Ruby Подсчет количества уникальных символов в строке

Question

Я работаю со строкой символов alphabet ="AABBBCCCCDDDDDEFGHIJKLMNOPQRSTUVWXYZZZZZZ" Я хочу создать def, который подсчитывает количество уникальных символов в строке и процент уникальных символов без необходимости использовать alphabet.count("A"), alphabet.count"("B"), alphabet.count("C"), etc etc, поэтому мне не нужно тратить время утомительно вводя каждый символ в методе .count().

Мне удалось в каком-то смысле получить желаемый вывод, но вывод повторяется каждый результат много раз из-за того, как я структурировал свой for l oop

Вот мой код:

alphabet ="AABBBCCCCDDDDDEFGHIJKLMNOPQRSTUVWXYZZZZZZ"

def count_num_of_uniq_chars(string)
  len = string.length
  len = len.to_f
  for i in 0..len-1

    uniq_char=string[i]
    puts "uniq_chars --> #{uniq_char}"

    count_of_uniq_char = string.count(string[i])

    puts "count_of_uniq_char--> #{count_of_uniq_char}"


    percent_of_uniq_char = ( (count_of_uniq_char / len) * 100 )
    percent_of_uniq_char=percent_of_uniq_char.to_f

    puts "there are #{count_of_uniq_char} letter '#{uniq_char}'s in the string which is #{percent_of_uniq_char}% of strings length "
    puts
  end # loop end

end #def end

count_num_of_uniq_chars(alphabet)

Выводится как:

uniq_chars --> A
count_of_uniq_char--> 2
there are 2 letter 'A's in the string which is 4.878048780487805% of strings length

uniq_chars --> A
count_of_uniq_char--> 2
there are 2 letter 'A's in the string which is 4.878048780487805% of strings length

uniq_chars --> B
count_of_uniq_char--> 3
there are 3 letter 'B's in the string which is 7.317073170731707% of strings length

uniq_chars --> B
count_of_uniq_char--> 3
there are 3 letter 'B's in the string which is 7.317073170731707% of strings length

uniq_chars --> B
count_of_uniq_char--> 3
there are 3 letter 'B's in the string which is 7.317073170731707% of strings length

uniq_chars --> C
count_of_uniq_char--> 4
there are 4 letter 'C's in the string which is 9.75609756097561% of strings length

uniq_chars --> C
count_of_uniq_char--> 4
there are 4 letter 'C's in the string which is 9.75609756097561% of strings length

uniq_chars --> C
count_of_uniq_char--> 4
there are 4 letter 'C's in the string which is 9.75609756097561% of strings length

uniq_chars --> C
count_of_uniq_char--> 4
there are 4 letter 'C's in the string which is 9.75609756097561% of strings length

uniq_chars --> D
count_of_uniq_char--> 5
there are 5 letter 'D's in the string which is 12.195121951219512% of strings length

uniq_chars --> D
count_of_uniq_char--> 5
there are 5 letter 'D's in the string which is 12.195121951219512% of strings length

uniq_chars --> D
count_of_uniq_char--> 5
there are 5 letter 'D's in the string which is 12.195121951219512% of strings length

uniq_chars --> D
count_of_uniq_char--> 5
there are 5 letter 'D's in the string which is 12.195121951219512% of strings length

uniq_chars --> D
count_of_uniq_char--> 5
there are 5 letter 'D's in the string which is 12.195121951219512% of strings length

uniq_chars --> E
count_of_uniq_char--> 1
there are 1 letter 'E's in the string which is 2.4390243902439024% of strings length

uniq_chars --> F
count_of_uniq_char--> 1
there are 1 letter 'F's in the string which is 2.4390243902439024% of strings length

uniq_chars --> G
count_of_uniq_char--> 1
there are 1 letter 'G's in the string which is 2.4390243902439024% of strings length

uniq_chars --> H
count_of_uniq_char--> 1
there are 1 letter 'H's in the string which is 2.4390243902439024% of strings length

uniq_chars --> I
count_of_uniq_char--> 1
there are 1 letter 'I's in the string which is 2.4390243902439024% of strings length

uniq_chars --> J
count_of_uniq_char--> 1
there are 1 letter 'J's in the string which is 2.4390243902439024% of strings length

uniq_chars --> K
count_of_uniq_char--> 1
there are 1 letter 'K's in the string which is 2.4390243902439024% of strings length

uniq_chars --> L
count_of_uniq_char--> 1
there are 1 letter 'L's in the string which is 2.4390243902439024% of strings length

uniq_chars --> M
count_of_uniq_char--> 1
there are 1 letter 'M's in the string which is 2.4390243902439024% of strings length

uniq_chars --> N
count_of_uniq_char--> 1
there are 1 letter 'N's in the string which is 2.4390243902439024% of strings length

uniq_chars --> O
count_of_uniq_char--> 1
there are 1 letter 'O's in the string which is 2.4390243902439024% of strings length

uniq_chars --> P
count_of_uniq_char--> 1
there are 1 letter 'P's in the string which is 2.4390243902439024% of strings length

uniq_chars --> Q
count_of_uniq_char--> 1
there are 1 letter 'Q's in the string which is 2.4390243902439024% of strings length

uniq_chars --> R
count_of_uniq_char--> 1
there are 1 letter 'R's in the string which is 2.4390243902439024% of strings length

uniq_chars --> S
count_of_uniq_char--> 1
there are 1 letter 'S's in the string which is 2.4390243902439024% of strings length

uniq_chars --> T
count_of_uniq_char--> 1
there are 1 letter 'T's in the string which is 2.4390243902439024% of strings length

uniq_chars --> U
count_of_uniq_char--> 1
there are 1 letter 'U's in the string which is 2.4390243902439024% of strings length

uniq_chars --> V
count_of_uniq_char--> 1
there are 1 letter 'V's in the string which is 2.4390243902439024% of strings length

uniq_chars --> W
count_of_uniq_char--> 1
there are 1 letter 'W's in the string which is 2.4390243902439024% of strings length

uniq_chars --> X
count_of_uniq_char--> 1
there are 1 letter 'X's in the string which is 2.4390243902439024% of strings length

uniq_chars --> Y
count_of_uniq_char--> 1
there are 1 letter 'Y's in the string which is 2.4390243902439024% of strings length

uniq_chars --> Z
count_of_uniq_char--> 6
there are 6 letter 'Z's in the string which is 14.634146341463413% of strings length

uniq_chars --> Z
count_of_uniq_char--> 6
there are 6 letter 'Z's in the string which is 14.634146341463413% of strings length

uniq_chars --> Z
count_of_uniq_char--> 6
there are 6 letter 'Z's in the string which is 14.634146341463413% of strings length

uniq_chars --> Z
count_of_uniq_char--> 6
there are 6 letter 'Z's in the string which is 14.634146341463413% of strings length

uniq_chars --> Z
count_of_uniq_char--> 6
there are 6 letter 'Z's in the string which is 14.634146341463413% of strings length

uniq_chars --> Z
count_of_uniq_char--> 6
there are 6 letter 'Z's in the string which is 14.634146341463413% of strings length

Обратите внимание, что оператор вывода для каждой буквы повторяется в зависимости от того, сколько раз эта буква встречается в строке. Как я могу заставить его выводить один раз за букву независимо от того, сколько вхождений в строке?

Answer 1

Я хочу создать def, который подсчитывает количество уникальных символов в строке [...]

Вы можете получить символы строки с помощью String#each_char и иметь Enumerable#tally подсчитывать вхождения: (tally требуется Ruby 2,7)

alphabet.each_char.tally
#=> {
#     "A"=>2, "B"=>3, "C"=>4, "D"=>5, "E"=>1, "F"=>1, "G"=>1,
#     "H"=>1, "I"=>1, "J"=>1, "K"=>1, "L"=>1, "M"=>1, "N"=>1,
#     "O"=>1, "P"=>1, "Q"=>1, "R"=>1, "S"=>1, "T"=>1, "U"=>1,
#     "V"=>1, "W"=>1, "X"=>1, "Y"=>1, "Z"=>6
#   }

Чтобы получить проценты, вы просто делите вхождения в символы на общее количество количество символов, например:

hash = alphabet.each_char.tally
hash.each do |char, count|
  q = count.quo(hash.size)
  puts format(" %s | %d | %4.1f%%", char, count, q * 100)
end

Вывод:

 A | 2 |  7.7%
 B | 3 | 11.5%
 C | 4 | 15.4%
 D | 5 | 19.2%
 E | 1 |  3.8%
 F | 1 |  3.8%
 G | 1 |  3.8%
 H | 1 |  3.8%
 I | 1 |  3.8%
 J | 1 |  3.8%
 K | 1 |  3.8%
 L | 1 |  3.8%
 M | 1 |  3.8%
 N | 1 |  3.8%
 O | 1 |  3.8%
 P | 1 |  3.8%
 Q | 1 |  3.8%
 R | 1 |  3.8%
 S | 1 |  3.8%
 T | 1 |  3.8%
 U | 1 |  3.8%
 V | 1 |  3.8%
 W | 1 |  3.8%
 X | 1 |  3.8%
 Y | 1 |  3.8%
 Z | 6 | 23.1%

Вместо hash.size (количество уникальных символов) вы также можете разделить на alphabet.size (количество символов в строка), в зависимости от того, что вы хотите.

Answer 2

Вот два способа сделать это.

Использовать форму метода класса Ха sh :: new , который принимает аргумент ноль (но без блока ), аргументом является значение ha sh по умолчанию

h = alphabet.each_char.with_object(Hash.new(0)) { |c,h| h[c] += 1 }
  #=> {"A"=>2, "B"=>3, "C"=>4,..., "Z"=>6}

h[c] += 1, расширяется до h[c] = h[c] + 1. Если h не имеет ключа c, h[c] справа от равенства возвращает значение по умолчанию, равное нулю, давая h[c] = 0 + 1.

Используйте метод Enumerable # group_by

h = alphabet.each_char.
             group_by(&:itself).
             transform_values(&:count)
  #=> <same as above>

См. Hash # transform_values ​​.

Шаги следующие:

enum = alphabet.each_char
  #=> #<Enumerator: "AABBB...ZZZ":each_char> 
a = enum.group_by(&:itself)
  #=> {"A"=>["A", "A"], "B"=>["B", "B", "B"],...,
  #          "Z"=>["Z", "Z", "Z", "Z", "Z", "Z"]} 
a.transform_values(&:count)
  #=> {"A"=>2, "B"=>3,..., "Z"=>6} 

Использование ха sh

Если у вас есть га sh, вы можете отображать информацию в режиме sh. Например:

n = alphabet.size
  #=> 41  
h.each { |k,v| puts "#{v} #{k}'s #{(100*v.fdiv(n)).round(2)}%" }
2 A's 4.88%
3 B's 7.32%
4 C's 9.76%
...
1 X's 2.44%
1 Y's 2.44%
6 Z's 14.63%

Answer 3

Вы можете использовать string.chars.uniq и избавиться от len, для инициализации l oop и uniq_char:

def count_num_of_uniq_chars(string)
  string.chars.uniq.each do |uniq_char|
    puts "uniq_chars --> #{uniq_char}"

    count_of_uniq_char = string.count(uniq_char)

    puts "count_of_uniq_char--> #{count_of_uniq_char}"


    percent_of_uniq_char = ( (count_of_uniq_char / string.length.to_f) * 100 )
    percent_of_uniq_char=percent_of_uniq_char.to_f

    puts "there are #{count_of_uniq_char} letter '#{uniq_char}'s in the string which is #{percent_of_uniq_char}% of strings length \n\n"
  end
end

См. String#chars и Array#uniq.

Обратите внимание, что percent_of_uniq_char рассчитывается как count_of_uniq_char, разделенное на длину строки, преобразованной в число с плавающей точкой. Если это проблема для этого случая, вы можете инициализировать ее за пределами l oop.

Answer 4

Это можно сделать так

def count_occ(str)
    d=Hash.new(0)
    str.split('').each do |ch|
            d[ch]=d[ch]+1
    end
    d.each do |key,value|
        count_ch=value
        percentage=count_ch/Float(str.length)
        puts "there are #{count_ch} letter '#{key}'s in the string which is #{percentage}% of strings length "
    end

end