Ruby在不使用正则表达式的序列中计算字符 [英] Ruby Counting chars in a sequence not using regex

问题描述

此代码在计数序列中的字符数方面需要帮助。

Need help with this code on counting chars in a sequence.

这就是我想要的：

word("aaabbcbbaaa") == [["a", 3], ["b", 2], ["c", 1], ["b", 2], ["a", 3]]
word("aaaaaaaaaa") == [["a", 10]]
word("") == []

这是我的代码：

def word(str)
words=str.split("")
count = Hash.new(0)

words.map {|char| count[char] +=1 }

return count
end

我得到了word（ aaabbcbbaaa）=> [[ a，6]，[ b，4]，[ c，1]]，这不是我想要的。我想计算每个序列。我更喜欢无正则表达式的解决方案。谢谢。

I got word("aaabbcbbaaa") => [["a", 6], ["b", 4], ["c", 1]], which is not what I want. I want to count each sequence. I prefer a none regex solution. Thanks.

推荐答案

按字符分割字符串，然后按字符分组，然后按块计数字符：

Split string by chars, then group chunks by char, then count chars in chunks:

def word str
  str
  .chars
  .chunk{ |e| e }
  .map{|(e,ar)| [e, ar.length] }
end

p word "aaabbcbbaaa"
p word("aaaaaaaaaa")
p word ""

结果：

[["a", 3], ["b", 2], ["c", 1], ["b", 2], ["a", 3]]
[["a", 10]]
[]

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