Question

编写一个方法，该方法返回字符串中使用的各种小写字母，大写字母，数字和特殊字符的No。利用范围。

输入=“ heLLo Every1”

我在提供的解决方案中使用范围和大小写方法。

解决方案：

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  }

代码有效。

Answer 1

是

class String
  def character_count
    counters = Hash.new(0)
    each_char do |item|
      case item
      when 'A'..'Z'
        counters[:uppercase] += 1
      when 'a'..'z'
        counters[:lowercase] += 1
      when '0'..'9'
        counters[:digit] += 1
      else
        counters[:special] += 1
      end
    end
    counters.values_at(:uppercase, :lowercase, :digit, :special)
  end
end

if ARGV.empty?
  puts 'Please provide an input'
else
  string = ARGV[0]
  uppercase, lowercase, digit, special = string.character_count
  puts "Lowercase characters = #{lowercase}"
  puts "Uppercase characters = #{uppercase}"
  puts "Numeric characters = #{digit}"
  puts "Special characters = #{special}"
end

Answer 2

您可以按照以下更好的方式使用regex，

type = { special: /[^0-9A-Za-z]/, numeric: /[0-9]/, uppercase: /[A-Z]/, lowercase: /[a-z]/ }

'Hello World'.scan(type[:special]).count
# => 1 
'Hello World'.scan(type[:numeric]).count
# => 0 
'Hello World'.scan(type[:uppercase]).count
# => 2
'Hello World'.scan(type[:lowercase]).count
# => 8

Answer 3

其他选项。

首先，将您的范围映射到哈希中：

mapping = { upper: ('A'..'Z'), lower: ('a'..'z'), digits: ('0'..'9'), specials: nil }

然后将收件人哈希初始化为默认的0：

res = Hash.new(0)

最后，映射输入的字符：

input = "heLLo Every1"

input.chars.each { |e| res[(mapping.find { |k, v| v.to_a.include? e } || [:specials]).first ] += 1 }

res
#=> {:upper=>3, :lower=>7, :digits=>1, :specials=>1}

Answer 4

str = "Agent 007 was on the trail of a member of SPECTRE"

str.each_char.with_object(Hash.new(0)) do |c,h|
  h[ case c
     when /\d/     then :digit
     when /\p{Lu}/ then :uppercase
     when /\p{Ll}/ then :downcase
     else               :special
     end
   ] += 1
  end
end
  #=> {:uppercase=>8, :downcase=>28, :special=>10, :digit=>3}

通过使用默认值初始化哈希的键，有什么其他方法可以编写下面的代码？

4 个答案: