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What is the most efficient and pretty way to map this:
{name:"cheese,test", uid:"1,2"}
to this:
[ {name:"cheese", uid:"1"}, {name:"test", uid:"2"} ]
should work dinamically for example with: { name:"cheese,test,third", uid:"1,2,3" } or {name:"cheese,test,third,fourth", uid:"1,2,3,4", age:"9,8,7,6" }
Finally I made this:
hash = {name:"cheese,test", uid:"1,2"}
results = []
length = hash.values.first.split(',').length
length.times do |i|
results << hash.map {|k,v| [k, v.split(',')[i]]}
end
results.map{|e| e.to_h}
It is working, but i am not pleased with it, has to be a cleaner and more 'rubyst' way to do this
def splithash(h)
# Transform each element in the Hash...
h.map do |k, v|
# ...by splitting the values on commas...
v.split(',').map do |vv|
# ...and turning these into individual { k => v } entries.
{ k => vv }
end
end.inject do |a,b|
# Then combine these by "zip" combining each list A to each list B...
a.zip(b)
# ...which will require a subsequent .flatten to eliminate nesting
# [ [ 1, 2 ], 3 ] -> [ 1, 2, 3 ]
end.map(&:flatten).map do |s|
# Then combine all of these { k => v } hashes into one containing
# all the keys with associated values.
s.inject(&:merge)
end
end
Which can be used like this:
splithash(name:"cheese,test", uid:"1,2", example:"a,b")
# => [{:name=>"cheese", :uid=>"1", :example=>"a"}, {:name=>"test", :uid=>"2", :example=>"b"}]
It looks a lot more convoluted at first glance, but this handles any number of keys.
I would likely use transpose and zip like so:
hash = {name:"cheese,test,third,fourth", uid:"1,2,3,4", age:"9,8,7,6" }
hash.values.map{|x| x.split(",")}.transpose.map{|v| hash.keys.zip(v).to_h}
#=> [{:name=>"cheese", :uid=>"1", :age=>"9"}, {:name=>"test", :uid=>"2", :age=>"8"}, {:name=>"third", :uid=>"3", :age=>"7"}, {:name=>"fourth", :uid=>"4", :age=>"6"}]
To break it down a bit (code slightly modified for operational clarity):
hash.values
#=> ["cheese,test,third,fourth", "1,2,3,4", "9,8,7,6"]
.map{|x| x.split(",")}
#=> [["cheese", "test", "third", "fourth"], ["1", "2", "3", "4"], ["9", "8", "7", "6"]]
.transpose
#=> [["cheese", "1", "9"], ["test", "2", "8"], ["third", "3", "7"], ["fourth", "4", "6"]]
.map do |v|
hash.keys #=> [[:name, :uid, :age], [:name, :uid, :age], [:name, :uid, :age], [:name, :uid, :age]]
.zip(v) #=> [[[:name, "cheese"], [:uid, "1"], [:age, "9"]], [[:name, "test"], [:uid, "2"], [:age, "8"]], [[:name, "third"], [:uid, "3"], [:age, "7"]], [[:name, "fourth"], [:uid, "4"], [:age, "6"]]]
.to_h #=> [{:name=>"cheese", :uid=>"1", :age=>"9"}, {:name=>"test", :uid=>"2", :age=>"8"}, {:name=>"third", :uid=>"3", :age=>"7"}, {:name=>"fourth", :uid=>"4", :age=>"6"}]
end
Input
hash={name:"cheese,test,third,fourth", uid:"1,2,3,4", age:"9,8,7,6" }
Code
p hash
.transform_values { |v| v.split(',') }
.map { |k, v_arr| v_arr.map { |v| [k, v] }
}
.transpose
.map { |array| array.to_h }
Output
[{:name=>"cheese", :uid=>"1", :age=>"9"}, {:name=>"test", :uid=>"2", :age=>"8"}, {:name=>"third", :uid=>"3", :age=>"7"}, {:name=>"fourth", :uid=>"4", :age=>"6"}]
We are given
h = { name: "cheese,test", uid: "1,2" }
Here are two ways to create the desired array. Neither construct arrays that are then converted to hashes.
#1
First compute
g = h.transform_values { |s| s.split(',') }
#=> {:name=>["cheese", "test"], :uid=>["1", "2"]}
then compute
g.first.last.size.times.map { |i| g.transform_values { |v| v[i] } }
#=> [{:name=>"cheese", :uid=>"1"}, {:name=>"test", :uid=>"2"}]
Note
a = g.first
#=> [:name, ["cheese", "test"]]
b = a.last
#=> ["cheese", "test"]
b.size
#=> 2
#2
This approach does not convert the values of the hash to arrays.
(h.first.last.count(',')+1).times.map do |i|
h.transform_values { |s| s[/(?:\w+,){#{i}}\K\w+/] }
end
#=> [{:name=>"cheese", :uid=>"1"}, {:name=>"test", :uid=>"2"}]
We have
a = h.first
#=> [:name, "cheese,test"]
s = a.last
#=> "cheese,test"
s.count(',')+1
#=> 2
We can express the regular expression in free-spacing mode to make it self-documenting.
/
(?: # begin a non-capture group
\w+, # match one or more word characters followed by a comma
) # end the non-capture group
{#{i}} # execute the preceding non-capture group i times
\K # discard all matches so far and reset the start of the match
\w+ # match one or more word characters
/x # invoke free-spacing regex definition mode
Hi i have some data that system send to me alternative like this:
"Screw:1,Bound:5,Hing:3"
"Bound:5,Screw:3,Hing:1"
"Bound:2,Screw:2"
how can i make this Hash?
{"Screw"=>6 ,"Bound"=>12, "Hing"=>4}
its probably add other key and value later , i hop solve this for me.
arr = [
"Screw:1,Bound:5,Hing:3",
"Bound:5,Screw:3,Hing:1",
"Bound:2,Screw:2"
]
arr.flat_map { |s| s.split(',') }
.each_with_object(Hash.new(0)) do |s,h|
k, v = s.split(':')
h[k] += v.to_i
end
#=> {"Screw"=>6, "Bound"=>12, "Hing"=>4}
Step 1
arr.flat_map { |s| s.split(',') }
#=>["Screw:1", "Bound:5", "Hing:3", "Bound:5", "Screw:3", "Hing:1",
# "Bound:2", "Screw:2"]
See the form of Hash::new that takes an argument and no block. The argument is called the default value, which is here zero. If h has been defined h = Hash.new(0), and h does not have a key k, h[k] returns the default value (and does not modify the hash). h[k] += v.to_i expands to
h[k] = h[k] + v.to_i
so if h does not have a key k this becomes
h[k] = 0 + v.to_i
Alternatively, one could write the following.
arr.flat_map { |s| s.split(/:|,/) }
.each_slice(2)
.with_object(Hash.new(0)) { |(k,v),h| h[k] += v.to_i }
#=> {"Screw"=>6, "Bound"=>12, "Hing"=>4}
Steps 1 and 2
a = arr.flat_map { |s| s.split(/:|,/) }
#=> ["Screw", "1", "Bound", "5", "Hing", "3", "Bound", "5",
# "Screw", "3", "Hing", "1", "Bound", "2", "Screw", "2"]
e = a.each_slice(2)
#=> #<Enumerator: ["Screw", "1", "Bound", "5", "Hing", "3",
# "Bound", "5", "Screw", "3", "Hing", "1",
# "Bound", "2", "Screw", "2"]:each_slice(2)>
The elements generated by the enumerator e can be seen as follows:
e.entries
#=> [["Screw", "1"], ["Bound", "5"], ["Hing", "3"], ["Bound", "5"],
# ["Screw", "3"], ["Hing", "1"], ["Bound", "2"], ["Screw", "2"]]
A good way would be too loop through all of the entries and update the hash depending on the entries that get found.
The following will do it for you.
str = "Screw:1,Bound:5,Hing:3"
output = Hash.new(0)
str.split(",").each do |entry|
key = entry.split(":")
output[key[0]] += key[1].to_i
end
Just modify it so that it handles multiple strings correctly, depending on how they are fed to you in the system.
Looks like the data is CSV so I'd opt to use a CSV parser to avoid possible encoding issues
require 'csv'
def parse input
Hash[CSV.parse_line(input).map { |pair| pair.split(":") }]
end
I have a hash:
hash = {"a_1_a" => "1", "a_1_b" => "2", "a_1_c" => "3", "a_2_a" => "3",
"a_2_b" => "4", "a_2_c" => "4"}
What's the best way to get the following sub-hashes:
[{"a_1_a" => "1", "a_1_b" => "2", "a_1_c" => "3"},
{"a_2_a" => "3", "a_2_b" => "4", "a_2_c" => "4"}]
I want them grouped by the key, based on the regexp /^a_(\d+)/. I'll have 50+ key/value pairs in the original hash, so something dynamic would work best, if anyone has any suggestions.
If you're only concerned about the middle component you can use group_by to get you most of the way there:
hash.group_by do |k,v|
k.split('_')[1]
end.values.map do |list|
Hash[list]
end
# => [{"a_1_a"=>"1", "a_1_b"=>"2", "a_1_c"=>"3"}, {"a_2_a"=>"3", "a_2_b"=>"4", "a_2_c"=>"4"}]
The final step is extracting the grouped lists and combining those back into the required hashes.
Code
def partition_hash(hash)
hash.each_with_object({}) do |(k,v), h|
key = k[/(?<=_).+(?=_)/]
h[key] = (h[key] || {}).merge(k=>v)
end.values
end
Example
hash = {"a_1_a"=>"1", "a_1_b"=>"2", "a_1_c"=>"3", "a_2_a"=>"3", "a_2_b"=>"4", "a_2_c"=>"4"}
partition_hash(hash)
#=> [{"a_1_a"=>"1", "a_1_b"=>"2", "a_1_c"=>"3"},
# {"a_2_a"=>"3", "a_2_b"=>"4", "a_2_c"=>"4"}]
Explanation
The steps are as follows.
enum = hash.each_with_object({})
#=> #<Enumerator: {"a_1_a"=>"1", "a_1_b"=>"2", "a_1_c"=>"3", "a_2_a"=>"3",
# "a_2_b"=>"4", "a_2_c"=>"4"}:each_with_object({})>
The first element of this enumerator is generated and passed to the block, and the block variables are computed using parallel assignment.
(k,v), h = enum.next
#=> [["a_1_a", "1"], {}]
k #=> "a_1_a"
v #=> "1"
h #=> {}
and the block calculation is performed.
key = k[/(?<=_).+(?=_)/]
#=> "1"
h[key] = (h[key] || {}).merge(k=>v)
#=> h["1"] = (h["1"] || {}).merge("a_1_a"=>"1")
#=> h["1"] = (nil || {}).merge("a_1_a"=>"1")
#=> h["1"] = {}.merge("a_1_a"=>"1")
#=> h["1"] = {"a_1_a"=>"1"}
so now
h #=> {"1"=>{"a_1_a"=>"1"}}
The next value of enum is now generated and passed to the block, and the following calculations are performed.
(k,v), h = enum.next
#=> [["a_1_b", "2"], {"1"=>{"a_1_a"=>"1"}}]
k #=> "a_1_b"
v #=> "2"
h #=> {"1"=>{"a_1_a"=>"1"}}
key = k[/(?<=_).+(?=_)/]
#=> "1"
h[key] = (h[key] || {}).merge(k=>v)
#=> h["1"] = (h["1"] || {}).merge("a_1_b"=>"2")
#=> h["1"] = ({"a_1_a"=>"1"}} || {}).merge("a_1_b"=>"2")
#=> h["1"] = {"a_1_a"=>"1"}}.merge("a_1_b"=>"2")
#=> h["1"] = {"a_1_a"=>"1", "a_1_b"=>"2"}
After the remaining four elements of enum have been passed to the block the following has is returned.
h #=> {"1"=>{"a_1_a"=>"1", "a_1_b"=>"2", "a_1_c"=>"3"},
# "2"=>{"a_2_a"=>"3", "a_2_b"=>"4", "a_2_c"=>"4"}}
The final step is simply to extract the values.
h.values
#=> [{"a_1_a"=>"1", "a_1_b"=>"2", "a_1_c"=>"3"},
# {"a_2_a"=>"3", "a_2_b"=>"4", "a_2_c"=>"4"}]
I have to make a multiplication of n arrays.
Example :
input = ["a", "b", "c"] * ["1", "2"] * ["&", "(", "$"]
output = ["a1&", "a1(", "a1$", "a2&", "a2(", "a2$", "b1&", "b1(", "b1$", "b2&", "b2(", "b2$, "c1&, "c1(, "c1$, "c2&", "c2(", "c2$"]
I have created an algorithm to do that, it works good.
# input
entries = ["$var1$", "$var2$", "$var3$"]
data = [["a", "b", "c"], ["1", "2"], ["&", "(", "$"]]
num_combinaison = 1
data.each { |item| num_combinaison = num_combinaison * item.length }
result = []
for i in 1..num_combinaison do
result.push entries.join()
end
num_repetition = num_combinaison
data.each_index do |index|
item = Array.new(data[index])
num_repetition = num_repetition / item.length
for i in 1..num_combinaison do
result[i-1].gsub!(entries[index], item[0])
if i % num_repetition == 0
item.shift
item = Array.new(data[index]) if item.length == 0
end
end
end
I'm sure there is a best way to do that, but I don't find it. I have tried to use product or flatten function without success.
Somebody see a best solution ?
Thanks for your help.
Eric
class Array
def * other; product(other).map(&:join) end
end
["a", "b", "c"] * ["1", "2"] * ["&", "(", "$"]
# =>
# ["a1&", "a1(", "a1$", "a2&", "a2(", "a2$", "b1&", "b1(", "b1$", "b2&",
# "b2(", "b2$", "c1&", "c1(", "c1$", "c2&", "c2(", "c2$"]
The best algorithm you can use is implemented by the Array#product method:
data = [["a", "b", "c"], ["1", "2"], ["&", "(", "$"]]
data.first.product(*entries.drop(1)).map(&:join)
# => ["a1&", "a1(", "a1$", "a2&", "a2(", "a2$", ...
Update
A safer alternative, my first solution raises a NoMethodError if data is emtpy:
data.reduce { |result, ary| result.product(ary).map(&:join) }
# => ["a1&", "a1(", "a1$", "a2&", "a2(", "a2$", ...
[].reduce { |r, a| r.product(a).map(&:join) }
# => nil
I've got a hash of the format:
{key1 => [a, b, c], key2 => [d, e, f]}
and I want to end up with:
{ a => key1, b => key1, c => key1, d => key2 ... }
What's the easiest way of achieving this?
I'm using Ruby on Rails.
UPDATE
OK I managed to extract the real object from the server log, it is being pushed via AJAX.
Parameters: {"status"=>{"1"=>["1", "14"], "2"=>["7", "12", "8", "13"]}}
hash = {:key1 => ["a", "b", "c"], :key2 => ["d", "e", "f"]}
first variant
hash.map{|k, v| v.map{|f| {f => k}}}.flatten
#=> [{"a"=>:key1}, {"b"=>:key1}, {"c"=>:key1}, {"d"=>:key2}, {"e"=>:key2}, {"f"=>:key2}]
or
hash.inject({}){|h, (k,v)| v.map{|f| h[f] = k}; h}
#=> {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2}
UPD
ok, your hash is:
hash = {"status"=>{"1"=>["1", "14"], "2"=>["7", "12", "8", "13"]}}
hash["status"].inject({}){|h, (k,v)| v.map{|f| h[f] = k}; h}
#=> {"12"=>"2", "7"=>"2", "13"=>"2", "8"=>"2", "14"=>"1", "1"=>"1"}
Lots of other good answers. Just wanted to toss this one in too for Ruby 2.0 and 1.9.3:
hash = {apple: [1, 14], orange: [7, 12, 8, 13]}
Hash[hash.flat_map{ |k, v| v.map{ |i| [i, k] } }]
# => {1=>:apple, 14=>:apple, 7=>:orange, 12=>:orange, 8=>:orange, 13=>:orange}
This is leveraging: Hash::[] and Enumerable#flat_map
Also in these new versions there is Enumerable::each_with_object which is very similar to Enumerable::inject/Enumerable::reduce:
hash.each_with_object(Hash.new){ |(k, v), inverse|
v.each{ |e| inverse[e] = k }
}
Performing a quick benchmark (Ruby 2.0.0p0; 2012 Macbook Air) using an original hash with 100 keys, each with 100 distinct values:
Hash::[] w/ Enumerable#flat_map
155.7 (±9.0%) i/s - 780 in 5.066286s
Enumerable#each_with_object w/ Enumerable#each
199.7 (±21.0%) i/s - 940 in 5.068926s
Shows that the each_with_object variant is faster for that data set.
Ok, let's guess. You say you have an array but I agree with Benoit that what you probably have is a hash. A functional approach:
h = {:key1 => ["a", "b", "c"], :key2 => ["d", "e", "f"]}
h.map { |k, vs| Hash[vs.map { |v| [v, k] }] }.inject(:merge)
#=> {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2}
Also:
h.map { |k, vs| Hash[vs.product([k])] }.inject(:merge)
#=> {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2}
In the case where a value corresponds to more than one key, like "c" in this example...
{ :key1 => ["a", "b", "c"], :key2 => ["c", "d", "e"]}
...some of the other answers will not give the expected result. We will need the reversed hash to store the keys in arrays, like so:
{ "a" => [:key1], "b" => [:key1], "c" => [:key1, :key2], "d" => [:key2], "e" => [:key2] }
This should do the trick:
reverse = {}
hash.each{ |k,vs|
vs.each{ |v|
reverse[v] ||= []
reverse[v] << k
}
}
This was my use case, and I would have defined my problem much the same way as the OP (in fact, a search for a similar phrase got me here), so I suspect this answer may help other searchers.
If you're looking to reverse a hash formatted like this, the following may help you:
a = {:key1 => ["a", "b", "c"], :key2 => ["d", "e", "f"]}
a.inject({}) do |memo, (key, values)|
values.each {|value| memo[value] = key }
memo
end
this returns:
{"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2}
new_hash={}
hash = {"key1" => ['a', 'b', 'c'], "key2" => ['d','e','f']}
hash.each_pair{|key, val|val.each{|v| new_hash[v] = key }}
This gives
new_hash # {"a"=>"key1", "b"=>"key1", "c"=>"key1", "d"=>"key2", "e"=>"key2", "f"=>"key2"}
If you want to correctly deal with duplicate values, then you should use the Hash#inverse
from Facets of Ruby
Hash#inverse preserves duplicate values,
e.g. it ensures that hash.inverse.inverse == hash
either:
use Hash#inverse from here: http://www.unixgods.org/Ruby/invert_hash.html
use Hash#inverse from FacetsOfRuby library 'facets'
usage like this:
require 'facets'
h = {:key1 => [:a, :b, :c], :key2 => [:d, :e, :f]}
=> {:key1=>[:a, :b, :c], :key2=>[:d, :e, :f]}
h.inverse
=> {:a=>:key1, :b=>:key1, :c=>:key1, :d=>:key2, :e=>:key2, :f=>:key2}
The code looks like this:
# this doesn't looks quite as elegant as the other solutions here,
# but if you call inverse twice, it will preserve the elements of the original hash
# true inversion of Ruby Hash / preserves all elements in original hash
# e.g. hash.inverse.inverse ~ h
class Hash
def inverse
i = Hash.new
self.each_pair{ |k,v|
if (v.class == Array)
v.each{ |x|
i[x] = i.has_key?(x) ? [k,i[x]].flatten : k
}
else
i[v] = i.has_key?(v) ? [k,i[v]].flatten : k
end
}
return i
end
end
h = {:key1 => [:a, :b, :c], :key2 => [:d, :e, :f]}
=> {:key1=>[:a, :b, :c], :key2=>[:d, :e, :f]}
h.inverse
=> {:a=>:key1, :b=>:key1, :c=>:key1, :d=>:key2, :e=>:key2, :f=>:key2}
One way to achieve what you're looking for:
arr = [{["k1"] => ["a", "b", "c"]}, {["k2"] => ["d", "e", "f"]}]
results_arr = []
arr.each do |hsh|
hsh.values.flatten.each do |val|
results_arr << { [val] => hsh.keys.first }···
end
end
Result: [{["a"]=>["k1"]}, {["b"]=>["k1"]}, {["c"]=>["k1"]}, {["d"]=>["k2"]}, {["e"]=>["k2"]}, {["f"]=>["k2"]}]