Ruby 算法删除元音，意外结果

我正在尝试构建一个算法，从字符串中删除元音。下面的代码是我到目前为止的代码

def shortcut(s)
  s = s.split("")
  for i in 0..s.length - 1 do
    if ["a","e","i","o","u"].include?(s[i])
      s.delete_at(i)
    end
  end
  s.join
end


puts shortcut("hello world, we are the champions")
# => hll wrld, w r th chmpons

---

为什么没有从字符串中删除

'o'

？

在对数组进行迭代时，正在从数组

s

中删除元素。这会给你带来意想不到的结果

改用：

---

通过删除字符，索引将无效

hello
  ^    before delete
01234
  v    after delete
hllo

您可以从头到脚迭代来解决它

def shortcut(s)
  s = s.split("")
  (s.length - 1).downto(0) do |i|
    if ["a","e","i","o","u"].include?(s[i])
      s.delete_at(i)
    end
  end
  s.join
end

但更好的解决方案是使用字符串替换方法，如

---

在从数组中删除元素时，迭代器i可以大于数组末尾之前的大小。

您的方法可以是

def shortcut(s)
  s = s.split("")
  for i in 0..s.length - 1 do
    s.delete(s[i]) if ["a","e","i","o","u"].include? s[i]
  end
  s.join
end


> shortcut("hello world, we are the champions")
=> "hll wrld, w r th chmpns"

另一种方式

> s = "hello world, we are the champions"
> (s.split("") - s.scan(/[aeiou]/)).join
=> "hll wrld, w r th chmpns"

最短路径

def shortcut(s)
  s.gsub(/[aeiou]/i, '')
end

shortcut('hello world, we are the champions')
#=> "hll wrld, w r th chmpns"

您也可以使用

---

如果研究以下简化示例，您将理解为什么删除要迭代的数组的元素会导致获得的结果

s = "two lions"
s = s.split("")
puts "s.split=#{s}"
for i in 0..s.length - 1 do
  puts "i=#{i}"
  str = s[i]
  puts "  str=/#{str}/"
  if ["a","e","i","o","u"].include?(s[i])
    puts "  str is a vowel"
    s.delete_at(i)
    puts "  s after delete_at(#{i})=#{s}"
  end
end
puts "s after loop=#{s}"
puts "s.join=#{s.join}"
s.join

印刷品

s.split=["t", "w", "o", " ", "l", "i", "o", "n", "s"]
i=0
  str=/t/
i=1
  str=/w/
i=2
  str=/o/
  str is a vowel
  s after delete_at(2)=["t", "w", " ", "l", "i", "o", "n", "s"]

跳过该空间

i=3
  str=/l/
i=4
  str=/i/
  str is a vowel
  s after delete_at(4)=["t", "w", " ", "l", "o", "n", "s"]

“o”现在位于索引4，“n”位于索引5

i=5
  str=/n/
i=6
  str=/s/
i=7
  str=//
i=8
  str=//
s after loop=["t", "w", " ", "l", "o", "n", "s"]
s.join=tw lons

---

我很感激你给我展示了解决这个问题的新方法。这很好，现在我知道如何用不同的方式来做了。但这仍然不能解释为什么我会得到意外的结果。@TodLazarov原因是我的第一句话：你在迭代数组时从数组

s

中删除元素。@YuHao严格地说，结果是预期的：实际字符串的第n个字母。这是正确的：）感谢您花时间回答我的问题。

compact

在这里完全没有意义，除非您期望二进制输入，否则您的解决方案无论如何都会失败。谢谢！我学到了一种新方法：）即使在这种情况下它不起作用。

s.split=["t", "w", "o", " ", "l", "i", "o", "n", "s"]
i=0
  str=/t/
i=1
  str=/w/
i=2
  str=/o/
  str is a vowel
  s after delete_at(2)=["t", "w", " ", "l", "i", "o", "n", "s"]

i=3
  str=/l/
i=4
  str=/i/
  str is a vowel
  s after delete_at(4)=["t", "w", " ", "l", "o", "n", "s"]

i=5
  str=/n/
i=6
  str=/s/
i=7
  str=//
i=8
  str=//
s after loop=["t", "w", " ", "l", "o", "n", "s"]
s.join=tw lons