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I'm trying to sort a hash within a hash. I'd like to sort the hash by the sub-key. I'm using ruby.
I've tried using the sort_by method and iterating the over the hash to reorganize the sub-key. I received the error "ArgumentError: comparison of Hash with Hash failed"
hash = {2012 => {"regularseason" => [game_1, game_2, game_3],
"post_season" => [game_4, game_5]
},
2013 => {"regularseason" => [game_6, game_7, game_8],
"post_season" => [game_9, game_10]
},
2014 => {"post_season" => [game_11, game_12, game_13],
"regularseason" => [game_14, game_15]
}
}
Desired Result:
I would like to sort this hash so sub-key post_season will always appear before sub-key regularseason.
Use Hash#transform_values to sort values:
hash.transform_values { |v| v.sort.to_h }
#⇒ {2012=>{"post_season"=>[:game_4, :game_5],
# "regularseason"=>[:game_1, :game_2, :game_3]},
# 2013=>{"post_season"=>[:game_9, :game_10],
# "regularseason"=>[:game_6, :game_7, :game_8]},
# 2014=>{"post_season"=>[:game_11, :game_12, :game_13],
# "regularseason"=>[:game_14, :game_15]}}
Hashes return keys in the order they are inserted so I believe you'll essentially need to rewrite the nested hash.
For example:
hash.each { |(key, nested_hash)| hash[key] = nested_hash.sort.to_h }
This is rather inefficient, however, so you'd be better to see if you can ensure they are always entered in that order or somehow extract them in the order you desire.
Given a hash with keys that include k, we can return a new hash with the same key/value pairs, with k being the first key inserted, and the remaining keys maintaining their original relative order, as follows:
def reorder_key(h, key)
{ key=>h[key] }.merge h
end
For example:
h = { 1=>2, :a=>:b, "c"=>"d" }
reorder_key(h, :a)
#=> {:a=>:b, 1=>2, "c"=>"d"}
We can use this method to obtain the desired hash in the present problem.
hash.transform_values { |h| reorder_key(h, "post_season") }
#=> {2012=>{"post_season" =>[:game_4, :game_5],
# "regularseason"=>[:game_1, :game_2, :game_3]},
# 2013=>{"post_season" =>[:game_9, :game_10],
# "regularseason"=>[:game_6, :game_7, :game_8]},
# 2014=>{"post_season" =>[:game_11, :game_12, :game_13],
# "regularseason"=>[:game_14, :game_15]}}
This approach does not depend on "post_season" coincidentally preceding "regularseason" lexicographically. If, for example, it were decided to add the key "spring_training" and make that the first key to appear for each year in the returned hash, it would only be necessary to change the value of the second argument of reorder_key to "spring_training".
I have read the xls and have formed these three hashes
hash1=[{'name'=>'Firstname',
'Locator'=>'id=xxx',
'Action'=>'TypeAndWait'},
{'name'=>'Password',
'Locator'=>'id=yyy',
'Action'=>'TypeAndTab'}]
Second Hash
hash2=[{'Test Name'=>'Example',
'TestNumber'=>'Test1'},
{'Test Name'=>'Example',
'TestNumber'=>'Test2'}]
My Thrid Hash
hash3=[{'name'=>'Firstname',
'Test1'=>'four',
'Test2'=>'Five',
'Test3'=>'Six'},
{'name'=>'Password',
'Test1'=>'Vicky',
'Test2'=>'Sujin',
'Test3'=>'Sivaram'}]
Now my resultant hash is
result={"Example"=>
{"Test1"=>
{'Firstname'=>
["id=xxx","four", "TypeAndWait"],
'Password'=>
["id=yyy","Vicky", "TypeAndTab"]},
"Test2"=>
{'Firstname'=>
["id=xxx","Five", "TypeAndWait"],
'Password'=>
["id=yyy","Sujin", "TypeAndTab"]}}}
I have gotten this result, but I had to write 60 lines of code in my program, but I don't think I have to write such a long program when I use Ruby, I strongly believe some easy way to achieve this. Can some one help me?
The second hash determines the which testcase has to be read, for an example, test3 is not present in the second testcase so resultant hash doesn't have test3.
We are given three arrays, which I've renamed arr1, arr2 and arr3. (hash1, hash2 and hash3 are not especially good names for arrays. :-))
arr1 = [{'name'=>'Firstname', 'Locator'=>'id=xxx', 'Action'=>'TypeAndWait'},
{'name'=>'Password', 'Locator'=>'id=yyy', 'Action'=>'TypeAndTab'}]
arr2 = [{'Test Name'=>'Example', 'TestNumber'=>'Test1'},
{'Test Name'=>'Example', 'TestNumber'=>'Test2'}]
arr3=[{'name'=>'Firstname', 'Test1'=>'four', 'Test2'=>'Five', 'Test3'=>'Six'},
{'name'=>'Password', 'Test1'=>'Vicky', 'Test2'=>'Sujin', 'Test3'=>'Sivaram'}]
The drivers are the values "Test1" and "Test2" in the hashes that are elements of arr2. Nothing else in that array is needed, so let's extract those values (of which there could be any number, but here there are just two).
a2 = arr2.map { |h| h['TestNumber'] }
#=> ["Test1", "Test2"]
Next we need to rearrange the information in arr3 by creating a hash whose keys are the elements of a2.
h3 = a2.each_with_object({}) { |test,h|
h[test] = arr3.each_with_object({}) { |f,g| g[f['name']] = f[test] } }
#=> {"Test1"=>{"Firstname"=>"four", "Password"=>"Vicky"},
# "Test2"=>{"Firstname"=>"Five", "Password"=>"Sujin"}}
Next we need to rearrange the content of arr1 by creating a hash whose keys match the keys of values of h3.
h1 = arr1.each_with_object({}) { |g,h| h[g['name']] = g.reject { |k,_| k == 'name' } }
#=> {"Firstname"=>{"Locator"=>"id=xxx", "Action"=>"TypeAndWait"},
# "Password"=>{"Locator"=>"id=yyy", "Action"=>"TypeAndTab"}}
It is now a simple matter of extracting information from these three objects.
{ 'Example'=>
a2.each_with_object({}) do |test,h|
h[test] = h3[test].each_with_object({}) do |(k,v),g|
f = h1[k]
g[k] = [f['Locator'], v, f['Action']]
end
end
}
#=> {"Example"=>
# {"Test1"=>{"Firstname"=>["id=xxx", "four", "TypeAndWait"],
# "Password"=>["id=yyy", "Vicky", "TypeAndTab"]},
# "Test2"=>{"Firstname"=>["id=xxx", "Five", "TypeAndWait"],
# "Password"=>["id=yyy", "Sujin", "TypeAndTab"]}}}
What do you call hash{1-2-3} are arrays in the first place. Also, I am pretty sure you have mistyped hash1#Locator and/or hash3#name. The code below works for this exact data, but it should not be hard to update it to reflect any changes.
hash2.
map(&:values).
group_by(&:shift).
map do |k, v|
[k, v.flatten.map do |k, v|
[k, hash3.map do |h3|
# lookup a hash from hash1
h1 = hash1.find do |h1|
h3['name'].start_with?(h1['Locator'])
end
# can it be nil btw?
[
h1['name'],
[
h3['name'][/.*(?=-id)/],
h3[k],
h1['Action']
]
]
end.to_h]
end.to_h]
end.to_h
# dictionary = {"cat"=>"Sam"}
This a return a key
#dictionary.key(x)
This returns a value
#dictionary[x]
How do I return the entire element
"cat"=>"Sam"
#dictionary
should do the trick for you
whatever is the last evaluated expression in ruby is the return value of a method.
If you want to return the hash as a whole. the last line of the method should look like the line I have written above
Your example is a bit (?) misleading in a sense it only has one pair (while not necessarily), and you want to get one pair. What you call a "dictionary" is actually a hashmap (called a hash among Rubyists).
A hashrocket (=>) is a part of hash definition syntax. It can't be used outside it. That is, you can't get just one pair without constructing a new hash. So, a new such pair would look as: { key => value }.
So in order to do that, you'll need a key and a value in context of your code somewhere. And you've specified ways to get both if you have one. If you only have a value, then:
{ #dictionary.key(x) => x }
...and if just a key, then:
{ x => #dictionary[x] }
...but there is no practical need for this. If you want to process each pair in a hash, use an iterator to feed each pair into some code as an argument list:
#dictionary.each do |key, value|
# do stuff with key and value
end
This way a block of code will get each pair in a hash once.
If you want to get not a hash, but pairs of elements it's constructed of, you can convert your hash to an array:
#dictionary.to_a
# => [["cat", "Sam"]]
# Note the double braces! And see below.
# Let's say we have this:
#dictionary2 = { 1 => 2, 3 => 4}
#dictionary2[1]
# => 2
#dictionary2.to_a
# => [[1, 2], [3, 4]]
# Now double braces make sense, huh?
It returns an array of pairs (which are arrays as well) of all elements (keys and values) that your hashmap contains.
If you wish to return one element of a hash h, you will need to specify the key to identify the element. As the value for key k is h[k], the key-value pair, expressed as an array, is [k, h[k]]. If you wish to make that a hash with a single element, use Hash[[[k, h[k]]]].
For example, if
h = { "cat"=>"Sam", "dog"=>"Diva" }
and you only wanted to the element with key "cat", that would be
["cat", h["cat"]] #=> ["cat", "Sam"]
or
Hash[[["cat", h["cat"]]]] #=> {"cat"=>"Sam"}
With Ruby 2.1 you could alternatively get the hash like this:
[["cat", h["cat"]]].to_h #=> {"cat"=>"Sam"}
Let's look at a little more interesting case. Suppose you have an array arr containing some or all of the keys of a hash h. Then you can get all the key-value pairs for those keys by using the methods Enumerable#zip and Hash#values_at:
arr.zip(arr.values_at(*arr))
Suppose, for example,
h = { "cat"=>"Sam", "dog"=>"Diva", "pig"=>"Petunia", "owl"=>"Einstein" }
and
arr = ["dog", "owl"]
Then:
arr.zip(h.values_at(*arr))
#=> [["dog", "Diva"], ["owl", "Einstein"]]
In steps:
a = h.values_at(*arr)
#=> h.values_at(*["dog", "owl"])
#=> h.values_at("dog", "owl")
#=> ["Diva", "Einstein"]
arr.zip(a)
#=> [["dog", "Diva"], ["owl", "Einstein"]]
To instead express as a hash:
Hash[arr.zip(h.values_at(*arr))]
#=> {"dog"=>"Diva", "owl"=>"Einstein"}
You can get the key and value in one go - resulting in an array:
#h = {"cat"=>"Sam", "dog"=>"Phil"}
key, value = p h.assoc("cat") # => ["cat", "Sam"]
Use rassoc to search by value ( .rassoc("Sam") )
I'm trying to increment all values of a hash by a given amount and return the hash. I am expecting:
add_to_value({"a" => 1, "c" => 2,"b"=> 3}, 1)
# => {"a" => 2, "c" => 3,"b"=> 4}
I'm thinking:
def add_to_value(hash, x)
hash.each {|key,value| value + x}
end
This returns:
{"a"=>1, "b"=>3, "c"=>2}
Why is hash sorted alphabetically?
You're super close, without any extra gems needed:
def add_to_value(hash, x)
hash.each {|key,value| hash[key] += x }
end
Just iterate the hash and update each value-by-key. #each returns the object being iterated on, so the result will be the original hash, which has been modified in place.
If you want a copy of the original hash, you can do that pretty easily, too:
def add_to_value(hash, x)
hash.each.with_object({}) {|(key, value), out| out[key] = value + x }
end
That'll define a new empty hash, pass it to the block, where it collects the new values. The new hash is returned from #with_object, and is thus returned out of add_to_value.
You can do the following to increment values:
hash = {}
{"a" => 1, "c" => 2,"b"=> 3}.each {|k,v| hash[k]=v+1}
hash
=>{"a"=>2, "c"=>3, "b"=>4}
And the hash will be sorted as you want.
The problem becomes trivial if you use certain gems, such as y_support. Type in your command line gem install y_support, and enjoy the extended hash iterators:
require 'y_support/core_ext/hash'
h = { "a"=>1, "c"=>3, "b"=>2 }
h.with_values do |v| v + 1 end
#=> {"a"=>2, "c"=>4, "b"=>3}
As for your sorting problem, I was unable to reproduce it.
Of course, a less elegant solution is possible without installing a gem:
h.each_with_object Hash.new do |(k, v), h| h[k] = v + 1 end
The gem also gives you Hash#with_keys (which modifies keys) and Hash#modify (which modifies both keys and values, kind of mapping from hash to hash), and banged versions Hash#with_values!, #with_keys! that modify the hash in place.
I'm trying to convert an array into a hash by using some matching. Before converting the array into a hash, I want to merge the values like this
"Desc,X1XXSC,C,CCCC4524,xxxs,xswd"
and create a hash from it. The rule is that, first value of the array is the key in Hash, in array there are repeating keys, for those keys I need to merge values and place it under one key. "Desc:" are keys. My program looks like this.
p 'test sample application'
str = "Desc:X1:C:CCCC:Desc:XXSC:xxxs:xswd:C:4524"
arr = Array.new
arr = str.split(":")
p arr
test_hash = Hash[*arr]
p test_hash
I could not find a way to figure it out. If any one can guide me, It will be thankful.
Functional approach with Facets:
require 'facets'
str.split(":").each_slice(2).map_by { |k, v| [k, v] }.mash { |k, vs| [k, vs.join] }
#=> {"Desc"=>"X1XXSC", "C"=>"CCCC4524", "xxxs"=>"xswd"}
Not that you cannot do it without Facets, but it's longer because of some basic abstractions missing in the core:
Hash[str.split(":").each_slice(2).group_by(&:first).map { |k, gs| [k, gs.map(&:last).join] }]
#=> {"Desc"=>"X1XXSC", "C"=>"CCCC4524", "xxxs"=>"xswd"}
A small variation on #Sergio Tulentsev's solution:
str = "Desc:X1:C:CCCC:Desc:XXSC:xxxs:xswd:C:4524"
str.split(':').each_slice(2).each_with_object(Hash.new{""}){|(k,v),h| h[k] += v}
# => {"Desc"=>"X1XXSC", "C"=>"CCCC4524", "xxxs"=>"xswd"}
str.split(':') results in an array; there is no need for initializing with arr = Array.new
each_slice(2) feeds the elements of this array two by two to a block or to the method following it, like in this case.
each_with_object takes those two elements (as an array) and passes them on to a block, together with an object, specified by:
(Hash.new{""}) This object is an empty Hash with special behaviour: when a key is not found then it will respond with a value of "" (instead of the usual nil).
{|(k,v),h| h[k] += v} This is the block of code which does all the work. It takes the array with the two elements and deconstructs it into two strings, assigned to k and v; the special hash is assigned to h. h[k] asks the hash for the value of key "Desc". It responds with "", to which "X1" is added. This is repeated until all elements are processed.
I believe you're looking for each_slice and each_with_object here
str = "Desc:X1:C:CCCC:Desc:XXSC:xxxs:xswd:C:4524"
hash = str.split(':').each_slice(2).each_with_object({}) do |(key, value), memo|
memo[key] ||= ''
memo[key] += value
end
hash # => {"Desc"=>"X1XXSC", "C"=>"CCCC4524", "xxxs"=>"xswd"}
Enumerable#slice_before is a good way to go.
str = "Desc:X1:C:CCCC:Desc:XXSC:xxxs:xswd:C:4524"
a = ["Desc","C","xxxs"] # collect the keys in a separate collection.
str.split(":").slice_before(""){|i| a.include? i}
# => [["Desc", "X1"], ["C", "CCCC"], ["Desc", "XXSC"], ["xxxs", "xswd"], ["C", "4524"]]
hsh = str.split(":").slice_before(""){|i| a.include? i}.each_with_object(Hash.new("")) do |i,h|
h[i[0]] += i[1]
end
hsh
# => {"Desc"=>"X1XXSC", "C"=>"CCCC4524", "xxxs"=>"xswd"}