Suppose I have the following hash table in ruby,
H = {"I"=>3, "Sam"=>2, "am"=>2, "do"=>1, "not"=>1, "eat"=> 1}
I want to construct the following hash table N from H
N = {"1" => 3, "2"=>2, "3"=>1}
where 3 in value signifies number of hash entires with value "1" in H (e.g. "do"=>1, "not"=>1, "eat"=> 1, therefore "1" => 3)
Is there any easy way in ruby to construct the N hash table from H??
Thanks in advance!!!
Hash[H.values.group_by{|i| i}.map {|k,v| [k, v.count]}]
Update: to get strings as keys and sorted by key:
Hash[h.values.group_by{|i| i}.map {|k,v| [k.to_s, v.count]}.sort]
Try this,
N = {}
H.values.each { |t| N["#{t}"] = (N["#{t}"] || 0) + 1}
Alternate solution:
H.values.sort.inject(Hash.new(0)) {|ac,e| ac[e.to_s]+=1;ac}
# {"1"=>3, "2"=>2, "3"=>1}
Related
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
#participants[id] = {nick: nick, points: 1}
=> {"1"=>{:nick=>"Test", :points=>3}, "30"=>{:nick=>"AnotherTest", :points=>5}, "20"=>{:nick=>"Newtest", :points=>3}}
I want my the lowest points (ID: 1 and 20). How do I get the lowest points go first and then ID 30 go last?
If you use Enumerable#max_by() or Enumerable#min_by() you can do following;
data = {
"1" => {nick: "U1", points: 3},
"30" => {nick: "U30", points: 5},
"20" => {nick: "U20", points: 3}
}
max_id, max_data = data.max_by {|k,v| v[:points]}
puts max_id # => 30
puts max_data # => {nick: "U30", points: 5}
Same thing works with #min_by() and if you want to get back Hash you do this:
minimal = Hash[*data.min_by {|k,v| v[:points]}]
puts minimal # => {"1"=>{:nick=>"U1", :points=>3}}
Functions min_by() and max_by() will always return one record. If you want to get all records with same points then you have to use min / max data an do another "lookup" like this:
min_id, min_data = data.min_by {|k,v| v[:points]}
all_minimal = data.select {|k,v| v[:points] == min_data[:points]}
puts all_minimal
# => {"1"=>{:nick=>"U1", :points=>3}, "20"=>{:nick=>"U20", :points=>3}}
Hashes are not a suitable data structure for this operation. They are meant when you need to get a value in O(1) complexity.
You are better off using a sorted array or a tree if you are interested in comparisons or Heap (in case you are interested in only maximum or minimum value) as #Vadim suggested
Use Enumerable#minmax_by:
h = { "1"=>{:nick=>"Test", :points=>3},
"30"=>{:nick=>"AnotherTest", :points=>5},
"20"=>{:nick=>"Newtest", :points=>3}}
h.minmax_by { |_,g| g[:points] }
#=> [[ "1", {:nick=>"Test", :points=>3}],
# ["30", {:nick=>"AnotherTest", :points=>5}]]
You can work around the fact that min_by and max_by are returning only one result by combining sort and chunk:
data.sort_by{|_,v| v[:points]}.chunk{|(_,v)| v[:points]}.first.last.map(&:first)
#=> ["1", "20"]
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 have a hash h:
h = {145=>1, 137=>2, 34=>3}
I want to convert it into an array of hashes of the form:
[{cid:145, qty:1}, {cid:137, qty:2}, {cid:34, qty:3}]
My first attempt a solution works for this example:
h.keys.zip(h.values).map { |cid, qty| {cid:cid, qty:qty} }
Evaluates to
[{:cid=>145, :qty=>1}, {:cid=>137, :qty=>2}, {:cid=>34, :qty=>3}]
My worry is that h.keys and h.values won't always align, since hashes aren't necessarily ordered.
How can I solve this problem with the guarantee that the keys of h will be paired with their corresponding values?
h = {145=>1, 137=>2, 34=>3}
h.map!{ |k, v| {:cid =>k, :qty => v} }
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"}