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...
Related
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
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.
I have a hash, which contains a hash, which contains a number of arrays, like this:
{ "bob" =>
{
"foo" => [1, 3, 5],
"bar" => [2, 4, 6]
},
"fred" =>
{
"foo" => [1, 7, 9],
"bar" => [8, 10, 12]
}
}
I would like to compare the arrays against the other arrays, and then alert me if they are duplicates. It is possible for hash["bob"]["foo"] and hash["fred"]["foo"] to have duplicates, but not for hash["bob"]["foo"] and hash["bob"]["bar"]. Same with hash["fred"].
I can't even figure out where to begin with this one. I suspect inject will be involved somewhere, but I could be wrong.
This snippet will return an array of duplicates for each key. Duplicates can only be generated for equal keys.
duplicates = (keys = h.values.map(&:keys).flatten.uniq).map do |key|
{key => h.values.map { |h| h[key] }.inject(&:&)}
end
This will return [{"foo"=>[1]}, {"bar"=>[]}] which indicates that the key foo was the only one containing a duplicate of 1.
The snippet above assume h is the variable name of your hash.
h = {
"bob" =>
{
"foo" => [1, 3, 5],
"bar" => [2, 4, 6]
},
"fred" =>
{
"foo" => [1, 7, 9],
"bar" => [1, 10, 12]
}
}
h.each do |k, v|
numbers = v.values.flatten
puts k if numbers.length > numbers.uniq.length
end
There are many ways to do it.
Here's one that should be easy to read.
It works in Ruby 1.9. It uses + to combine two arrays and then uses the uniq! operator to figure out whether there is a duplicate number.
h = { "bob" =>
{
"foo" => [1, 3, 5],
"bar" => [2, 4, 6]
},
"fred" =>
{
"foo" => [1, 7, 12],
"bar" => [8, 10, 12]
}
}
h.each do |person|
if (person[1]["foo"] + person[1]["bar"]).uniq! != nil
puts "Duplicate in #{person[1]}"
end
end
I'm not sure what exactly you are looking for. But at look at a possible solution, perhaps you can reuse something.
outer_hash.each do |person, inner_hash|
seen_arrays = Hash.new
inner_hash.each do |inner_key, array|
other = seen_arrays[array]
if other
raise "array #{person}/#{inner_key} is a duplicate of #{other}"
end
seen_arrays[array] = "#{person}/#{inner_key}"
end
end
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"]
How do you count duplicates in a ruby array?
For example, if my array had three a's, how could I count that
Another version of a hash with a key for each element in your array and value for the count of each element
a = [ 1, 2, 3, 3, 4, 3]
h = Hash.new(0)
a.each { | v | h.store(v, h[v]+1) }
# h = { 3=>3, 2=>1, 1=>1, 4=>1 }
Given:
arr = [ 1, 2, 3, 2, 4, 5, 3]
My favourite way of counting elements is:
counts = arr.group_by{|i| i}.map{|k,v| [k, v.count] }
# => [[1, 1], [2, 2], [3, 2], [4, 1], [5, 1]]
If you need a hash instead of an array:
Hash[*counts.flatten]
# => {1=>1, 2=>2, 3=>2, 4=>1, 5=>1}
This will yield the duplicate elements as a hash with the number of occurences for each duplicate item. Let the code speak:
#!/usr/bin/env ruby
class Array
# monkey-patched version
def dup_hash
inject(Hash.new(0)) { |h,e| h[e] += 1; h }.select {
|k,v| v > 1 }.inject({}) { |r, e| r[e.first] = e.last; r }
end
end
# unmonkeey'd
def dup_hash(ary)
ary.inject(Hash.new(0)) { |h,e| h[e] += 1; h }.select {
|_k,v| v > 1 }.inject({}) { |r, e| r[e.first] = e.last; r }
end
p dup_hash([1, 2, "a", "a", 4, "a", 2, 1])
# {"a"=>3, 1=>2, 2=>2}
p [1, 2, "Thanks", "You're welcome", "Thanks",
"You're welcome", "Thanks", "You're welcome"].dup_hash
# {"You're welcome"=>3, "Thanks"=>3}
Simple.
arr = [2,3,4,3,2,67,2]
repeats = arr.length - arr.uniq.length
puts repeats
arr = %w( a b c d c b a )
# => ["a", "b", "c", "d", "c", "b", "a"]
arr.count('a')
# => 2
Another way to count array duplicates is:
arr= [2,2,3,3,2,4,2]
arr.group_by{|x| x}.map{|k,v| [k,v.count] }
result is
[[2, 4], [3, 2], [4, 1]]
requires 1.8.7+ for group_by
ary = %w{a b c d a e f g a h i b}
ary.group_by{|elem| elem}.select{|key,val| val.length > 1}.map{|key,val| key}
# => ["a", "b"]
with 1.9+ this can be slightly simplified because Hash#select will return a hash.
ary.group_by{|elem| elem}.select{|key,val| val.length > 1}.keys
# => ["a", "b"]
To count instances of a single element use inject
array.inject(0){|count,elem| elem == value ? count+1 : count}
arr = [1, 2, "a", "a", 4, "a", 2, 1]
arr.group_by(&:itself).transform_values(&:size)
#=> {1=>2, 2=>2, "a"=>3, 4=>1}
Ruby >= 2.7 solution here:
A new method .tally has been added.
Tallies the collection, i.e., counts the occurrences of each element. Returns a hash with the elements of the collection as keys and the corresponding counts as values.
So now, you will be able to do:
["a", "b", "c", "b"].tally #=> {"a"=>1, "b"=>2, "c"=>1}
What about a grep?
arr = [1, 2, "Thanks", "You're welcome", "Thanks", "You're welcome", "Thanks", "You're welcome"]
arr.grep('Thanks').size # => 3
Its Easy:
words = ["aa","bb","cc","bb","bb","cc"]
One line simple solution is:
words.each_with_object(Hash.new(0)) { |word,counts| counts[word] += 1 }
It works for me.
Thanks!!
I don't think there's a built-in method. If all you need is the total count of duplicates, you could take a.length - a.uniq.length. If you're looking for the count of a single particular element, try
a.select {|e| e == my_element}.length.
Improving #Kim's answer:
arr = [1, 2, "a", "a", 4, "a", 2, 1]
Hash.new(0).tap { |h| arr.each { |v| h[v] += 1 } }
# => {1=>2, 2=>2, "a"=>3, 4=>1}
Ruby code to get the repeated elements in the array:
numbers = [1,2,3,1,2,0,8,9,0,1,2,3]
similar = numbers.each_with_object([]) do |n, dups|
dups << n if seen.include?(n)
seen << n
end
print "similar --> ", similar
Another way to do it is to use each_with_object:
a = [ 1, 2, 3, 3, 4, 3]
hash = a.each_with_object({}) {|v, h|
h[v] ||= 0
h[v] += 1
}
# hash = { 3=>3, 2=>1, 1=>1, 4=>1 }
This way, calling a non-existing key such as hash[5] will return nil instead of 0 with Kim's solution.
I've used reduce/inject for this in the past, like the following
array = [1,5,4,3,1,5,6,8,8,8,9]
array.reduce (Hash.new(0)) {|counts, el| counts[el]+=1; counts}
produces
=> {1=>2, 5=>2, 4=>1, 3=>1, 6=>1, 8=>3, 9=>1}