使用Ruby的句子中的字母数

Question

假设我有一个字符串"I am a good boy".我想要字符串中每个字母的总数.字母区分大小写.也就是说,D和d需要被视为两个不同的字符.

Answer 1

"I am a good boy".scan(/\w/).inject(Hash.new(0)){|h, c| h[c] += 1; h}
# => {"I"=>1, "a"=>2, "m"=>1, "g"=>1, "o"=>3, "d"=>1, "b"=>1, "y"=>1}

Answer 2

a = "I am a good boy"

a.chars.group_by(&:chr).map { |k, v| [k, v.size] }

Answer 3

这不是一个答案,只是对现有答案的补充.

因为讨论了性能,所以这里有一些数据.

require 'benchmark'

s0 = "I am a good boy"
s = s0 * 1
N = 10000

Benchmark.bm(20) do | x |
  x.report('sawa') do
    N.times { s.scan(/\w/).inject(Hash.new(0)){|h, c| h[c] += 1; h} }
  end

  x.report('digitalross') do 
    N.times { s.chars.to_a.sort.group_by(&:chr).map { |k, v| [k, v.size] } }
  end

  x.report("digitalross'") do 
    N.times { s.chars.group_by(&:chr).map { |k, v| [k, v.size] } }
  end

  x.report('rubylovely') do
    N.times { s.gsub(/\s/,'').chars.with_object({}) {|c,ob| ob[c] = s.count(c)} }
  end
end

给(ruby 1.9.3p392在我的机器上)

                           user     system      total        real
sawa                   0.600000   0.000000   0.600000 (  0.601734)
digitalross            0.790000   0.000000   0.790000 (  0.806674)
digitalross'           0.640000   0.010000   0.650000 (  0.651802)
rubylovely             0.570000   0.000000   0.570000 (  0.572501)

有了s = s0 * 1000,N = 10我得到了

                           user     system      total        real
sawa                   0.340000   0.000000   0.340000 (  0.340617)
digitalross            0.380000   0.000000   0.380000 (  0.411393)
digitalross'           0.230000   0.010000   0.240000 (  0.243389)
rubylovely             6.530000   0.000000   6.530000 (  6.603198)

所以对于非常短的字符串,RubyLovely解决方案的多次计数并没有受到伤害.实际上,它确实如此.

Answer 4

我用的是:

str = "I am a good boy"
str.scan(/[[:alpha:]]/i).each_with_object(Hash.new(0)) { |c, h| h[c] += 1 }

哪个回报:

{
    "I" => 1,
    "a" => 2,
    "m" => 1,
    "g" => 1,
    "o" => 3,
    "d" => 1,
    "b" => 1,
    "y" => 1
}

我更喜欢使用,scan因为它使用的正则表达式立即确定允许计算哪些字符.如果输入的字符串包含"我是个好孩子".其他一些解决方案会失败,因为它们对输入字符串太具体,并且每次遇到意外字符时都需要调整.我们知道输入很少是无效的,在现实世界中,这种代码将被用于一般句子.预先忽略不需要的角色很重要.

'Français'.scan(/[[:alpha:]]/i).each_with_object(Hash.new(0)) { |c, h| h[c] += 1 }

哪个回报:

{
    "F" => 1,
    "r" => 1,
    "a" => 2,
    "n" => 1,
    "ç" => 1,
    "i" => 1,
    "s" => 1
}