I frequently write something like this:
a_hash['x'] ? a_hash['x'] += ' some more text' : a_hash['x'] = 'first text'
There ought to be a better way to do this, but I can't find it.
There are two ways to create initial values with for a Hash.
One is to pass a single object in to Hash.new. This works well in many situations, especially if the object is a frozen value, but if the object has internal state, this may have unexpected side-effects. Since the same object is shared between all keys without an assigned value, modifying the internal state for one will show up in all.
a_hash = Hash.new "initial value"
a_hash['a'] #=> "initial value"
# op= methods don't modify internal state (usually), since they assign a new
# value for the key.
a_hash['b'] += ' owned by b' #=> "initial value owned by b"
# other methods, like #<< and #gsub modify the state of the string
a_hash['c'].gsub!(/initial/, "c's")
a_hash['d'] << " modified by d"
a_hash['e'] #=> "c's value modified by d"
Another initialization method is to pass Hash.new a block, which is invoked each time a value is requested for a key that has no value. This allows you to use a distinct value for each key.
another_hash = Hash.new { "new initial value" }
another_hash['a'] #=> "new initial value"
# op= methods still work as expected
another_hash['b'] += ' owned by b'
# however, if you don't assign the modified value, it's lost,
# since the hash rechecks the block every time an unassigned key's value is asked for
another_hash['c'] << " owned by c" #=> "new initial value owned by c"
another_hash['c'] #=> "new initial value"
The block is passed two arguments: the hash being asked for a value, and the key used. This gives you the option of assigning a value for that key, so that the same object will be presented each time a particular key is given.
yet_another_hash = Hash.new { |hash, key| hash[key] = "#{key}'s initial value" }
yet_another_hash['a'] #=> "a's initial value"
yet_another_hash['b'] #=> "b's initial value"
yet_another_hash['c'].gsub!('initial', 'awesome')
yet_another_hash['c'] #=> "c's awesome value"
yet_another_hash #=> { "a" => "a's initial value", "b" => "b's initial value", "c" => "c's awesome value" }
This last method is the one I most often use. It's also useful for caching the result of an expensive calculation.
You can specify the initial value when you create your hash:
a_hash = { 'x' => 'first text' }
// ...
a_hash['x'] << ' some more text'
Since you are using the hash to collect strings, I assume you simply want to make sure that you don't get an error when appending. Therefore, I'd make the hash default the empty string. Then you can append without error, no matter the hash key.
a_hash = Hash.new {|h,k| h[k]=""}
texts = ['first text', ' some more text']
texts.each do |text|
a_hash['x'] << text
end
puts a_hash['x'] #=> 'first text some more text'
The constructor of Hash, in its first argument, have a default value for the keys. This way
>> a_hash = Hash.new "first text"
=> {}
>> a_hash['a']
=> "first text"
>> a_hash['b'] += ", edit"
=> "first text, edit"
In case you do not want to mess with the default= value (existing hash), you can shorten your code by using fetch(key [, default]) as a look-up with a default value:
a_hash['x'] = a_hash.fetch('x', 'first_text') + ' some more text'
Related
I have a hash
h = Hash.new{|hsh,key| hsh[key] = [] }
And its values are stored as an array
Iv added to the value array for the key like so:
h[#name] << #age
h[#name] << #grade
and im trying to access the age like this
puts h[:#name][0]
But it doesn't work?
Is there a better way of doing this?
What im trying to do is create is a hash where there is a key which has loads of values:
For example key=>name and values equal age, address, gender etc
IMHO your idea is ok.
The only mistake is.. how you access the hash. No need to add extra colon : before # sign.
Remove colon and it should work as you expect:
puts h[#name][0]
A Hash is a collection of key-value pairs like this: "employee" => "salary". It is similar to an Array, except that indexing is done via arbitrary keys of any object type, not an integer index.
The order in which you traverse a hash by either key or value may seem arbitrary and will generally not be in the insertion order. If you attempt to access a hash with a key that does not exist, the method will return nil.
A hash is used to stored large (or small) amounts of data and access it efficiently. For example lets say you have this as a hash:
prices = {
'orange' => 3.15,
'apple' => 2.25,
'pear' => 3.50
}
Now you want to call the keyword apple and get the prices of those items from some users input:
print 'Enter an item to verify price: '
item = gets.chomp
puts "The price of an #{item}: #{prices[item]}"
# <= The price of an apple: 2.25
That's a basic hash, now lets get into what you're doing, using an Array as a key.
prices = {
'apple' => ['Granny Smith', 'Red'],
'orange' => ['Good', 'Not good'],
'pear' => ['Big', 'Small']
}
print 'Enter an item for a list: '
item = gets.chomp
puts "We have the following #{item}'s available: #{prices[item]}"
# <= We have the following apple's available: ["Granny Smith", "Red"]
Now if we wanted to grab one of the types:
puts prices[item][0]
# <= Granny Smith
puts prices[item][1]
#<= Red
Now lets get into more advanced techniques like you are doing above, you're idea is great and all, but what you need to do is append the information into the hash and when you call #name don't try to call it as a symbol:
h = Hash.new{|hsh,key| hsh[key] = [] }
h[#name] = []
#<= []
h[#name] << ['apple', 'pear']
#<= [["apple", "pear"]]
h[#name] << ['orange', 'apple']
#<= [["apple", "pear"], ["orange", "apple"]]
h[#name].flatten[0]
#<= "apple"
h[#name].flatten[1]
#<= "pear"
h[#name].flatten[1, 2]
#<= ["pear", "orange"]
Alright so what did we do?
h = Hash.new{|hsh,key| hsh[key] = [] }
Created a hash with a value as an empty array.
h[#name] = []
Initialized #name to the empty array
h[#name] << ['apple', 'pear']
Appended an array containing apple, pear to the #name key.
h[#name] << ['orange', 'apple']
Appended a second array containing orange, apple to the array, so when we call h[#name][1] right now it will output the first array appended to it.
h[#name].flatten[0]
Flattened the array into a single array and called the first element of the array.
When you call your key (#name) you don't call it as a symbol, because it's already contained inside of a variable. So all you have to do is call that variable and the value for that key will be output successfully. Hopefully this clarifies a few, things, for more information on hashes check this out: http://www.tutorialspoint.com/ruby/ruby_hashes.htm
So let's say I do the following:
lph = Hash.new([]) #=> {}
lph["passed"] << "LCEOT" #=> ["LCEOT"]
lph #=> {} <-- Expected that to have been {"passed" => ["LCEOT"]}
lph["passed"] #=> ["LCEOT"]
lph["passed"] = lph["passed"] << "HJKL"
lph #=> {"passed"=>["LCEOT", "HJKL"]}
I'm surprised by this. A couple questions:
Why does it not get set until I push the second string on to the array? What is happening in the background?
What is the more idiomatic ruby way to essentially say. I have a hash, a key, and a value I want to to end up in the array associated with the key. How do I push the value in an array associated with a key into a hash the first time. In all future uses of the key, I just want to addd to the array.
Read the Ruby Hash.new documentation carefully - "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".
new(obj) → new_hash
...If obj is specified, this single object will be used for all default values.
In your example you attempt to push something onto the value associated with a key which does not exist, so you end up mutating the same anonymous array you used to construct the hash initially.
the_array = []
h = Hash.new(the_array)
h['foo'] << 1 # => [1]
# Since the key 'foo' was not found
# ... the default value (the_array) is returned
# ... and 1 is pushed onto it (hence [1]).
the_array # => [1]
h # {} since the key 'foo' still has no value.
You probably want to use the block form:
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.
For example:
h = Hash.new { |hash, key| hash[key] = [] } # Assign a new array as default for missing keys.
h['foo'] << 1 # => [1]
h['foo'] << 2 # => [1, 2]
h['bar'] << 3 # => [3]
h # => { 'foo' => [1, 2], 'bar' => [3] }
Why does it not get set until I push the second string on to the array?
In short; because you don't set anything in the hash until the point, where you also add the second string to the array.
What is happening in the background?
To see what's happening in the background, let's take this one line at a time:
lph = Hash.new([]) #=> {}
This creates an empty hash, configured to return the [] object whenever a non-existing key is accessed.
lph["passed"] << "LCEOT" #=> ["LCEOT"]
This can be written as
value = lph["passed"] #=> []
value << "LCEOT" #=> ["LCEOT"]
We see that lph["passed"] returns [] as expected, and we then proceed to append "LCEOT" to [].
lph #=> {}
lph is still an empty Hash. At no point have we added anything to the Hash. We have added something to its default value, but that doesn't change lph itself.
lph["passed"] #=> ["LCEOT"]
This is where it gets interesting. Remember above when we did value << ["LCEOT"]. That actually changed the default value that lph returns when a key isn't found. The default value is no longer [], but has become ["LCEOT"]. That new default value is returned here.
lph["passed"] = lph["passed"] << "HJKL"
This is our first change to lph. And what we actually assign to lph["passed"] is the default value (because "passed" is still a non-existing key in lph) with "HJKL" appended. Before this, the default value was ["LCEOT"], after this it is ["LCEOT", "HJKL"].
In other words lph["passed"] << "HJKL" returns ["LCEOT", "HJKL"] which is then assigned to lph["passed"].
What is the more idiomatic Ruby way
Using <<=:
>> lph = Hash.new { [] }
=> {}
>> lph["passed"] <<= "LCEOT"
=> ["LCEOT"]
>> lph
=> {"passed"=>["LCEOT"]}
Also note the change in how the Hash is initialized, using a block instead of a verbatim array. This ensures a new, blank array is created and returned whenever a new key is accessed, as opposed to the same array being used every time.
Stuck on a Code Wars Challenge: Complete the solution so that it takes an array of keys and a default value and returns a hash with all keys set to the default value.
My answer results in a parse error:
def solution([:keys, :default_value])
return { :keys => " ", :default_value => " " }
end
Am I missing something to do with returning a hash key with all the keys set to the default value?
Do as below :
def solution(keys,default_val)
Hash[keys.product([default_val])]
end
solution([:key1,:key2],12) # => {:key1=>12, :key2=>12}
Read Array#product and Kernel#Hash.
I'd advise amending your solution to this:
def solution(keys, default_value)
hash = {}
keys.each do |key|
value = default_value.dup rescue default_value
hash[key] = value
end
hash
end
The dup is to work around the nasty case where default_value is a string and you then do e.g.:
hash[:foo] << 'bar'
… with your version, this would modify multiple values in place instead of a single one.
I am currently learning Ruby and I'm trying to write a simple Ruby grocery_list method. Here are the instructions:
We want to write a program to help keep track of a grocery list. It takes a grocery item (like "eggs") as an argument, and returns the grocery list (that is, the item names with the quantities of each item). If you pass the same argument twice, it should increment the quantity.
def grocery_list(item)
array = []
quantity = 1
array.each {|x| quantity += x }
array << "#{quantity}" + " #{item}"
end
puts grocery_list("eggs", "eggs")
so I'm trying to figure out here how to return "2 eggs" by passing eggs twice
To help you count the different items you can use as Hash. A Hash is similar to an Array, but with Strings instead of Integers als an Index:
a = Array.new
a[0] = "this"
a[1] = "that"
h = Hash.new
h["sonja"] = "asecret"
h["brad"] = "beer"
In this example the Hash might be used for storing passwords for users. But for your
example you need a hash for counting. Calling grocery_list("eggs", "beer", "milk", "eggs")
should lead to the following commands being executed:
h = Hash.new(0) # empty hash {} created, 0 will be default value
h["eggs"] += 1 # h is now {"eggs"=>1}
h["beer"] += 1 # {"eggs"=>1, "beer"=>1}
h["milk"] += 1 # {"eggs"=>1, "beer"=>1, "milk"=>1}
h["eggs"] += 1 # {"eggs"=>2, "beer"=>1, "milk"=>1}
You can work through all the keys and values of a Hash with the each-loop:
h.each{|key, value| .... }
and build up the string we need as a result, adding
the number of items if needed, and the name of the item.
Inside the loop we always add a comma and a blank at the end.
This is not needed for the last element, so after the
loop is done we are left with
"2 eggs, beer, milk, "
To get rid of the last comma and blank we can use chop!, which "chops off"
one character at the end of a string:
output.chop!.chop!
One more thing is needed to get the complete implementation of your grocery_list:
you specified that the function should be called like so:
puts grocery_list("eggs", "beer", "milk","eggs")
So the grocery_list function does not know how many arguments it's getting. We can handle
this by specifying one argument with a star in front, then this argument will
be an array containing all the arguments:
def grocery_list(*items)
# items is an array
end
So here it is: I did your homework for you and implemented grocery_list.
I hope you actually go to the trouble of understanding the implementation,
and don't just copy-and-paste it.
def grocery_list(*items)
hash = Hash.new(0)
items.each {|x| hash[x] += 1}
output = ""
hash.each do |item,number|
if number > 1 then
output += "#{number} "
end
output += "#{item}, "
end
output.chop!.chop!
return output
end
puts grocery_list("eggs", "beer", "milk","eggs")
# output: 2 eggs, beer, milk
def grocery_list(*item)
item.group_by{|i| i}
end
p grocery_list("eggs", "eggs","meat")
#=> {"eggs"=>["eggs", "eggs"], "meat"=>["meat"]}
def grocery_list(*item)
item.group_by{|i| i}.flat_map{|k,v| [k,v.length]}
end
p grocery_list("eggs", "eggs","meat")
#=>["eggs", 2, "meat", 1]
def grocery_list(*item)
Hash[*item.group_by{|i| i}.flat_map{|k,v| [k,v.length]}]
end
grocery_list("eggs", "eggs","meat")
#=> {"eggs"=>2, "meat"=>1}
grocery_list("eggs", "eggs","meat","apple","apple","apple")
#=> {"eggs"=>2, "meat"=>1, "apple"=>3}
or as #Lee said:
def grocery_list(*item)
item.each_with_object(Hash.new(0)) {|a, h| h[a] += 1 }
end
grocery_list("eggs", "eggs","meat","apple","apple","apple")
#=> {"eggs"=>2, "meat"=>1, "apple"=>3}
Use a Hash Instead of an Array
When you want an easy want to count things, you can use a hash key to hold the name of the thing you want to count, and the value of that key is the quantity. For example:
#!/usr/bin/env ruby
class GroceryList
attr_reader :list
def initialize
# Specify hash with default quantity of zero.
#list = Hash.new(0)
end
# Increment the quantity of each item in the #list, using the name of the item
# as a hash key.
def add_to_list(*items)
items.each { |item| #list[item] += 1 }
#list
end
end
if $0 == __FILE__
groceries = GroceryList.new
groceries.add_to_list('eggs', 'eggs')
puts 'Grocery list correctly contains 2 eggs.' if groceries.list['eggs'] == 2
end
Here's a more verbose, but perhaps more readable solutions to your challenge.
def grocery_list(*items) # Notice the asterisk in front of items. It means "put all the arguments into an array called items"
my_grocery_hash = {} # Creates an empty hash
items.each do |item| # Loops over the argument array and passes each argument into the loop as item.
if my_grocery_hash[item].nil? # Returns true of the item is not a present key in the hash...
my_grocery_hash[item] = 1 # Adds the key and sets the value to 1.
else
my_grocery_hash[item] = my_grocery_hash[item] + 1 # Increments the value by one.
end
end
my_grocery_hash # Returns a hash object with the grocery name as the key and the number of occurences as the value.
end
This will create an empty hash (called dictionaries or maps in other languages) where each grocery is added as a key with the value set to one. In case the same grocery appears multiple times as a parameter to your method, the value is incremented.
If you want to create a text string and return that instead of the hash object and you can do like this after the iteration:
grocery_list_string = "" # Creates an empty string
my_grocery_hash.each do |key, value| # Loops over the hash object and passes two local variables into the loop with the current entry. Key being the name of the grocery and value being the amount.
grocery_list_string << "#{value} units of #{key}\n" # Appends the grocery_list_string. Uses string interpolation, so #{value} becomes 3 and #{key} becomes eggs. The remaining \n is a newline character.
end
return grocery_list_string # Explicitly declares the return value. You can ommit return.
Updated answer to comment:
If you use the first method without adding the hash iteration you will get a hash object back which can be used to look up the amount like this.
my_hash_with_grocery_count = grocery_list("Lemonade", "Milk", "Eggs", "Lemonade", "Lemonade")
my_hash_with_grocery_count["Milk"]
--> 1
my_hash_with_grocery_count["Lemonade"]
--> 3
Enumerable#each_with_object can be useful for things like this:
def list_to_hash(*items)
items.each_with_object(Hash.new(0)) { |item, list| list[item] += 1 }
end
def hash_to_grocery_list_string(hash)
hash.each_with_object([]) do |(item, number), result|
result << (number > 1 ? "#{number} #{item}" : item)
end.join(', ')
end
def grocery_list(*items)
hash_to_grocery_list_string(list_to_hash(*items))
end
p grocery_list('eggs', 'eggs', 'bread', 'milk', 'eggs')
# => "3 eggs, bread, milk"
It iterates an array or hash to enable building another object in a convenient way. The list_to_hash method uses it to build a hash from the items array (the splat operator converts the method arguments to an array); the hash is created so that each value is initialized to 0. The hash_to_grocery_list_string method uses it to build an array of strings that is joined to a comma-separated string.
I'm going through Ruby Koans, and I hit #41 which I believe is this:
def test_default_value_is_the_same_object
hash = Hash.new([])
hash[:one] << "uno"
hash[:two] << "dos"
assert_equal ["uno","dos"], hash[:one]
assert_equal ["uno","dos"], hash[:two]
assert_equal ["uno","dos"], hash[:three]
assert_equal true, hash[:one].object_id == hash[:two].object_id
end
It could not understand the behavior so I Googled it and found Strange ruby behavior when using Hash default value, e.g. Hash.new([]) that answered the question nicely.
So I understand how that works, my question is, why does a default value such as an integer that gets incremented not get changed during use? For example:
puts "Text please: "
text = gets.chomp
words = text.split(" ")
frequencies = Hash.new(0)
words.each { |word| frequencies[word] += 1 }
This will take user input and count the number of times each word is used, it works because the default value of 0 is always used.
I have a feeling it has to do with the << operator but I'd love an explanation.
The other answers seem to indicate that the difference in behavior is due to Integers being immutable and Arrays being mutable. But that is misleading. The difference is not that the creator of Ruby decided to make one immutable and the other mutable. The difference is that you, the programmer decided to mutate one but not the other.
The question is not whether Arrays are mutable, the question is whether you mutate it.
You can get both the behaviors you see above, just by using Arrays. Observe:
One default Array with mutation
hsh = Hash.new([])
hsh[:one] << 'one'
hsh[:two] << 'two'
hsh[:nonexistent]
# => ['one', 'two']
# Because we mutated the default value, nonexistent keys return the changed value
hsh
# => {}
# But we never mutated the hash itself, therefore it is still empty!
One default Array without mutation
hsh = Hash.new([])
hsh[:one] += ['one']
hsh[:two] += ['two']
# This is syntactic sugar for hsh[:two] = hsh[:two] + ['two']
hsh[:nonexistant]
# => []
# We didn't mutate the default value, it is still an empty array
hsh
# => { :one => ['one'], :two => ['two'] }
# This time, we *did* mutate the hash.
A new, different Array every time with mutation
hsh = Hash.new { [] }
# This time, instead of a default *value*, we use a default *block*
hsh[:one] << 'one'
hsh[:two] << 'two'
hsh[:nonexistent]
# => []
# We *did* mutate the default value, but it was a fresh one every time.
hsh
# => {}
# But we never mutated the hash itself, therefore it is still empty!
hsh = Hash.new {|hsh, key| hsh[key] = [] }
# This time, instead of a default *value*, we use a default *block*
# And the block not only *returns* the default value, it also *assigns* it
hsh[:one] << 'one'
hsh[:two] << 'two'
hsh[:nonexistent]
# => []
# We *did* mutate the default value, but it was a fresh one every time.
hsh
# => { :one => ['one'], :two => ['two'], :nonexistent => [] }
It is because Array in Ruby is mutable object, so you can change it internal state, but Fixnum isn't mutable. So when you increment value using += internally it get that (assume that i is our reference to Fixnum object):
get object referenced by i
get it internal value (lets name it raw_tmp)
create new object that internal value is raw_tmp + 1
assign reference to created object to i
So as you can see, we created new object, and i reference now to something different than at the beginning.
In the other hand, when we use Array#<< it works that way:
get object referenced by arr
to it's internal state append given element
So as you can see it is much simpler, but it can cause some bugs. One of them you have in your question, another one is thread race when booth are trying simultaneously append 2 or more elements. Sometimes you can end with only some of them and with thrashes in memory, when you use += on arrays too, you will get rid of both of these problems (or at least minimise impact).
From the doc, setting a default value has the following behaviour:
Returns the default value, the value that would be returned by hsh if key did not exist in hsh. See also Hash::new and Hash#default=.
Therefore, every time frequencies[word] is not set, the value for that individual key is set to 0.
The reason for the discrepancy between the two code blocks is that arrays are mutable in Ruby, while integers are not.