I'm trying to achieve the following JSON results:
{
"movie" =>
[{
"title": "Thor",
"year" : 2011,
},
{
"title": "Iron Man",
"year" : 2008,
}],
"tv" =>
[{
"title": "Parks and Recreation"
"year": 2009
},
{
"title": "Friends"
"year": 1994
}]
}
With JavaScript, I would loop through my results and do something like:
results['movie'].push(item);
results['tv'].push(item);
With Ruby code, the farthest I've gone is this:
#results = Hash.new
results['Search'].each do |r|
if r['Type'] == 'movie'
#results['movie'] << {
'title' => r['Title'],
'year' => r['Year']
}
elsif r['Type'] == 'series'
#results['tv'] << {
'title' => r['Title'],
'year' => r['Year']
}
end
end
What am I missing here?
I think you can get what you want by using Enumerable#each_with_object and assigning a default value to the hash.
def group_search_results(items)
results = Hash.new { |hash, key| hash[key] = [] }
items.each_with_object(results) do |item|
results[item['Type']] << {'title' => item['Title'], 'year' => item['Year']}
end
end
describe "search_results" do
it "groups into an object" do
items = [
{'Type' => 'movie', 'Title' => 'Thor', 'Year' => 2011},
{'Type' => 'movie', 'Title' => 'Iron Man', 'Year' => 2008},
{'Type' => 'series', 'Title' => 'Parks and Recreation', 'Year' => 2009},
{'Type' => 'series', 'Title' => 'Friends', 'Year' => 1994},
]
results = group_search_results(items)
expect(results).to eq({
'movie' => [
{'title' => 'Thor', 'year' => 2011},
{'title' => 'Iron Man', 'year' => 2008},
],
'series' => [
{'title' => 'Parks and Recreation', 'year' => 2009},
{'title' => 'Friends', 'year' => 1994},
],
})
end
end
I believe the problem has to do with the initialization of your hash. The movie and tv keys aren't currently an array. You can initialize your hash like this:
#results = { 'movie' => [], 'tv' => [] }
Here's how it looks with the rest of your code:
#results = { 'movie' => [], 'tv' => [] }
results['Search'].each do |r|
if r['Type'] == 'movie'
#results['movie'] << {
'title' => r['Title'],
'year' => r['Year']
}
elsif r['Type'] == 'series'
#results['tv'] << {
'title' => r['Title'],
'year' => r['Year']
}
end
end
results = {
search: {
movie: [
{ title: 'Thor', year: 2011 },
{ title: 'Iron Man', year: 2008 },
],
tv: [
{ title: 'Parks and Recreation', year: 2009 },
{ title: 'Friends', year: 1994 },
]
}
}
#results = Hash.new{|k, v| k[v] = []}
results[:search].each do |type, array|
#results[type].push(*array)
end
results[:search].each_with_object(Hash.new{|k, v| k[v] = []}) do |(type, array), hash|
hash[type].push(*array)
end
Let's say I want to access the values of a hash like this:
munsters = {
"Herman" => { "age" => 32, "gender" => "male" },
"Lily" => { "age" => 30, "gender" => "female" },
"Grandpa" => { "age" => 402, "gender" => "male" },
"Eddie" => { "age" => 10, "gender" => "male" },
"Marilyn" => { "age" => 23, "gender" => "female"}
}
I could use #each with two parameters:
munsters.each do |key, value|
puts "#{name} is a #{value["age"]}-year-old #{value["gender"]}."
end
Or I could use #each_pair with two parameters:
munsters.each_pair do |key, value|
puts "#{name} is a #{value["age"]}-year-old #{value["gender"]}."
end
Perhaps the difference between the two is not borne out in this simple example, but can someone help me to understand the advantage of using #each_pair over #each ?
Because Hash is an Enumerable, it has to have an each method. each_pair may be a clearer name, since it strongly suggests that two-element arrays containing key-value pairs are passed to the block.
They are aliases for each other: they share the same source code.
I am working with a large data set with multiple arrays of hashes which all have a key-value pair in common ("date" & the date value) as the first element of the hash.
The array of hashes I need to parse (#data["snapshot"]) is in the following format. Note that #data["snapshot"][0], #data["snapshot"][1], and #data["snapshot"][2] are in identical format with identical dates but their total's differ. In the resulting hash I need to have a key-value pair which identifies where the data came from.
#data["snapshot"][0] is as follows:
[{"date"=>"1455672010", "total"=>"**817**", "I"=>"1", "L"=>"3", "M"=>"62", "H"=>"5", "C"=>"0"},
{"date"=>"1455595298", "total"=>"**40**", "I"=>"8", "L"=>"5", "M"=>"562", "H"=>"125", "C"=>"0"},
{"date"=>"1455336016", "total"=>"**555**", "I"=>"10", "L"=>"1", "M"=>"93", "H"=>"121", "C"=>"0"}]
#data["snapshot"][1] is as follows:
[{"date"=>"1455672010", "total"=>"**70**", "I"=>"1", "L"=>"9", "M"=>"56", "H"=>"25", "C"=>"0"},
{"date"=>"1455595298", "total"=>"**54**", "I"=>"8", "L"=>"2", "M"=>"5", "H"=>"5", "C"=>"0"},
{"date"=>"1455336016", "total"=>"**25**", "I"=>"0", "L"=>"9", "M"=>"93", "H"=>"12", "C"=>"0"}]
#data["snapshot"][2] is as follows:
[{"date"=>"1455672010", "total"=>"**70**", "I"=>"12", "L"=>"5", "M"=>"5662", "H"=>"125", "C"=>"0"},
{"date"=>"1455595298", "total"=>"**43212**", "I"=>"56", "L"=>"6", "M"=>"5662", "H"=>"125", "C"=>"0"},
{"date"=>"1455336016", "total"=>"**55525**", "I"=>"100", "L"=>"19", "M"=>"5593", "H"=>"121", "C"=>"0"}]
My Question Is Ultimately:
How do I convert (flatten?) the 3 existing array of hashes (#data["snapshot"][0], #data["snapshot"][1], and #data["snapshot"][2]) into a single array of hashes in the following format?
[{"date"=>"1455672010", "CameFromDataSource0"=>"817", "CameFromDataSource1"=>"70", "CameFromDataSource2"=>"70"},
{"date"=>"1455595298", "CameFromDataSource0"=>"40", "CameFromDataSource1"=>"54", "CameFromDataSource2"=>"43212"},
{"date"=>"1455336016", "CameFromDataSource0"=>"555", "CameFromDataSource1"=>"25", "CameFromDataSource2"=>"55525"}]
TL;DR
snapshots.each_with_object(Hash.new {|hsh, date| hsh[date] = { "date" => date } })
.with_index do |(snapshot, hsh), i|
snapshot["data"].each {|datum| hsh[datum["date"]]["data#{i}"] = datum["total"] }
end.values
How it works
I'll break it down so you see how each part works. Here's our data (extraneous keys elided for clarity):
snapshots = [
{ "dataSourceID" => "152970",
"data" => [ { "date" => "1455672010", "total" => "817" },
{ "date" => "1455595298", "total" => "40" },
{ "date" => "1455336016", "total" => "555" } ]
}
{ "dataSourceID" => "33151",
"data" => [ { "date" => "1455672010", "total" => "70" },
{ "date" => "1455595298", "total" => "54" },
{ "date" => "1455336016", "total" => "25" } ]
},
{ "dataSourceID" => "52165",
"data" => [ { "date" => "1455672010", "total" => "70" },
{ "date" => "1455595298", "total" => "43212" },
{ "date" => "1455336016", "total" => "55525" } ]
}
]
Most of the magic is here:
result_hash = Hash.new {|hsh, date| hsh[date] = { "date" => date } }
Here we're using the Hash's default proc to automatically initialize new keys in the following way:
result_hash = Hash.new {|hsh, date| hsh[date] = { "date" => date } }
p result_hash["1455672010"]
# => { "date" => "1455672010" }
p result_hash
# => { "1455672010" => { "date" => "1455672010" } }
Simply accessing result_hash[foo] creates the hash { "date" => foo } and assigns it to result_hash[foo]. This enables the following:
result_hash["1455672010"]["data0"] = "817"
p result_hash
# => { "1455672010" => { "date" => "1455672010", "data0" => "817" } }
Magic!
Now suppose we have the following data:
data = [ { "date" => "1455672010", "total" => "817" },
{ "date" => "1455595298", "total" => "40" },
{ "date" => "1455336016", "total" => "555" } ]
Using our magic result_hash, we can do this:
data.each do |datum|
result_hash[datum["date"]]["data0"] = datum["total"]
end
p result_hash
# => { "1455672010" => { "date" => "1455672010", "data0" => "817" },
# "1455595298" => { "date" => "1455595298", "data0" => "40" },
# "1455336016" => { "date" => "1455336016", "data0" => "555" } }
See where I'm going with this? Here's all of our data, finally:
snapshots = [
{ "dataSourceID" => "152970",
"data" => [ { "date" => "1455672010", "total" => "817" },
{ "date" => "1455595298", "total" => "40" },
{ "date" => "1455336016", "total" => "555" } ]
}
{ "dataSourceID" => "33151",
"data" => [ { "date" => "1455672010", "total" => "70" },
{ "date" => "1455595298", "total" => "54" },
{ "date" => "1455336016", "total" => "25" } ]
},
{ "dataSourceID" => "52165",
"data" => [ { "date" => "1455672010", "total" => "70" },
{ "date" => "1455595298", "total" => "43212" },
{ "date" => "1455336016", "total" => "55525" } ]
}
]
Instead of hard-coding "data0", we can iterate over the snapshots hashes using each_with_index and build that key ("data0", then "data1", and so on) for each iteration. Inside that loop we can do exactly what we did above but with the "data" array from each snapshots hash:
result_hash = Hash.new {|hsh, date| hsh[date] = { "date" => date } }
snapshots.each_with_index do |snapshot, i|
data_key = "data#{i}"
snapshot["data"].each do |datum|
date = datum["date"]
result_hash[date][data_key] = datum["total"]
end
end
p result_hash.values
# => [ { "date" => "1455672010", "data0" => "817", "data1" => "70", "data2" => "70" },
# { "date" => "1455595298", "data0" => "40", "data1" => "54", "data2" => "43212" },
# { "date" => "1455336016", "data0" => "555", "data1" => "25", "data2" => "55525" } ]
Of course, this can be condensed some, which I've done in TL;DR above.
This is one way to do it.
Code
def convert(data)
data.each_with_object({}) { |a,h|
a.each { |g| h.update(g["date"]=>[g["total"][/\d+/]]) { |_,o,n| o+n } } }.
map { |date, arr| arr.each_with_index.with_object({"date"=>date}) { |(e,i),h|
h["key#{i}"] = e } }
end
Example
convert(data)
#=> [{"date"=>"1455672010", "key0"=>"817", "key1"=>"70", "key2"=>"70"},
# {"date"=>"1455595298", "key0"=>"40", "key1"=>"54", "key2"=>"43212"},
# {"date"=>"1455336016", "key0"=>"555", "key1"=>"25", "key2"=>"55525"}]
Two steps
You can see that I've done this in two steps. First construct a hash:
f = data.each_with_object({}) { |a,h| a.each { |g|
h.update(g["date"]=>[g["total"][/\d+/]]) { |_,o,n| o+n } } }
#=> {"1455672010"=>["817", "70", "70"],
# "1455595298"=>["40", "54", "43212"],
# "1455336016"=>["555", "25", "55525"]}
Here I have used the form of Hash#update (aka merge!) that employs a block ({ |_,o,n| o+n }) to determine the values of keys that are present in both hashes being merged.
Then convert the hash to the desired format:
f.map { |date, arr| arr.each_with_index.with_object({"date"=>date}) { |(e,i),h|
h["key#{i}"] = e } }
#=> [{"date"=>"1455672010", "key0"=>"817", "key1"=>"70", "key2"=>"70"},
# {"date"=>"1455595298", "key0"=>"40", "key1"=>"54", "key2"=>"43212"},
# {"date"=>"1455336016", "key0"=>"555", "key1"=>"25", "key2"=>"55525"}]
array = [{ name:'Joe', class:'foo' },
{ name:'Bob', class:'bar' },
{ name:'Hal', class:'baz' },
{ name:'Kim', class:'qux' },
{ name:'Zoe', class:'bar' }
]
What is a good way to sort by class in the following order: qux, bar, foo, baz?
Like this, for example:
array = [{ name:'Joe', class:'foo' },
{ name:'Bob', class:'bar' },
{ name:'Hal', class:'baz' },
{ name:'Kim', class:'qux' },
{ name:'Zoe', class:'bar' }
]
order = %w[qux bar foo baz]
sorted = array.sort_by{|el| order.index(el[:class])}
sorted # => [{:name=>"Kim", :class=>"qux"},
# {:name=>"Bob", :class=>"bar"},
# {:name=>"Zoe", :class=>"bar"},
# {:name=>"Joe", :class=>"foo"},
# {:name=>"Hal", :class=>"baz"}]
order = %w[qux bar foo baz]
array.sort_by{|h| order.index(h[:class])}
gives:
[
{
:name => "Kim",
:class => "qux"
},
{
:name => "Bob",
:class => "bar"
},
{
:name => "Zoe",
:class => "bar"
},
{
:name => "Joe",
:class => "foo"
},
{
:name => "Hal",
:class => "baz"
}
]
a = {"rows" => [{"id" => "231-z", "name" => 'jon', "age"=> 27, "state" => 'AL'},
{"id" => "4121-x", "name" => 'ton', "age"=> 37, "state" => 'VA'}
]
}
b = {"rows" => [{"key" => ["xyz","4121-x"], "value" =>{"sum" => 12312, "realage" => 29}},
{"key" => ["xyz","231-z"], "value" =>{"sum" => 1212, "realage" => 33}}
]
}
In hash a, age is incorrect
In hash b, realage is correct. Also in hash b id is the second value in the first array that maps to id of hash a . Those are 4121-x, 231-z correspond to hash a
I want to correct the age in hash a and swap it with the realage of hash b
I can do it in multiple steps, but is it possible to do it in one liner or very short? So finally correct hash a should look like
a = {"rows" => [{"id" => "231-z", "name" => 'jon', "age"=> 33, "state" => 'AL'},
{"id" => "4121-x", "name" => 'ton', "age"=> 29, "state" => 'VA'}
]
}
does this look reasonable?
a['rows'].each_with_index do |ah, i|
(bh = b['rows'].select {|h| h['key'].last == ah['id'] }.first) &&
a['rows'][i] = ah.update('age' => bh['value']['realage'])
end
p a
{
"rows" => [
[0] {
"id" => "231-z",
"name" => "jon",
"age" => 33,
"state" => "AL"
},
[1] {
"id" => "4121-x",
"name" => "ton",
"age" => 29,
"state" => "VA"
}
]
}
Please note it will update a only if corresponding id found in b.
Also, the rows order does not matter, nor matter the rows number, it is only important b to have a row with same id as processed row in a
Here is a Working Demo