wherever you're making this array and change it to aggregate values like this.
Functional approach using some handy abstractions -no need to reinvent the wheel- from facets:
require 'facets'
r.map_by { |h| h.to_a }.mash { |k, vs| [k, vs.frequency] }
#=> {:A=>{:X=>4, :Y=>1, :Z=>1}, :B=>{:X=>2, :Z=>1}, :C=>{:X=>1, :Y=>3}}
Hash.each returns an array [key, value],
but if I want a hash?
Example: {:key => value }
I'm assuming you meant "yields" where you said "return" because Hash#each already returns a hash (the receiver).
To answer your question: If you need a hash with the key and the value you can just create one. Like this:
hash.each do |key, value|
kv_hash = {key => value}
do_something_with(kv_hash)
end
There is no alternative each method that yields hashs, so the above is the best you can do.
I think you are trying to transform the hash somehow, so I will give you my solution to this problem, which may be not exactly the same. To modify a hash, you have to .map them and construct a new hash.
This is how I reverse key and values:
h = {:a => 'a', :b => 'b'}
Hash[h.map{ |k,v| [v, k] }]
# => {"a"=>:a, "b"=>:b}
Call .each with two parameters:
>> a = {1 => 2, 3 => 4}
>> a.each { |b, c|
?> puts "#{b} => #{c}"
>> }
1 => 2
3 => 4
=> {1=>2, 3=>4}
You could map the hash to a list of single-element hashes, then call each on the list:
h = {:a => 'a', :b => 'b'}
h.map{ |k,v| {k => v}}.each{ |x| puts x }
I'd like to save some hash objects to a collection (in the Java world think of it as a List). I search online to see if there is a similar data structure in Ruby and have found none. For the moment being I've been trying to save hash a[] into hash b[], but have been having issues trying to get data out of hash b[].
Are there any built-in collection data structures on Ruby? If not, is saving a hash in another hash common practice?
If it's accessing the hash in the hash that is the problem then try:
>> p = {:name => "Jonas", :pos => {:x=>100.23, :y=>40.04}}
=> {:pos=>{:y=>40.04, :x=>100.23}, :name=>"Jonas"}
>> p[:pos][:x]
=> 100.23
There shouldn't be any problem with that.
a = {:color => 'red', :thickness => 'not very'}
b = {:data => a, :reason => 'NA'}
Perhaps you could explain what problems you're encountering.
The question is not completely clear, but I think you want to have a list (array) of hashes, right?
In that case, you can just put them in one array, which is like a list in Java:
a = {:a => 1, :b => 2}
b = {:c => 3, :d => 4}
list = [a, b]
You can retrieve those hashes like list[0] and list[1]
Lists in Ruby are arrays. You can use Hash.to_a.
If you are trying to combine hash a with hash b, you can use Hash.merge
EDIT: If you are trying to insert hash a into hash b, you can do
b["Hash a"] = a;
All the answers here so far are about Hash in Hash, not Hash plus Hash, so for reasons of completeness, I'll chime in with this:
# Define two independent Hash objects
hash_a = { :a => 'apple', :b => 'bear', :c => 'camel' }
hash_b = { :c => 'car', :d => 'dolphin' }
# Combine two hashes with the Hash#merge method
hash_c = hash_a.merge(hash_b)
# The combined hash has all the keys from both sets
puts hash_c[:a] # => 'apple'
puts hash_c[:c] # => 'car', not 'camel' since B overwrites A
Note that when you merge B into A, any keys that A had that are in B are overwritten.