I have two hashes like below,
h1 = {"a" => 1, "b" => 2, "c" => 3}
h2 = {"a" => 2, "b" => 2, "d" => 3}
I want to iterate over hash1 and hash2 and find matching keys and their values and print it on console.
Example here it should return output "b" => 2 .its not working with below code,
h1.each do |key1, value1|
h2.each do |key2, value2|
if ((h2.include? key1) && (h2.include? value1))
puts "matching h2 key #{h2[key2]}and h1 key #{h1[key1]}"
else
puts " don not match h2 key #{h2[key2]}and h1 key #{h1[key1]}"
end
end
end
I am from basically C++ and Java background and its very easy to do using for loops and iterators, but using Ruby, it is very difficult.
h1.merge(h2) { |k,o,n| puts "#{k}=>#{o}" if o == n }
"b" => 2
This uses the form of Hash#merge that employs a bock to determine the values of keys that are present in both hashes being merged. See the doc for details.
The second loop is not required.
h1.each do |key1, value1|
if (h2.include? key1) and (h2[key1] == value1)
puts "Match #{key1} with value #{value1}"
else
puts "#{key1} does not match"
end
end
You might write something like
h1.each do |k,v|
if h2[k] == v
puts "matched key = #{k} and value = #{v}"
else
puts "NOT matched key = #{k} and value = #{v}"
end
end
Output
NOT matched key = a and value = 1
matched key = b and value = 2
NOT matched key = c and value = 3
If the result is the main objective, select can work too:
h1.select{|k,v| h2[k] == v }
Related
I'm working on this problem:
Write a method retrieve_values that takes in two hashes and a key. The method should return an array containing the values from the two hashes that correspond with the given key.
def retrieve_values(hash1, hash2, key)
end
dog1 = {"name"=>"Fido", "color"=>"brown"}
dog2 = {"name"=>"Spot", "color"=> "white"}
print retrieve_values(dog1, dog2, "name") #=> ["Fido", "Spot"]
puts
print retrieve_values(dog1, dog2, "color") #=> ["brown", "white"]
puts
I came up with a working solution:
def retrieve_values(hash1, hash2, key)
arr = []
hash1.each { |key| } && hash2.each { |key| }
if key == "name"
arr << hash1["name"] && arr << hash2["name"]
elsif key == "color"
arr << hash1["color"] && arr << hash2["color"]
end
return arr
end
I then looked at the 'official' solution:
def retrieve_values(hash1, hash2, key)
val1 = hash1[key]
val2 = hash2[key]
return [val1, val2]
end
What is wrong with my code? Or is it an acceptable "different" approach?
Line with hash1.each { |key| } && hash2.each { |key| } just does nothing it is not needed even in your solution.
This part a bit difficult to read arr << hash1["name"] && arr << hash2["name"]. It mutates the array two times in one line, this kind of style could lead to bugs.
Also, your code sticks only to two keys name and color:
dog1 = {"name"=>"Fido", "color"=>"brown", "age" => 1}
dog2 = {"name"=>"Spot", "color"=> "white", "age" => 2}
> retrieve_values(dog1, dog2, "age")
=> []
The official solution will return [1, 2].
You don't need here to explicitly use return keyword, any block of code returns the last evaluated expression. But it is a matter of style guide.
It is possible to simplify even the official solution:
def retrieve_values(hash1, hash2, key)
[hash1[key], hash2[key]]
end
My apologies for the potentially stupid question, I'm an absolute beginner to Ruby and code in general.
I have set up a hash with some predetermined values. I want to ask the user for input, if that input matches an existing key, I want the corresponding value to be updated (+ 1, in this case). Then I want to print all the current up-to-date values.
hash = {"apple": 6, "banana": 2, "carrot": 3}
order = gets.chomp.downcase
hash.each do |key, value|
if key.to_s == order
value += 1
puts "Your order includes: #{value} #{key}."
end
end
My problem is that I only know how to print a single key value pair.
E.g. if the user inputs "apple", I'd like the output to say "Your order includes: 7 apple, 2 banana, 3 carrot."
hash = {apple: 6, banana: 2, carrot: 3}
order = gets.chomp.downcase.to_sym
hash[order] = hash.fetch(order, 0) + 1
puts "Your order includes: " + hash.map { |k, v| "#{v} #{k}" }.join(", ")
Some notes:
your hash initialization hash = {"apple": 6, "banana": 2, "carrot": 3}. the keys of your hash seem strings, but if you use that syntax with the colon, they become symbols. So, you have two choice. this syntax:
hash = {"apple" => 6, "banana" => 2, "carrot" => 3}
or you can use symbols as I did and convert the user input in a symbol
what's really cool about hash is that you don't need to iterate through the elements to find what you're looking for. There's a mapping between keys and values, so it's easy find and update a value
in the third row, I'm dealing with the fact that the key could not be in the hash, I used fetch to have 0 in that case. then, I increment and I assign back to that key
The question does not specify if you want to mutate the initial hash, so I suppose you do. Then the following will do.
hash = Hash.new(0).merge(apple: 6, banana: 2, carrot: 3)
hash[gets.chomp.downcase.to_sym] += 1
puts "Your order includes: " <<
hash.map { |k, v| [v, k].join(' ') }.join(', ')
or:
puts hash.reduce("Your order includes: ") { |acc, (k, v)|
acc << "#{v} #{k}, "
}[0..-3]
Consider to initialize the hash providing a default value (Hash#default)
basket = {'apple' => 6, 'banana' => 2, 'carrot' => 3}
basket.default = 0 # <-- set default 0 for start counting new keys
Define a method to present the data:
def show(basket)
puts "Your order includes:"
basket.each{ |key, value| puts "#{value}: #{key}" }
end
Capture user input in a loop (explanation in comments):
loop do
puts "Place your order:"
order = gets.downcase.chomp # <-- format the input
break if order == '0' # <-- breaks the input loop if this contition is matched
next unless basket.has_key? order # <-- skip to next loop no matches with hash keys or remove this line if you want to count also non initialised keys
basket[order] += 1 # <-- increment by one the key
show(basket) # <-- call the metod to show the basket
end
show(basket)
If I have a string like this
str =<<END
7312357006,1.121
3214058234,3456
7312357006,1234
1324958723,232.1
3214058234,43.2
3214173443,234.1
6134513494,23.2
7312357006,11.1
END
If a number in the first value shows up again, I want to add their second values together. So the final string would look like this
7312357006,1246.221
3214058234,3499.2
1324958723,232.1
3214173443,234.1
6134513494,23.2
If the final output is an array that's fine too.
There are lots of ways to do this in Ruby. One particularly terse way is to use String#scan:
str = <<END
7312357006,1.121
3214058234,3456
7312357006,1234
1324958723,232.1
3214058234,43.2
3214173443,234.1
6134513494,23.2
7312357006,11.1
END
data = Hash.new(0)
str.scan(/(\d+),([\d.]+)/) {|k,v| data[k] += v.to_f }
p data
# => { "7312357006" => 1246.221,
# "3214058234" => 3499.2,
# "1324958723" => 232.1,
# "3214173443" => 234.1,
# "6134513494" => 23.2 }
This uses the regular expression /(\d+),([\d.]+)/ to extract the two values from each line. The block is called with each pair as arguments, which are then merged into the hash.
This could also be written as a single expression using each_with_object:
data = str.scan(/(\d+),([\d.]+)/)
.each_with_object(Hash.new(0)) {|(k,v), hsh| hsh[k] += v.to_f }
# => (same as above)
There are likewise many ways to print the result, but here are a couple I like:
puts data.map {|kv| kv.join(",") }.join("\n")
# => 7312357006,1246.221
# 3214058234,3499.2
# 1324958723,232.1
# 3214173443,234.1
# 6134513494,23.2
# or:
puts data.map {|k,v| "#{k},#{v}\n" }.join
# => (same as above)
You can see all of these in action on repl.it.
Edit: Although I don't recommend either of these for the sake of readability, here's more just for kicks (requires Ruby 2.4+):
data = str.lines.group_by {|s| s.slice!(/(\d+),/); $1 }
.transform_values {|a| a.sum(&:to_f) }
...or, to going straight to a string:
puts str.lines.group_by {|s| s.slice!(/(\d+),/); $1 }
.map {|k,vs| "#{k},#{vs.sum(&:to_f)}\n" }.join
Since repl.it is stuck on Ruby 2.3: Try it online!
You could achieve this using each_with_object, as below:
str = "7312357006,1.121
3214058234,3456
7312357006,1234
1324958723,232.1
3214058234,43.2
3214173443,234.1
6134513494,23.2
7312357006,11.1"
# convert the string into nested pairs of floats
# to briefly summarise the steps: split entries by newline, strip whitespace, split by comma, convert to floats
arr = str.split("\n").map(&:strip).map { |el| el.split(",").map(&:to_f) }
result = arr.each_with_object(Hash.new(0)) do |el, hash|
hash[el.first] += el.last
end
# => {7312357006.0=>1246.221, 3214058234.0=>3499.2, 1324958723.0=>232.1, 3214173443.0=>234.1, 6134513494.0=>23.2}
# You can then call `to_a` on result if you want:
result.to_a
# => [[7312357006.0, 1246.221], [3214058234.0, 3499.2], [1324958723.0, 232.1], [3214173443.0, 234.1], [6134513494.0, 23.2]]
each_with_object iterates through each pair of data, providing them with access to an accumulator (in this the hash). By following this approach, we can add each entry to the hash, and add together the totals if they appear more than once.
Hope that helps - let me know if you've any questions.
def combine(str)
str.each_line.with_object(Hash.new(0)) do |s,h|
k,v = s.split(',')
h.update(k=>v.to_f) { |k,o,n| o+n }
end.reduce('') { |s,kv_pair| s << "%s,%g\n" % kv_pair }
end
puts combine str
7312357006,1246.22
3214058234,3499.2
1324958723,232.1
3214173443,234.1
6134513494,23.2
Notes:
using String#each_line is preferable to str.split("\n") as the former returns an enumerator whereas the latter returns a temporary array. Each element generated by the enumerator is line of str that (unlike the elements of str.split("\n")) ends with a newline character, but that is of no concern.
see Hash::new, specifically when a default value (here 0) is used. If a hash has been defined h = Hash.new(0) and h does not have a key k, h[k] returns the default value, zero (h is not changed). When Ruby encounters the expression h[k] += 1, the first thing she does is expand it to h[k] = h[k] + 1. If h has been defined with a default value of zero, and h does not have a key k, h[k] on the right of the equality (syntactic sugar1 for h.[](k)) returns zero.
see Hash#update (aka merge!). h.update(k=>v.to_f) is syntactic sugar for h.update({ k=>v.to_f })
see Kernel#sprint for explanations of the formatting directives %s and %g.
the receiver for the expression reduce('') { |s,kv_pair| s << "%s,%g\n" % kv_pair } (in the penultimate line), is the following hash.
{"7312357006"=>1246.221, "3214058234"=>3499.2, "1324958723"=>232.1,
"3214173443"=>234.1, "6134513494"=>23.2}
1 Syntactic sugar is a shortcut allowed by Ruby.
Implemented this solution as hash was giving me issues:
d = []
s.split("\n").each do |line|
x = 0
q = 0
dup = false
line.split(",").each do |data|
if x == 0 and d.include? data then dup = true ; q = d.index(data) elsif x == 0 then d << data end
if x == 1 and dup == false then d << data end
if x == 1 and dup == true then d[q+1] = "#{'%.2f' % (d[q+1].to_f + data.to_f).to_s}" end
if x == 2 and dup == false then d << data end
x += 1
end
end
x = 0
s = ""
d.each do |val|
if x == 0 then s << "#{val}," end
if x == 1 then s << "#{val}\n ; x = 0" end
x += 1
end
puts(s)
I want to take a hash with nested hashes and arrays and flatten it out into a single hash with unique values. I keep trying to approach this from different angles, but then I make it way more complex than it needs to be and get myself lost in what's happening.
Example Source Hash:
{
"Name" => "Kim Kones",
"License Number" => "54321",
"Details" => {
"Name" => "Kones, Kim",
"Licenses" => [
{
"License Type" => "PT",
"License Number" => "54321"
},
{
"License Type" => "Temp",
"License Number" => "T123"
},
{
"License Type" => "AP",
"License Number" => "A666",
"Expiration Date" => "12/31/2020"
}
]
}
}
Example Desired Hash:
{
"Name" => "Kim Kones",
"License Number" => "54321",
"Details_Name" => "Kones, Kim",
"Details_Licenses_1_License Type" => "PT",
"Details_Licenses_1_License Number" => "54321",
"Details_Licenses_2_License Type" => "Temp",
"Details_Licenses_2_License Number" => "T123",
"Details_Licenses_3_License Type" => "AP",
"Details_Licenses_3_License Number" => "A666",
"Details_Licenses_3_Expiration Date" => "12/31/2020"
}
For what it's worth, here's my most recent attempt before giving up.
def flattify(hashy)
temp = {}
hashy.each do |key, val|
if val.is_a? String
temp["#{key}"] = val
elsif val.is_a? Hash
temp.merge(rename val, key, "")
elsif val.is_a? Array
temp["#{key}"] = enumerate val, key
else
end
print "=> #{temp}\n"
end
return temp
end
def rename (hashy, str, n)
temp = {}
hashy.each do |key, val|
if val.is_a? String
temp["#{key}#{n}"] = val
elsif val.is_a? Hash
val.each do |k, v|
temp["#{key}_#{k}#{n}"] = v
end
elsif val.is_a? Array
temp["#{key}"] = enumerate val, key
else
end
end
return flattify temp
end
def enumerate (ary, str)
temp = {}
i = 1
ary.each do |x|
temp["#{str}#{i}"] = x
i += 1
end
return flattify temp
end
Interesting question!
Theory
Here's a recursive method to parse your data.
It keeps track of which keys and indices it has found.
It appends them in a tmp array.
Once a leaf object has been found, it gets written in a hash as value, with a joined tmp as key.
This small hash then gets recursively merged back to the main hash.
Code
def recursive_parsing(object, tmp = [])
case object
when Array
object.each.with_index(1).with_object({}) do |(element, i), result|
result.merge! recursive_parsing(element, tmp + [i])
end
when Hash
object.each_with_object({}) do |(key, value), result|
result.merge! recursive_parsing(value, tmp + [key])
end
else
{ tmp.join('_') => object }
end
end
As an example:
require 'pp'
pp recursive_parsing(data)
# {"Name"=>"Kim Kones",
# "License Number"=>"54321",
# "Details_Name"=>"Kones, Kim",
# "Details_Licenses_1_License Type"=>"PT",
# "Details_Licenses_1_License Number"=>"54321",
# "Details_Licenses_2_License Type"=>"Temp",
# "Details_Licenses_2_License Number"=>"T123",
# "Details_Licenses_3_License Type"=>"AP",
# "Details_Licenses_3_License Number"=>"A666",
# "Details_Licenses_3_Expiration Date"=>"12/31/2020"}
Debugging
Here's a modified version with old-school debugging. It might help you understand what's going on:
def recursive_parsing(object, tmp = [], indent="")
puts "#{indent}Parsing #{object.inspect}, with tmp=#{tmp.inspect}"
result = case object
when Array
puts "#{indent} It's an array! Let's parse every element:"
object.each_with_object({}).with_index(1) do |(element, result), i|
result.merge! recursive_parsing(element, tmp + [i], indent + " ")
end
when Hash
puts "#{indent} It's a hash! Let's parse every key,value pair:"
object.each_with_object({}) do |(key, value), result|
result.merge! recursive_parsing(value, tmp + [key], indent + " ")
end
else
puts "#{indent} It's a leaf! Let's return a hash"
{ tmp.join('_') => object }
end
puts "#{indent} Returning #{result.inspect}\n"
result
end
When called with recursive_parsing([{a: 'foo', b: 'bar'}, {c: 'baz'}]), it displays:
Parsing [{:a=>"foo", :b=>"bar"}, {:c=>"baz"}], with tmp=[]
It's an array! Let's parse every element:
Parsing {:a=>"foo", :b=>"bar"}, with tmp=[1]
It's a hash! Let's parse every key,value pair:
Parsing "foo", with tmp=[1, :a]
It's a leaf! Let's return a hash
Returning {"1_a"=>"foo"}
Parsing "bar", with tmp=[1, :b]
It's a leaf! Let's return a hash
Returning {"1_b"=>"bar"}
Returning {"1_a"=>"foo", "1_b"=>"bar"}
Parsing {:c=>"baz"}, with tmp=[2]
It's a hash! Let's parse every key,value pair:
Parsing "baz", with tmp=[2, :c]
It's a leaf! Let's return a hash
Returning {"2_c"=>"baz"}
Returning {"2_c"=>"baz"}
Returning {"1_a"=>"foo", "1_b"=>"bar", "2_c"=>"baz"}
Unlike the others, I have no love for each_with_object :-). But I do like passing a single result hash around so I don't have to merge and remerge hashes over and over again.
def flattify(value, result = {}, path = [])
case value
when Array
value.each.with_index(1) do |v, i|
flattify(v, result, path + [i])
end
when Hash
value.each do |k, v|
flattify(v, result, path + [k])
end
else
result[path.join("_")] = value
end
result
end
(Some details adopted from Eric, see comments)
Non-recursive approach, using BFS with an array as a queue. I keep the key-value pairs where the value isn't an array/hash, and push array/hash contents to the queue (with combined keys). Turning arrays into hashes (["a", "b"] ↦ {1=>"a", 2=>"b"}) as that felt neat.
def flattify(hash)
(q = hash.to_a).select { |key, value|
value = (1..value.size).zip(value).to_h if value.is_a? Array
!value.is_a?(Hash) || !value.each { |k, v| q << ["#{key}_#{k}", v] }
}.to_h
end
One thing I like about it is the nice combination of keys as "#{key}_#{k}". In my other solution, I could've also used a string path = '' and extended that with path + "_" + k, but that would've caused a leading underscore that I'd have to avoid or trim with extra code.
I need to compare 2 hashes in ruby where one of the hash contains the numeric values in quoted string, which makes that a string. Consider the following 2 hashes:
hash1 = {"A"=>"0", "B"=>"1", "SVHTID"=>"VH", "D"=>"0", "E"=>"19930730", "F"=>"TEST - DEPOSIT", "G"=>"2.25000000"}
hash2 = {"a"=>"0", "b"=>1, "c"=>"VH", "d"=>0,"e"=>19930730, "f"=>"TEST - DEPOSIT", "g"=>2.25}
Now the code i have written so far is as follows:
hash2 = Hash[hash2.map {|key, value| [key.upcase, value] }]
hash1.each{|k,v| hash1[k] = hash1[k].to_i if hash1[k].match(/-?\d+(?:\.\d+)?/)}
hash1.keys.select { |key| hash1[key] != hash2[key] }.each { |key|
puts "expected #{key} => #{hash1[key]}, but found #{key} => #{hash2[key]}"
}
what is does is that it also converts the float value to integer and the original value is lost.
What i want is that when the above 2 hashes are compared the output should contain only G as mismatched type and following should be printed:
Expected: G=>2.25000000 but found G=>2.25
# normalization
h1, h2 = [hash1, hash2].map do |h|
h.map do |k, v|
[k.to_s.downcase, v.to_s.to_i.to_s == v.to_s ? v.to_i : v]
end.to_h
end
Now we are ready to compare:
h1.reject { |k, v| h2[k].nil? || h2[k] == v }
#⇒ { "g" => "2.25000000" }
This might be printed, formatted etc as you want.