To find the integer (Fixnum) values in ruby array - ruby

I have an array [1, 2, "3", "4", "1a", "abc", "a"] with
pure integers (1, 2),
string formatted integers ("1", "2"),
strings ("a", "b"), and
mixed string numbers ("1a", "2s").
From this, I need to pick up only the integers (including string formatted) 1, 2, "3", "4".
First I tried with to_i:
arr = [1, 2, "3", "4", "1a", "abc", "a"]
arr.map {|x| x.to_i}
# => [1, 2, 3, 4, 1, 0, 0]
but this one converts "1a" to 1, which I don't expect.
Then I tried Integer(item):
arr.map {|x| Integer(x) } # and it turned out to be
# => ArgumentError: invalid value for Integer(): "1a"
Now I am out of straight conversion options here. Finally, I decided to do this way, which converts the value to_i and to_s. So "1" == "1".to_i.to_s is an integer, but not "1a" == "1a".to_i.to_s and "a" == "a".to_i.to_s
arr = arr.map do |x|
if (x == x.to_i.to_s)
x.to_i
else
x
end
end
and
ids, names= arr.partition { |item| item.kind_of? Fixnum }
Now I got the arrays of integers and strings. Is there a simple way to do this?


Similar solution as provided by #maerics, but a bit slimmer:
arr.map {|x| Integer(x) rescue nil }.compact


class Array
def to_i
self.map {|x| begin; Integer(x); rescue; nil; end}.compact
end
end
arr = [1, 2, "3", "4", "1a", "abc", "a"]
arr.to_i # => [1, 2, 3, 4]


something like this:
a = [1,2,"3","4","1a","abc","a"]
irb(main):005:0> a.find_all { |e| e.to_s =~ /^\d+$/ }.map(&:to_i)
=> [1, 2, 3, 4]


Hey, thanks awakening my ruby. Here is my go at this problem:
arr=[1,2,"3","4","1a","abc","a"]
arr.map {|i| i.to_s}.select {|s| s =~ /^[0-9]+$/}.map {|i| i.to_i}
//=> [1, 2, 3, 4]


I noticed most of the answer so far changes the value of "3" and "4" to actual integers.
>> array=[1, 2, "3", "4", "1a", "abc", "a", "a13344a" , 10001, 3321]
=> [1, 2, "3", "4", "1a", "abc", "a", "a13344a", 10001, 3321]
>> array.reject{|x| x.to_s[/[^0-9]/] }
=> [1, 2, "3", "4", 10001, 3321]
#OP, I have not tested my solution exhaustively, but so far it seems to work (of course its done according to provided sample ), so please test thoroughly yourself.


How about this?
[1,2,"3","4","1a","abc","a"].select{|x| x.to_i.to_s == x.to_s}
# => [1, 2, "3", "4"]


Looks pretty simple
arr.select{ |b| b.to_s =~ /\d+$/ }
# or
arr.select{ |b| b.to_s[/\d+$/] }
#=> [1, 2, "3", "4"]

Related

How to detect duplicate keys in hash and add prefix to the duplicate?

I have two arrays and I am creating a key-value-pair using hash in Ruby. How can I detect a duplicate key when zipping two arrays into key-value-pair and adding a prefix like "A-" in front of the key name for the duplicates?
I am using .zip to merge two arrays and making one a key and other one a value
[0] = "David"
[1] = "John"
[2] = "Alex"
[3] = "Sam"
[4] = "Caleb"
[5] = "David"
[6] = "John"
[7] = "Alex"
[8] = "Sam"
[0] = "1"
[1] = "2"
[2] = "3"
[3] = "4"
[4] = "5"
[5] = "6"
[6] = "7"
[7] = "8"
[8] = "9"
name_number_key_value_pair_hash = first_names.zip(numbers).to_h
puts(name_number_key_value_pair_hash)
Expected:
{"David"=>"1", "John"=>"2", "Alex"=>"3", "Sam"=>"4", "Caleb"=>"5", "A-David"=>"6", "A-John"=>"7", "A-Alex"=>"8", "A-Sam"=>"9"}
Actual:
{"David"=>"6", "John"=>"7", "Alex"=>"8", "Sam"=>"9", "Caleb"=>"5"}

It seems straight forward Have attached code snippet
names = %w[David John Alex Sam Caleb David John Alex Sam]
numbers = %w[1 2 3 4 5 6 7 8 9]
key_pair = {}
names.each_with_index do |name, index|
name = "A-#{name}" if key_pair[name]
key_pair[name] = numbers[index]
end
It generates the expected output:
{"David"=>"1", "John"=>"2", "Alex"=>"3", "Sam"=>"4", "Caleb"=>"5", "A-David"=>"6", "A-John"=>"7", "A-Alex"=>"8", "A-Sam"=>"9"}

You basically just need to keep track of the state of the hash as you build it and, when you find a conflict, create a new key instead. This captures the general approach:
def hash_with_prefixes(a, b, prefixes)
kv_pairs = a.zip(b)
prefixes = prefixes.to_enum
result_hash = {}
kv_pairs.each do |initial_key, value|
final_key = initial_key
while result_hash.include? final_key
final_key = "#{pfx.next}-#{initial_key}"
end
prefixes.rewind
result_hash[final_key] = value
end
result_hash
rescue StopIteration
fail "Insufficient prefixes to provide unique keys for input lists."
end
At the slight expense of clarity, you can also write it in a rather shorter form:
def hash_with_prefixes(a, b, prefixes)
pi = Hash[a.map {|k| [k, prefixes.lazy.map {|p| "#{p}-#{k}"}]}]
a.zip(b).inject({}) {|h, kv| h[h.include?(kv[0]) ? pi[kv[0]].next : kv[0]] = kv[1]; h}
rescue StopIteration
fail "Insufficient prefixes to provide unique keys for input lists."
end
(Don't do this.)

This is really very simple.
names = ["John","John", "John", "David", "David", "Susan", "Sue"]
numbers = ["1", "2", "3", "4", "5", "6","7"]
def uniq_hash_keys(names, numbers)
hash = {}
names.each_with_index do |name,i|
if hash[name]
prefix = 'A1-'
key = prefix + name
while hash[key]
version = prefix.match(/A(\d+)-.*/i)[1].to_i
prefix = "A#{version + 1}-"
key = prefix + name
end
name = key
end
hash[name] = numbers[i]
end
hash
end
This function produces:
{
"John"=>"1",
"A1-John"=>"2",
"A2-John"=>"3",
"David"=>"4",
"A1-David"=>"5",
"Susan"=>"6",
"Sue"=>"7"
}
Notice that there are 3 Johns, this is why the while loop is inside the function.

This is one way to create the desired hash. Note that in arr1 "John" appears three times.
arr1 = ["David", "John", "Alex", "Sam", "Caleb",
"David", "John", "Alex", "John", "Sam"]
arr2 = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
prefixes =
arr1.each_with_object({}) do |s,h|
if h.key?(s)
prefix = "A-"
(h[s].size-1).times { prefix = prefix.next }
h[s] << prefix
else
h[s] = ['']
end
end
#=> {"David"=>["", "A-"], "John"=>["", "A-", "B-"],
# "Alex"=>["", "A-"], "Sam"=>["", "A-"],
# "Caleb"=>[""]}
arr1.map { |s| "#{prefixes[s].shift}#{s}" }.zip(arr2).to_h
#=> {"David"=>"1", "John"=>"2", "Alex"=>"3", "Sam"=>"4",
# "Caleb"=>"5", "A-David"=>"6", "A-John"=>"7",
# "A-Alex"=>"8", "B-John"=>"9", "A-Sam"=>"10"}
Note that "A-".next #=> "B-" and "Z-".next #=> "AA-".
Alternative data structure
You may wish to consider a different data structure, one that returns
{"David"=>["1", "6"], "John"=>["2", "7", "9"],
"Alex" =>["3", "8"], "Sam" =>["4", "10"], "Caleb"=>["5"]}
You could do that as follows.
arr1.each_with_index.
group_by(&:first).
transform_values { |v| arr2.values_at(*v.map(&:last)) }
#=> {"David"=>["1", "6"], "John"=>["2", "7", "9"],
# "Alex" =>["3", "8"], "Sam" =>["4", "10"],
# "Caleb"=>["5"]}
See Enumerable#each_with_index, Enumerable#group_by, Hash#transform_values1 and Array#values_at. v.map(*:last) is here the same as v.map { |arr| arr.last }.
The steps are as follows.
a = arr1.each_with_index
#=> #<Enumerator: ["David", "John", "Alex", "Sam",
# "Caleb", "David", "John", "Alex", "John", "Sam"]:
# each_with_index>
We can see the values that will be generated by this enumerator by converting it to an array.
a.to_a
#=> [["David", 0], ["John", 1], ["Alex", 2], ["Sam", 3],
# ["Caleb", 4], ["David", 5], ["John", 6], ["Alex", 7],
# ["John", 8], ["Sam", 9]]
Continuing,
b = a.group_by(&:first)
#=> {"David"=>[["David", 0], ["David", 5]],
# "John"=> [["John", 1], ["John", 6], ["John", 8]],
# "Alex"=> [["Alex", 2], ["Alex", 7]],
# "Sam"=> [["Sam", 3], ["Sam", 9]],
# "Caleb"=>[["Caleb", 4]]}
b.transform_values { |v| arr2.values_at(*v.map(&:last)) }
#=> {"David"=>["1", "6"], "John"=>["2", "7", "9"],
# "Alex"=> ["3", "8"], "Sam"=> ["4", "10"], "Caleb"=>["5"]}
For the last step, the first value of the hash b is passed to the block and the block variable is assigned to that value.
v = b.values.first
#=> [["David", 0], ["David", 5]]
The block calculations are then as follows.
c = v.map(&:last)
#=> [0, 5]
arr2.values_at(*c)
#=> arr2.values_at(0, 5)
#=> ["1", "6"]
The calculations are similar for each of the remaining values of b that are passed to the block.
1. New in Ruby MRI v2.4.

This code is less readable but compact and functional-style.
It conceptually the same as rahul mishra code https://stackoverflow.com/a/54697573/2109121
names = %w[David John Alex Sam Caleb David John Alex Sam]
numbers = %w[1 2 3 4 5 6 7 8 9]
result = names.zip(numbers).reduce({}) { |a, (b, c)| a.merge(a.key?(b) ? "A-#{b}" : b => c) }

Using zip and each_with_object
names = %w[David John Alex Sam Caleb David John Alex Sam]
numbers = %w[1 2 3 4 5 6 7 8 9]
names.zip(numbers).each_with_object({}) do |(name, number), hash|
key = hash.key?(name) ? "A-#{name}" : name
hash[key] = number
end

Sort hash by key which is a string

Assuming I get back a string:
"27,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,12,17,17,41,17,17,17,17,17,17,17,17,17,17,17,17,17,26,26,26,26,26,26,26,26,26,29,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,40,48,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,34,34,34,34,34,34,36,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,40,40,40,40,40,40,40,40,41,41,41,41,41,41,41,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,43,43,43,43,43,43,43,43,43,43,43,43,43,44,44,44,44,48,49,29,41,6,30,11,29,29,36,29,29,36,29,43,1,29,29,29,1,41"
I turn that into an array by calling
str.split(',')
Then turning it into a hash by calling
arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
I would get back a hash that looks like
{"1"=>2, "6"=>1, "39"=>23, "36"=>23, "34"=>39, "32"=>31, "30"=>18, "3"=>8, "2"=>10, "28"=>36, "29"=>21, "26"=>41, "27"=>48, "49"=>1, "44"=>4, "43"=>14, "42"=>34, "48"=>2, "40"=>9, "41"=>10, "11"=>1, "17"=>15, "12"=>1}
However, I'd like to sort that hash by key.
I've tried the solutions listed here.
I believe my problem is related to the fact they keys are strings.
The closest I got was using
Hash[h.sort_by{|k,v| k.to_i}]

Hashes shouldn't be treated as a sorted data structure. They have other advantages and use case as to return their values sequentially. As Mladen Jablanović already pointed out a array of tuples might be the better data structure when you need a sorted key/value pair.
But in current versions of Ruby there actually exists a certain order in which key/value pairs are returned when you call for example each on a hash and that is the order of insertion. Using this behavior you can just build a new hash and insert all key/value pairs into that new hash in the order you want them to be. But keep in mind that the order will break when you add more entries later on.
string = "27,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,12,17,17,41,17,17,17,17,17,17,17,17,17,17,17,17,17,26,26,26,26,26,26,26,26,26,29,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,40,48,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,34,34,34,34,34,34,36,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,40,40,40,40,40,40,40,40,41,41,41,41,41,41,41,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,43,43,43,43,43,43,43,43,43,43,43,43,43,44,44,44,44,48,49,29,41,6,30,11,29,29,36,29,29,36,29,43,1,29,29,29,1,41"
sorted_number_count_tupels = string.split(',').
group_by(&:itself).
map { |k, v| [k, v.size] }.
sort_by { |(k, v)| k.to_i }
#=> [["1",2],["2",10],["3",8],["6",1],["11",1],["12",1],["17",15],["26",41],["27",48],["28",36],["29",21],["30",18],["32",31],["34",39],["36",23],["39",23],["40",9],["41",10],["42",34],["43",14],["44",4],["48",2],["49",1]]
sorted_number_count_hash = sorted_number_count_tupels.to_h
#=> { "1" => 2, "2" => 10, "3" => 8, "6" => 1, "11" => 1, "12" => 1, "17" => 15, "26" => 41, "27" => 48, "28" => 36, "29" => 21, "30" => 18, "32" => 31, "34" => 39, "36" => 23, "39" => 23, "40" => 9, "41" => 10, "42" => 34, "43" => 14, "44" => 4, "48" => 2, "49" => 1}

Suppose you started with
str = "27,2,2,2,41,26,26,26,48,48,41,6,11,1,41"
and created the following hash
h = str.split(',').inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
#=> {"27"=>1, "2"=>3, "41"=>3, "26"=>3, "48"=>2, "6"=>1, "11"=>1, "1"=>1}
I removed compact because the array str.split(',') contains only (possibly empty) strings, no nils.
Before continuing, you may want to change this last step to
h = str.split(/\s*,\s*/).each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
#=> {"27"=>1, "2"=>3, "41"=>3, "26"=>3, "48"=>2, "6"=>1, "11"=>1, "1"=>1}
Splitting on the regex allows for the possibility of one or more spaces before or after each comma, and Enumerable#each_with_object avoids the need for that pesky ; h. (Notice the block variables are reversed.)
Then
h.sort_by { |k,_| k.to_i }.to_h
#=> {"1"=>1, "2"=>3, "6"=>1, "11"=>1, "26"=>3, "27"=>1, "41"=>3, "48"=>2}
creates a new hash that contains h's key-value pairs sorted by the integer representations of the keys. See Hash#sort_by.
Notice we've created two hashes. Here's a way to do that by modifying h in place.
h.keys.sort_by(&:to_i).each { |k| h[k] = h.delete(k) }
#=> ["1", "2", "6", "11", "26", "27", "41", "48"] (each always returns the receiver)
h #=> {"1"=>1, "2"=>3, "6"=>1, "11"=>1, "26"=>3, "27"=>1, "41"=>3, "48"=>2}
Lastly, another alternative is to sort str.split(',') before creating the hash.
str.split(',').sort_by(&:to_i).each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
#=> {"1"=>1, "2"=>3, "6"=>1, "11"=>1, "26"=>3, "27"=>1, "41"=>3, "48"=>2}

Notes
compact
String#split cannot return a nil element. compact won't be useful, here. split might return an empty string, though :
p "1,,2,3".split(',')
# ["1", "", "2", "3"]
p "1,,2,3".split(',').compact
# ["1", "", "2", "3"]
p "1,,2,3".split(',').reject(&:empty?)
# ["1", "2", "3"]
inject
If you have to use two statements inside inject block, each_with_object might be a better idea :
arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
can be rewritten :
arr.compact.each_with_object(Hash.new(0)) { |e, h| h[e] += 1 }
Hash or Array?
If you need to sort results, an Array of pairs might be more suitable than a Hash.
String or Integer?
If you accept to have an integer as key, it might make your code easier to write.
Refactoring
Here's a possibility to rewrite your code :
str.split(',')
.reject(&:empty?)
.map(&:to_i)
.group_by(&:itself)
.map { |k, v| [k, v.size] }
.sort
It outputs :
[[1, 2], [2, 10], [3, 8], [6, 1], [11, 1], [12, 1], [17, 15], [26, 41], [27, 48], [28, 36], [29, 21], [30, 18], [32, 31], [34, 39], [36, 23], [39, 23], [40, 9], [41, 10], [42, 34], [43, 14], [44, 4], [48, 2], [49, 1]]
If you really want a Hash, you can add .to_h :
{1=>2, 2=>10, 3=>8, 6=>1, 11=>1, 12=>1, 17=>15, 26=>41, 27=>48, 28=>36, 29=>21, 30=>18, 32=>31, 34=>39, 36=>23, 39=>23, 40=>9, 41=>10, 42=>34, 43=>14, 44=>4, 48=>2, 49=>1}

You can assign the arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h } to a variable and sort it by key:
num = arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
num.keys.sort
That would sort the hash by key.

A Ruby hash will keep the order of keys added. If the array is small enough to sort I would just change
str.split(',').
to
str.split(',').sort_by(&:to_i)
in order to get the values, and therefore also you hash sorted...

Make the first row as the keys for hash for the next rows?

I am having a hard time figuring out how to make the next rows a hash with the key from the first row.
I have an array structured like this:
[["id", "name", "address"], [1, "James", "...."], [2, "John", "...."] ]
To be:
[{ id : 1, name: "James", address: "..."}, ...]
I used a gem "simple_xlsx_reader", I am extracting out only the first sheet.
wb.sheets.first.row
and got a similar array output from above.
thanks!

arr = [["id", "name"], [1, "Jack"], [2, "Jill"]]
[arr.first].product(arr.drop 1).map { |a| a.transpose.to_h }
#=> [{"id"=>1, "name"=>"Jack"}, {"id"=>2, "name"=>"Jill"}]
The steps:
b = [arr.first]
#=> [["id", "name"]]
c = arr.drop 1
#=> [[1, "Jack"], [2, "Jill"]]
d = b.product(c)
#=> [[["id", "name"], [1, "Jack"]], [["id", "name"], [2, "Jill"]]]
d.map { |a| a.transpose.to_h }
#=> [{"id"=>1, "name"=>"Jack"}, {"id"=>2, "name"=>"Jill"}]
The first element of d passed to map's block is:
a = d.first
[["id", "name"], [1, "Jack"]]
The block calculation is therefore:
e = a.transpose
#=> [["id", 1], ["name", "Jack"]]
e.to_h
#=> {"id"=>1, "name"=>"Jack"}

This is what you're looking for:
arr = [["id", "name", "address"], [1, "James", "...."], [2, "John", "...."] ]
keys, *values = arr
values.map {|vals| keys.zip(vals).to_h }
Enumerable#zip takes two arrays (the receiver and the argument) and "zips" them together, producing an array of tuples (two-element arrays) e.g.:
keys = [ "foo", "bar", "baz" ]
values = [ 1, 2, 3 ]
p keys.zip(values)
# => [ [ "foo", 1 ], [ "bar", 2 ], [ "baz", 3 ] ]
Array#to_h takes an array of tuples and turns it into a hash.
If you're using a version of Ruby earlier than 2.1 you'll need to use Hash[ *keys.zip(vals) ] instead.
P.S. If you want symbol keys instead of string keys you'll want to perform that conversion before the map, e.g.:
keys = keys.map(&:to_sym)
Or, if you don't mind modifying the original array:
keys.map!(&:to_sym)

You can try this very simple one line that make your work
arr =[["id", "name", "address"], [1, "James", "add 1"], [2, "John", "add2"] ]
arr.map {|a| arr.first.zip(a).to_h unless a == arr.first }.compact

Multiply all even indexed integers by two

Wanting to take a fixnum of integers and multiply all even(indexed) integers by two. I figured the best way to do this is first turn fixnum into an array. So lets say the following number of 16 digits: a = 4408041234567901
I know I could:
a.to_s.split('')
Which will return 'a' to an array of 'stringed' numbers. But then I cant follow up with:
a.map!.with_index {|i,n| i.even? n*2}
Guess I'm kinda stuck on how to create a method to do this. So my question may even be how to turn that group of numbers into an array of fixnums/integers instead of strings.

I would prefer to remove the conditional altogether from the loop, by creating an Enumerator that contains the coefficients you want to multiply by (2 for even indexes and 1 for odd.)
coef = [2, 1].cycle
This essentially creates an Enumerator that alternately returns 2 and 1 when next is called on it. You can then use this to simplify your map to:
a.to_s.each_char.map { |v| v.to_i * coef.next }

To change it to an Array, you could do
a = 4408041234567901
arr = a.to_s.chars.map(&:to_i)
# => [4, 4, 0, 8, 0, 4, 1, 2, 3, 4, 5, 6, 7, 9, 0, 1]
You can also multiply alternate numbers by 2
arr = a.to_s.chars.map.with_index {|n,i| i.even? ? n.to_i * 2 : n.to_i }
# => [8, 4, 0, 8, 0, 4, 2, 2, 6, 4, 10, 6, 14, 9, 0, 1]
Improving a little bit, you can use a Hash to find the number to be multiplied.
h = {true => 2, false => 1}
a.to_s.each_char.map.with_index {|n,i| n.to_i * h[i.even?]}
EDIT
I can explain each step, But it will be better if you can try to figure it out on your own. Open irb, type a.to_s and check the output. Then type a.to_s.chars and inspect the output and so on..

a = 4408041234567901
even_odd = [:even, :odd].cycle
#=> #<Enumerator: [:even, :odd]:cycle>
If the indexing starts with the highest-order (leftmost) digit:
a.to_s.each_char.map { |d|
(even_odd.next == :even) ? 2*d.to_i : d.to_i }
#=> [8, 4, 0, 8, 0, 4, 2, 2, 6, 4, 10, 6, 14, 9, 0, 1]
If the indexing starts with the ones digit:
s = a.to_s
even_odd.next if s.size.even?
s.each_char.map { |d| ( even_odd.next == :even) ? 2*d.to_i : d.to_i }
#=> [4, 8, 0, 16, 0, 8, 1, 4, 3, 8, 5, 12, 7, 18, 0, 2]
Here are the steps for the example when zero-based indexing starts with the highest-order digit.
Array#cycle converts the array [:even, :odd] to an enumerator:
even_odd = [:even, :odd].cycle
even_odd.next #=> :even
even_odd.next #=> :odd
even_odd.next #=> :even
even_odd.next #=> :odd
...
b = a.to_s
#=> "4408041234567901"
enum0 = b.each_char
#=> #<Enumerator: "4408041234567901":each_char>
The enumerator enum0 passes the digits of b to map. I could have instead written:
b = a.to_s.chars
# => ["4", "4", "0", "8", "0", "4", "1", "2",
# "3", "4", "5", "6", "7", "9", "0", "1"]
but that creates an intermediate array. The enumerator does not and therefore is more efficient. Continuing...
enum1 = enum0.map
#=> #<Enumerator: #<Enumerator: "4408041234567901":each_char>:map>
You can think of this as a "compound enumerator". We can see its contents by converting it to an array:
enum1.to_a
#=> ["4", "4", "0", "8", "0", "4", "1", "2",
# "3", "4", "5", "6", "7", "9", "0", "1"]
The method each will pass each element of the enumerator into the block. Proof:
enum1.each { |d| (enum.next == :even) ? 2*d.to_i : d.to_i }
# => [8, 4, 0, 8, 0, 4, 2, 2, 6, 4, 10, 6, 14, 9, 0, 1]
We can manually step through the elements of enum1 by using Enumerator#next. We will assign the value to the block variable d and perform the calculation in the block to map the digit d:
d = enum1.next
#=> "4"
(enum.next == :even) ? 2*d.to_i : d.to_i
#=> (:even == :even) ? 2*"4".to_i : "4".to_i
#=> (true) ? 8 : 4
#=> 8 ("4" is mapped to 8)
d = enum1.next
#=> "4"
(enum.next == :even) ? 2*d.to_i : d.to_i
#=> (:odd == :even) ? 2*"4".to_i : "4".to_i
#=> (false) ? 8 : 4
#=> 4 ("4" is mapped to 4)
d = enum1.next
#=> "0"
#=> (:even == :even) ? 2*"0".to_i : "0".to_i
#=> (true) ? 0 : 0
#=> 8 ("0" is mapped to 0)
and so on.

How to invert a hash, maintaining duplicate keys

From an initial hash t:
t = {"1"=>1, "2"=>2, "3"=>2, "6"=>3, "5"=>4, "4"=>1, "8"=>2, "9"=>2, "0"=>1, "7"=>1}
I need to swap the keys and values as follows:
t = {"1"=>1, "2"=>2, "3"=>2, "6"=>3, "5"=>4, "1"=>4, "8"=>2, "9"=>2, "1"=>0, "1"=>7}
While maintaining the structure of the hash (ie, without collapsing duplicate keys).
Then I'll make an array out of this hash.
Is there a way to do this? I tried this:
t.find_all{ |key,value| value == 1 } # pluck all elements with values of 1
#=> [["1", 1], ["4", 1], ["0", 1], ["7", 1]]
But it returns a new array, and the initial hash isn't changed.
The following doesn't work either:
t.invert.find_all{ |key,value| value == 1 }
#=> []

Here's a way to do this:
>> t = {"1" => 1, "2" => 2, "3" => 2, "6" => 3, "5" => 4, "4" => 1, "8" => 2, "9" => 2, "0" => 1, "7" => 1}
Hash#compare_by_identity allows for keys that are duplicates by value but unique by object id:
>> h = Hash.new.compare_by_identity
>> t.each_pair{ |k,v| h[v.to_s] = v.to_i }
The inverse hash of t:
>> h
#=> {"1" => 1, "2" => 2, "2" => 3, "3" => 6, "4" => 5, "1" => 4, "2" => 8, "2" => 9, "1" => 0, "1" => 7}
You can then use find_all to retrieve an array of elements without mutating h:
>> h.find_all{ |k,_| k == "1" }
#=> [["1", 1], ["1", 1], ["1", 1], ["1", 1]]
or keep_if to return the mutated h:
>> h.keep_if{ |k,_| k == "1" }
#=> {"1"=>1, "1"=>1, "1"=>1, "1"=>1}
>> h
#=> {"1"=>1, "1"=>1, "1"=>1, "1"=>1}
Note that this solution assumes you want to maintain the pattern of string keys and integer values in your hash. If you require integer keys, compare_by_identity won't be helpful to you.

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