I am really new at ruby. I have created a function to count occurrences of words in a string. However, I am getting NoMethodError for + all the time. I searched, tried different variations, but couldn't solve the problem. Here is the code:
def count_words(str)
str_down = str.downcase
arr = str_down.scan(/([\w]+)/).flatten
hash = Hash[]
arr.each {|x| hash[x] += 1 }
(hash.sort_by {|key, value| value}.reverse)
end
Here is the error:
NoMethodError: undefined method `+' for nil:NilClass
from ./***.rb:14:in `count_words'
from ./***.rb:14:in `each'
from ./***.rb:14:in `count_words'
from (irb):137
Change
hash = Hash[]
arr.each {|x| hash[x] += 1 }
To
hash = {}
arr.each {|x| hash[x] =0 unless hash[x]; hash[x] += 1 }
OR
hash = Hash.new(0)
arr.each {|x| hash[x] += 1 }
EXPLAINATION
hash = {}
hash[1] = "example1" #ASSIGNMENT gives hash = {1: "example1"}
p hash[2] #This gives `nil` by default, as key is not present in hash
To give default value to the key which is not present in hash we have to do the following:
hash = Hash.new("new value")
p hash #Gives {}
p hash[4] #gives "new value"
In the first iteration, h[x] is nil. Trying to add 1 to nil throws error. Setting the initial value of h[x] to 0 will solve the issue.
arr.each {|x| hash[x]||=0; hash[x] += 1 }
instead of
arr.each {|x| hash[x] += 1 }
Related
Trying to get the most occurring letter in a string.
So far:
puts "give me a string"
words = gets.chomp.split
counts = Hash.new(0)
words.each do |word|
counts[word] += 1
end
Does not run further than asking for a string. What am I doing wrong?
If you're running this in irb, then the computer may think that the ruby code you're typing in is the text to analyse:
irb(main):001:0> puts "give me a string"
give me a string
=> nil
irb(main):002:0> words = gets.chomp.split
counts = Hash.new(0)
words.each do |word|
counts[word] += 1
end=> ["counts", "=", "Hash.new(0)"]
irb(main):003:0> words.each do |word|
irb(main):004:1* counts[word] += 1
irb(main):005:1> end
NameError: undefined local variable or method `counts' for main:Object
from (irb):4:in `block in irb_binding'
from (irb):3:in `each'
from (irb):3
from /Users/agrimm/.rbenv/versions/2.2.1/bin/irb:11:in `<main>'
irb(main):006:0>
If you wrap it in a block of some sort, you won't get that confusion:
begin
puts "give me a string"
words = gets.chomp.split
counts = Hash.new(0)
words.each do |word|
counts[word] += 1
end
counts
end
gives
irb(main):001:0> begin
irb(main):002:1* puts "give me a string"
irb(main):003:1> words = gets.chomp.split
irb(main):004:1> counts = Hash.new(0)
irb(main):005:1> words.each do |word|
irb(main):006:2* counts[word] += 1
irb(main):007:2> end
irb(main):008:1> counts
irb(main):009:1> end
give me a string
foo bar
=> {"foo"=>1, "bar"=>1}
Then you can work on the fact that split by itself isn't what you want. :)
This should work:
puts "give me a string"
result = gets.chomp.split(//).reduce(Hash.new(0)) { |h, v| h.store(v, h[v] + 1); h }.max_by{|k,v| v}
puts result.to_s
Output:
#Alan ➜ test rvm:(ruby-2.2#europa) ruby test.rb
give me a string
aa bbb cccc ddddd
["d", 5]
Or in irb:
:008 > 'This is some random string'.split(//).reduce(Hash.new(0)) { |h, v| h.store(v, h[v] + 1); h }.max_by{|k,v| v}
=> ["s", 4]
Rather than getting a count word by word, you can process the whole string immediately.
str = gets.chomp
hash = Hash.new(0)
str.each_char do |c|
hash[c] += 1 unless c == " " #used to filter the space
end
After getting the number of letters, you can then find the letter with highest count with
max = hash.values.max
Then match it to the key in the hash and you're done :)
puts hash.select{ |key| hash[key] == max }
Or to simplify the above methods
hash.max_by{ |key,value| value }
The compact form of this is :
hash = Hash.new(0)
gets.chomp.each_char { |c| hash[c] += 1 unless c == " " }
puts hash.max_by{ |key,value| value }
This returns the highest occurring character within a given string:
puts "give me a string"
characters = gets.chomp.split("").reject { |c| c == " " }
counts = Hash.new(0)
characters.each { |character| counts[character] += 1 }
print counts.max_by { |k, v| v }
[1,2,3,3] - [1,2,3] produces the empty array []. Is it possible to retain duplicates so it returns [3]?
I am so glad you asked. I would like to see such a method added to the class Array in some future version of Ruby, as I have found many uses for it:
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
A description of the method and links to some of its applications are given here.
By way of example:
a = [1,2,3,4,3,2,4,2]
b = [2,3,4,4,4]
a - b #=> [1]
a.difference b #=> [1,2,3,2]
Ruby v2.7 gave us the method Enumerable#tally, allowing us to replace the first line of the method with
h = other.tally
As far as I know, you can't do this with a built-in operation. Can't see anything in the ruby docs either. Simplest way to do this would be to extend the array class like this:
class Array
def difference(array2)
final_array = []
self.each do |item|
if array2.include?(item)
array2.delete_at(array2.find_index(item))
else
final_array << item
end
end
end
end
For all I know there's a more efficient way to do this, also
EDIT:
As suggested by user2864740 in question comments, using Array#slice! is a much more elegant solution
def arr_sub(a,b)
a = a.dup #if you want to preserve the original array
b.each {|del| a.slice!(a.index(del)) if a.include?(del) }
return a
end
Credit:
My original answer
def arr_sub(a,b)
b = b.each_with_object(Hash.new(0)){ |v,h| h[v] += 1 }
a = a.each_with_object([]) do |v, arr|
arr << v if b[v] < 1
b[v] -= 1
end
end
arr_sub([1,2,3,3],[1,2,3]) # a => [3]
arr_sub([1,2,3,3,4,4,4],[1,2,3,4,4]) # => [3, 4]
arr_sub([4,4,4,5,5,5,5],[4,4,5,5,5,5,6,6]) # => [4]
Would anybody have an idea, why following code returns error:
stock = {"M9788375085969"=>5, "M9788392289760"=>5, "M9788389371461"=>1, "M9788389371447"=>3, "M9788392289761"=>2}
add = {"M9788375085969"=>1, "M9788392289760"=>2, "NEW9788392289753"=>1 }
add.each do |key, value|
stock[key] += value
end
NoMethodError: undefined method `+' for nil:NilClass
while similar thing works fine:
key = "M9788375085969"
value = 1
stock[key] += value
=> 6
There is one key in your add hash that is missing in your stock hash : "NEW9788392289753".
When executing stock["NEW9788392289753"], nil is returned, as the key is not mapped.
The key "NEW9788392289753" is not present in the Hash stock,but present in add hash. See below :
stock = {"M9788375085969"=>5, "M9788392289760"=>5, "M9788389371461"=>1, "M9788389371447"=>3, "M9788392289761"=>2}
stock['NEW9788392289753'] # => nil
nil.respond_to?(:+) # => false # means NilClass don't has method called :+
Thus nil.+(value) throwing a valid error. Do as below :
stock = {"M9788375085969"=>5, "M9788392289760"=>5, "M9788389371461"=>1, "M9788389371447"=>3, "M9788392289761"=>2}
add = {"M9788375085969"=>1, "M9788392289760"=>2, "NEW9788392289753"=>1 }
add.each do |key, value|
p stock[key] += value if stock.has_key?(key) # it will take care of the error.
end
output
6
7
As per OP's comment I would do as :
add.each do |key, value|
if stock.has_key?(key)
stock[key] += value
else
stock[key] = value
end
end
because the key NEW9788392289753 from add is not contained in stock.
Another way of treating non-existent keys is providing a default of zero:
stock = {"M9788375085969"=>5, "M9788392289760"=>5, "M9788389371461"=>1, "M9788389371447"=>3, "M9788392289761"=>2}
add = {"M9788375085969"=>1, "M9788392289760"=>2, "NEW9788392289753"=>1 }
stock.default = 0
add.each do |key, value|
stock[key] += value
end
p stock #=> {"M9788375085969"=>6, "M9788392289760"=>7, "M9788389371461"=>1, "M9788389371447"=>3, "M9788392289761"=>2, "NEW9788392289753"=>1}
How can I have a hash of hash of hash?
My test returns
undefined method `[]' for nil:NilClass (NoMethodError)
Any tips?
found = Hash.new()
x = 1;
while x < 4 do
found[x] = Hash.new()
y = 1
while y < 4 do
found[x][y] = Hash.new()
found[x][y]['name1'] = 'abc1'
found[x][y]['name2'] = 'abc2'
found[x][y]['name3'] = 'abc3'
y += 1
end
x += 1
end
found.each do |k, v, y|
puts "k : #{k}"
puts " : #{v[y['name1']]}"
puts " : #{v[y['name2']]}"
puts " : #{v[y['name3']]}"
puts
end
I think you want something like this:
First of all create the data structure. You want nested hashes so you need to define default values for each hash key.
found = Hash.new do |hash,key|
hash[key] = Hash.new do |hash,key|
hash[key] = Hash.new
end
end
Run the search
(1..3).each do |x|
(1..3).each do |y|
found[x][y]['name1'] = 'abc1'
found[x][y]['name2'] = 'abc1'
found[x][y]['name3'] = 'abc1'
end
end
Then display the results
found.each do |x, y_hash|
y_hash.each do |y, name_hash|
name_hash.each do |name, value|
puts "#{x} => #{y} => #{name} => #{value}"
end
end
end
The way you build the hash seems to be functional. What probably causes the error is this loop:
found.each do |k, v, y|
Hash#each yields key/value pairs, so y will be assigned nil, thus causing the error two lines below. What you probably meant is a nested loop like
found.each do |x, h1|
h1.each do |y, h2|
puts h2['name1']
end
end
You should also be aware that you can write these kinds of counting loops more concisely in Ruby:
found = Hash.new { |h,k| h[k] = {} }
1.upto(3) do |x|
1.upto(3) do |y|
found[x][y] = {
'name1' => 'abc1',
'name2' => 'abc2',
'name3' => 'abc3',
}
end
end
In order to implement auto-vivification of Ruby hash, one can employ the following class
class AutoHash < Hash
def initialize(*args)
super()
#update, #update_index = args[0][:update], args[0][:update_key] unless
args.empty?
end
def [](k)
if self.has_key?k
super(k)
else
AutoHash.new(:update => self, :update_key => k)
end
end
def []=(k, v)
#update[#update_index] = self if #update and #update_index
super
end
def few(n=0)
Array.new(n) { AutoHash.new }
end
end
This class allows to do the following things
a = AutoHash.new
a[:a][:b] = 1
p a[:c] # => {} # key :c has not been created
p a # => {:a=>{:b=>1}} # note, that it does not have key :c
a,b,c = AutoHash.new.few 3
b[:d] = 1
p [a,b,c] # => [{}, {:d=>1}, {}] # hashes are independent
There is a bit more advanced definition of this class proposed by Joshua, which is a bit hard for me to understand.
Problem
There is one situation, where I think the new class can be improved. The following code fails with the error message NoMethodError: undefined method '+' for {}:AutoHash
a = AutoHash.new
5.times { a[:sum] += 10 }
What would you do to handle it? Can one define []+= operator?
Related questions
Is auto-initialization of multi-dimensional hash array possible in Ruby, as it is in PHP?
Multiple initialization of auto-vivifying hashes using a new operator in Ruby
ruby hash initialization r
still open: How to create an operator for deep copy/cloning of objects in Ruby?
There is no way to define a []+= method in ruby. What happens when you type
x[y] += z
is
x[y] = x[y] + z
so both the [] and []= methods are called on x (and + is called on x[y], which in this case is an AutoHash). I think that the best way to handle this problem would be to define a + method on AutoHash, which will just return it's argument. This will make AutoHash.new[:x] += y work for just about any type of y, because the "empty" version of y.class ('' for strings, 0 for numbers, ...) plus y will almost always equal y.
class AutoHash
def +(x); x; end
end
Adding that method will make both of these work:
# Numbers:
a = AutoHash.new
5.times { a[:sum] += 10 }
a[:sum] #=> 50
# Strings:
a = AutoHash.new
5.times { a[:sum] += 'a string ' }
a[:sum] #=> "a string a string a string a string a string "
And by the way, here is a cleaner version of your code:
class AutoHash < Hash
def initialize(args={})
super
#update, #update_index = args[:update], args[:update_key]
end
def [](k)
if has_key? k
super(k)
else
AutoHash.new :update => self, :update_key => k
end
end
def []=(k, v)
#update[#update_index] = self if #update and #update_index
super
end
def +(x); x; end
def self.few(n)
Array.new(n) { AutoHash.new }
end
end
:)
What I think you want is this:
hash = Hash.new { |h, k| h[k] = 0 }
hash['foo'] += 3
# => 3
That will return 3, then 6, etc. without an error, because the the new value is default assigned 0.
require 'xkeys' # on rubygems.org
a = {}.extend XKeys::Hash
a[:a, :b] = 1
p a[:c] # => nil (key :c has not been created)
p a # => { :a => { :b => 1 } }
a.clear
5.times { a[:sum, :else => 0] += 10 }
p a # => { :sum => 50 }