Related
I'm new to ruby, I am solving a problem that involves hashes and key. The problem asks me to Implement a method, #pet_types, that accepts a hash as an argument. The hash uses people's # names as keys, and the values are arrays of pet types that the person owns. My question is about using Hash#each method to iterate through each num inside the array. I was wondering if there's any difference between solving the problem using hash#each or hash.sort.each?
I spent several hours coming up different solution and still to figure out what are the different approaches between the 2 ways of solving the problem below.
I include my code in repl.it: https://repl.it/H0xp/6 or you can see below:
# Pet Types
# ------------------------------------------------------------------------------
# Implement a method, #pet_types, that accepts a hash as an argument. The hash uses people's
# names as keys, and the values are arrays of pet types that the person owns.
# Example input:
# {
# "yi" => ["dog", "cat"],
# "cai" => ["dog", "cat", "mouse"],
# "venus" => ["mouse", "pterodactyl", "chinchilla", "cat"]
# }
def pet_types(owners_hash)
results = Hash.new {|h, k| h[k] = [ ] }
owners_hash.sort.each { |k, v| v.each { |pet| results[pet] << k } }
results
end
puts "-------Pet Types-------"
owners_1 = {
"yi" => ["cat"]
}
output_1 = {
"cat" => ["yi"]
}
owners_2 = {
"yi" => ["cat", "dog"]
}
output_2 = {
"cat" => ["yi"],
"dog" => ["yi"]
}
owners_3 = {
"yi" => ["dog", "cat"],
"cai" => ["dog", "cat", "mouse"],
"venus" => ["mouse", "pterodactyl", "chinchilla", "cat"]
}
output_3 = {
"dog" => ["cai", "yi"],
"cat" => ["cai", "venus", "yi"],
"mouse" => ["cai", "venus"],
"pterodactyl" => ["venus"],
"chinchilla" => ["venus"]
}
# method 2
# The 2nd and 3rd method should return a hash that uses the pet types as keys and the values should
# be a list of the people that own that pet type. The names in the output hash should
# be sorted alphabetically
# switched_hash = Hash.new()
# owners_hash.each do |owner, pets_array|
# pets_array.each do |pet|
# select_owners = owners_hash.select { |owner, pets_array|
owners_hash[owner].include?(pet) }
# switched_hash[pet] = select_owners.keys.sort
# end
# end
# method 3
#switched_hash
# pets = Hash.new {|h, k| h[k] = [ ] } # WORKS SAME AS: pets = Hash.new( Array.new )
# owners = owners_hash.keys.sort
# owners.each do |owner|
# owners_hash[owner].each do |pet|
# pets[pet] << owner
# end
# end
# pets
# Example output:
# output_3 = {
# "dog" => ["cai", "yi"],
# "cat" => ["cai", "venus", "yi"], ---> (sorted alphabetically!)
# "mouse" => ["cai", "venus"],
# "pterodactyl" => ["venus"],
# "chinchilla" => ["venus"]
# }
I used a hash data structure in my program to first solve this problem. Then I tried to rewrite it using the pet_hash. And my final codes is the following:
def pet_types(owners_hash)
pets_hash = Hash.new { |k, v| v = [] }
owners_hash.each do |owner, pets|
pets.each do |pet|
pets_hash[pet] += [owner]
end
end
pets_hash.values.each(&:sort!)
pets_hash
end
puts "-------Pet Types-------"
owners_1 = {
"yi" => ["cat"]
}
output_1 = {
"cat" => ["yi"]
}
owners_2 = {
"yi" => ["cat", "dog"]
}
output_2 = {
"cat" => ["yi"],
"dog" => ["yi"]
}
owners_3 = {
"yi" => ["dog", "cat"],
"cai" => ["dog", "cat", "mouse"],
"venus" => ["mouse", "pterodactyl", "chinchilla", "cat"]
}
output_3 = {
"dog" => ["cai", "yi"],
"cat" => ["cai", "venus", "yi"],
"mouse" => ["cai", "venus"],
"pterodactyl" => ["venus"],
"chinchilla" => ["venus"]
}
puts pet_types(owners_1) == output_1
puts pet_types(owners_2) == output_2
puts pet_types(owners_3) == output_3
Hash#sort has the same effect (at least for my basic test) as Hash#to_a followed by Array#sort.
hash = {b: 2, a: 1}
hash.to_a.sort # => [[:a, 1, [:b, 2]]
hash.sort # => the same
Now let's look at #each, both on Hash and Array.
When you provide two arguments to the block, that can handle both cases. For the hash, the first argument will be the key and the second will be the value. For the nested array, the values essentially get splatted out to the args:
[[:a, 1, 2], [:b, 3, 4]].each { |x, y, z| puts "#{x}-#{y}-#{z}" }
# => a-1-2
# => b-3-4
So basically, you should think of Hash#sort to be a shortcut to Hash#to_a followed by Array#sort, and recognize that #each will work the same on a hash as a hash converted to array (a nested array). In this case, it doesn't matter which approach you take. Clearly if you need to sort iteration by the keys then you should use sort.
I have a set of activities that I want to sort based on an average rating, and how many times the activity has been rated.
activities = {
:one => { :avg_rating => 5, :total_ratings => 23 },
:two => { :avg_rating => 5, :total_ratings => 18 },
:three => { :avg_rating => 5, :total_ratings => 54 }
}
EDIT - updated so the results I was expecting was correct
The result of the sort would be in order of :three, :one, :two
Thanks!
activities.sort_by { |k,v|
[v[:avg_rating], v[:total_ratings]]
}.reverse
Whether you need the element names only:
activities.sort_by { |k,v|
[v[:avg_rating], v[:total_ratings]]
}.reverse.map &:first
#⇒ [
# [0] :three,
# [1] :one,
# [2] :two
#]
I'm pretty new in Ruby programming. In Ruby there are plenty ways to write elegant code. Is there any elegant way to link two arrays with objects of the same type by attribute value?
It's hard to explain. Let's look at the next example:
a = [ { :id => 1, :value => 1 }, { :id => 2, :value => 2 }, { :id => 3, :value => 3 } ]
b = [ { :id => 1, :value => 2 }, { :id => 3, :value => 4 } ]
c = link a, b
# Result structure after linkage.
c = {
"1" => {
:a => { :id => 1, :value => 1 },
:b => { :id => 1, :value => 1 }
},
"3" => {
:a => { :id => 3, :value => 3 },
:b => { :id => 3, :value => 4 }
}
}
So the basic idea is to get pairs of objects from different arrays by their common ID and construct a hash, which will give this pair by ID.
Thanks in advance.
If you want to take an adventure through Enumerable, you could say this:
(a.map { |h| [:a, h] } + b.map { |h| [:b, h] })
.group_by { |_, h| h[:id] }
.select { |_, a| a.length == 2 }
.inject({}) { |h, (n, v)| h.update(n => Hash[v]) }
And if you really want the keys to be strings, say n.to_s => Hash[v] instead of n => Hash[v].
The logic works like this:
We need to know where everything comes from we decorate the little hashes with :a and :b symbols to track their origins.
Then add the decorated arrays together into one list so that...
group_by can group things into almost-the-final-format.
Then find the groups of size two since those groups contain the entries that appeared in both a and b. Groups of size one only appeared in one of a or b so we throw those away.
Then a little injection to rearrange things into their final format. Note that the arrays we built in (1) just somehow happen to be in the format that Hash[] is looking for.
If you wanted to do this in a link method then you'd need to say things like:
link :a => a, :b => b
so that the method will know what to call a and b. This hypothetical link method also easily generalizes to more arrays:
def link(input)
input.map { |k, v| v.map { |h| [k, h] } }
.inject(:+)
.group_by { |_, h| h[:id] }
.select { |_, a| a.length == input.length }
.inject({}) { |h, (n, v)| h.update(n => Hash[v]) }
end
link :a => [...], :b => [...], :c => [...]
I assume that, for any two elements h1 and h2 of a (or of b), h1[:id] != h2[:id].
I would do this:
def convert(arr) Hash[arr.map {|h| [h[:id], h]}] end
ah, bh = convert(a), convert(b)
c = ah.keys.each_with_object({}) {|k,h|h[k]={a: ah[k], b: bh[k]} if bh.key?(k)}
# => {1=>{:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}},
# 3=>{:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}}
Note that:
ah = convert(a)
# => {1=>{:id=>1, :value=>1}, 2=>{:id=>2, :value=>2}, 3=>{:id=>3, :value=>3}}
bh = convert(b)
# => {1=>{:id=>1, :value=>2}, 3=>{:id=>3, :value=>4}}
Here's a second approach. I don't like it as well, but it represents a different way of looking at the problem.
def sort_by_id(a) a.sort_by {|h| h[:id]} end
c = Hash[*sort_by_id(a.select {|ha| b.find {|hb| hb[:id] == ha[:id]}})
.zip(sort_by_id(b))
.map {|ha,hb| [ha[:id], {a: ha, b: hb}]}
.flatten]
Here's what's happening. The first step is to select only the elements ha of a for which there is an element hb of b for which ha[:id] = hb[id]. Then we sort both (what's left of) a and b on h[:id], zip them together and then make the hash c.
r1 = a.select {|ha| b.find {|hb| hb[:id] == ha[:id]}}
# => [{:id=>1, :value=>1}, {:id=>3, :value=>3}]
r2 = sort_by_id(r1)
# => [{:id=>1, :value=>1}, {:id=>3, :value=>3}]
r3 = sort_by_id(b)
# => [{:id=>1, :value=>2}, {:id=>3, :value=>4}]
r4 = r2.zip(r3)
# => [[{:id=>1, :value=>1}, {:id=>1, :value=>2}],
# [{:id=>3, :value=>3}, {:id=>3, :value=>4}]]
r5 = r4.map {|ha,hb| [ha[:id], {a: ha, b: hb}]}
# => [[1, {:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}}],
# [3, {:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}]]
r6 = r5.flatten
# => [1, {:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}},
# 3, {:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}]
c = Hash[*r6]
# => {1=>{:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}},
# 3=>{:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}}
Ok, I've found the answer by myself. Here is a quite short line of code, which should do the trick:
Hash[a.product(b)
.select { |pair| pair[0][:id] == pair[1][:id] }
.map { |pair| [pair[0][:id], { :a => pair[0], :b => pair[1] }] }]
The product method gives us all possible pairs, then we filter them by equal IDs of pair elements. And then we map pairs to the special form, which will produce a Hash we are looking for.
So Hash[["key1", "value1"], ["key2", "value2"]] returns { "key1" => "value1", "key2" => "value2" }. And I use this to get the answer on my question.
Thanks.
P.S.: you can use pair.first instead of pair[0] and pair.last instead of pair[1] for better readability.
UPDATE
As Cary pointed out, it is better to replace |pair| with |ha, hb| to avoid these ugly indices:
Hash[a.product(b)
.select { |ha, hb| ha[:id] == hb[:id] }
.map { |ha, hb| [ha[:id], { :a => ha, :b => hb }] }]
I would like to combine arrrays #a, #b, and #c into a single array with multiple data elements, for example OpenStruct:
#a = ["my", "foo", "bar"]
#b = ["yan", "can", "cook"]
#c = ["in", "your", "dreams"]
the output would be like:
[
{ :a => "my", :b => "yan", :c => "in" },
{ :a => "foo", :b => "can", :c => "your" },
{ :a => bar, :b => "cook", :c => "dreams" }
]
What is the fastest way to do this? Should I consider another class?
Here is one solution, I am not quite convinced it is the neatest though:
#a.zip(#b, #c).map {|t| {:a => t[0], :b => t[1], :c => t[2]}}
Functional approach:
[#a, #b, #c].transpose.map { |xs| Hash[[:a, :b, :c].zip(xs)] }
#=> [{:a=>"my", :b=>"yan", :c=>"in"}, {:a=>"foo", :b=>"can", :c=>"your"}, {:a=>"bar", :b=>"cook", :c=>"dreams"}]
The best way to know the fastest way is do a benchmark. Based on previous answers:
require 'benchmark'
#a = ["my", "foo", "bar"]
#b = ["yan", "can", "cook"]
#c = ["in", "your", "dreams"]
$n = 500_000
Benchmark.bmbm do |x|
x.report("Boris Strandjev") do $n.times do
#a.zip(#b, #c).map {|t| {:a => t[0], :b => t[1], :c => t[2]}}
end end
x.report("tokland") do $n.times do
[#a, #b, #c].transpose.map { |xs| Hash[[:a, :b, :c].zip(xs)] }
end end
x.report("mu is too short") do $n.times do
(0 ... [#a, #b, #c].max_by(&:length).length).map { |i| { :a => #a[i], :b => #b[i], :c => #c[i] } }
end end
x.report("KL-7") do $n.times do
#a.each_with_index.map { |a, i| { :a => a, :b => #b[i], :c => #c[i] } }
end end
end
Output:
Rehearsal ---------------------------------------------------
Boris Strandjev 4.540000 0.015000 4.555000 ( 4.571261)
tokland 7.145000 0.000000 7.145000 ( 7.268415)
mu is too short 5.304000 0.047000 5.351000 ( 5.560318)
KL-7 4.914000 0.000000 4.914000 ( 5.030287)
----------------------------------------- total: 21.965000sec
user system total real
Boris Strandjev 4.462000 0.016000 4.478000 ( 4.553260)
tokland 7.129000 0.031000 7.160000 ( 7.309418)
mu is too short 5.366000 0.031000 5.397000 ( 5.447312)
KL-7 4.898000 0.016000 4.914000 ( 4.997286)
If there is no need to worry about arrays having different length I think that's the fastest way (as it iterate over array only once):
#a.each_with_index.map { |a, i| { :a => a, :b => #b[i], :c => #c[i] } }
Something like this should work:
(0 ... [#a, #b, #c].max_by(&:length).length).map { |i| { :a => #a[i], :b => #b[i], :c => #c[i] } }
That doesn't assume they're all the same length but you'll get nil values if they're not.
We have the following datastructures:
{:a => ["val1", "val2"], :b => ["valb1", "valb2"], ...}
And I want to turn that into
[{:a => "val1", :b => "valb1"}, {:a => "val2", :b => "valb2"}, ...]
And then back into the first form. Anybody with a nice looking implementation?
This solution works with arbitrary numbers of values (val1, val2...valN):
{:a => ["val1", "val2"], :b => ["valb1", "valb2"]}.inject([]){|a, (k,vs)|
vs.each_with_index{|v,i| (a[i] ||= {})[k] = v}
a
}
# => [{:a=>"val1", :b=>"valb1"}, {:a=>"val2", :b=>"valb2"}]
[{:a=>"val1", :b=>"valb1"}, {:a=>"val2", :b=>"valb2"}].inject({}){|a, h|
h.each_pair{|k,v| (a[k] ||= []) << v}
a
}
# => {:a=>["val1", "val2"], :b=>["valb1", "valb2"]}
Using a functional approach (see Enumerable):
hs = h.values.transpose.map { |vs| h.keys.zip(vs).to_h }
#=> [{:a=>"val1", :b=>"valb1"}, {:a=>"val2", :b=>"valb2"}]
And back:
h_again = hs.first.keys.zip(hs.map(&:values).transpose).to_h
#=> {:a=>["val1", "val2"], :b=>["valb1", "valb2"]}
Let's look closely what the data structure we are trying to convert between:
#Format A
[
["val1", "val2"], :a
["valb1", "valb2"], :b
["valc1", "valc2"] :c
]
#Format B
[ :a :b :c
["val1", "valb1", "valc1"],
["val2", "valb2", "valc3"]
]
It is not diffculty to find Format B is the transpose of Format A in essential , then we can come up with this solution:
h={:a => ["vala1", "vala2"], :b => ["valb1", "valb2"], :c => ["valc1", "valc2"]}
sorted_keys = h.keys.sort_by {|a,b| a.to_s <=> b.to_s}
puts sorted_keys.inject([]) {|s,e| s << h[e]}.transpose.inject([]) {|r, a| r << Hash[*sorted_keys.zip(a).flatten]}.inspect
#[{:b=>"valb1", :c=>"valc1", :a=>"vala1"}, {:b=>"valb2", :c=>"valc2", :a=>"vala2"}]
m = {}
a,b = Array(h).transpose
b.transpose.map { |y| [a, y].transpose.inject(m) { |m,x| m.merge Hash[*x] }}
My attempt, perhaps slightly more compact.
h = { :a => ["val1", "val2"], :b => ["valb1", "valb2"] }
h.values.transpose.map { |s| Hash[h.keys.zip(s)] }
Should work in Ruby 1.9.3 or later.
Explanation:
First, 'combine' the corresponding values into 'rows'
h.values.transpose
# => [["val1", "valb1"], ["val2", "valb2"]]
Each iteration in the map block will produce one of these:
h.keys.zip(s)
# => [[:a, "val1"], [:b, "valb1"]]
and Hash[] will turn them into hashes:
Hash[h.keys.zip(s)]
# => {:a=>"val1", :b=>"valb1"} (for each iteration)
This will work assuming all the arrays in the original hash are the same size:
hash_array = hash.first[1].map { {} }
hash.each do |key,arr|
hash_array.zip(arr).each {|inner_hash, val| inner_hash[key] = val}
end
You could use inject to build an array of hashes.
hash = { :a => ["val1", "val2"], :b => ["valb1", "valb2"] }
array = hash.inject([]) do |pairs, pair|
pairs << { pair[0] => pair[1] }
pairs
end
array.inspect # => "[{:a=>["val1", "val2"]}, {:b=>["valb1", "valb2"]}]"
Ruby documentation has a few more examples of working with inject.