-1

我想做的比我的问题要复杂一些，但我无法用语言表达我想要做什么，而且我的术语可能是错误的。对不起。如果我之后可以编辑我的问题的标题以更好地代表我想做的事情，我会这样做。

我有一个看起来像这样的数组：

array = [
  ["bnb ><b>(BASE CABINET)</b>", "bnb ><b>(BASE CABINET)</b>", "1", "22\"", "22\"", "22\""]
  ["bnb >Edgebanding (sides)", "bnb >Edgebanding (sides)", "2", "22\"", "1/2\"", "1/2\""]
  ["bnb >Edgebanding (top/bottom)", "bnb >Edgebanding (top/bottom)", "2", "21\"", "1/2\"", "1/2\""]
  ["bnb >Panel (back)", "bnb >Panel (back)", "1", "21\"", "21\"", "1/2\""]
  ["bnb >Panel (sides)", "bnb >Panel (sides)", "2", "21 1/2\"", "21\"", "1/2\""]
  ["bnb >Panel (top/bottom)", "bnb >Panel (top/bottom)", "1", "22\"", "21 1/2\"", "1/2\""]
  ["bnb >Top (front/back)", "bnb >Top (front/back)", "2", "22\"", "3\"", "1/2\""]
  ["bnb >Top (sides)", "bnb >Top (sides)", "2", "15 1/2\"", "3\"", "1/2\""]
]

我想要做的是得到这个结果，基本上是>在每个子数组的第一个元素中删除所有内容，并>在子数组的第二个元素中删除之前的所有内容：

array = [
  ["bnb", "<b>(BASE CABINET)</b>", "1", "22\"", "22\"", "22\""]
  ["bnb", "Edgebanding (sides)", "2", "22\"", "1/2\"", "1/2\""]
  ["bnb", "Edgebanding (top/bottom)", "2", "21\"", "1/2\"", "1/2\""]
  ["bnb", "Panel (back)", "1", "21\"", "21\"", "1/2\""]
  ["bnb", "Panel (sides)", "2", "21 1/2\"", "21\"", "1/2\""]
  ["bnb", "Panel (top/bottom)", "1", "22\"", "21 1/2\"", "1/2\""]
  ["bnb", "Top (front/back)", "2", "22\"", "3\"", "1/2\""]
  ["bnb", "Top (sides)", "2", "15 1/2\"", "3\"", "1/2\""]
]

到目前为止，这是我所拥有的代码：

val = 0
@test_list = []
test.each do |comp|
  new_list = test[val]
  new_list.map! { |element| element.gsub(/\s\>.*/, '') }
  comp = comp[0]
  @test_list.push(new_list)
  val = val + 1
end

它删除了之后的所有内容，>但它在第一个和第二个元素中都这样做了。

4

2 回答 2

1

遇到问题时无需使用正则表达式:) 有时一个简单的String#split就可以了。

source_array = [
  ["bnb ><b>(BASE CABINET)</b>", "bnb ><b>(BASE CABINET)</b>", "1", "22\"", "22\"", "22\""],
  ["bnb >Edgebanding (sides)", "bnb >Edgebanding (sides)", "2", "22\"", "1/2\"", "1/2\""],
  ["bnb >Edgebanding (top/bottom)", "bnb >Edgebanding (top/bottom)", "2", "21\"", "1/2\"", "1/2\""],
  ["bnb >Panel (back)", "bnb >Panel (back)", "1", "21\"", "21\"", "1/2\""],
  ["bnb >Panel (sides)", "bnb >Panel (sides)", "2", "21 1/2\"", "21\"", "1/2\""],
  ["bnb >Panel (top/bottom)", "bnb >Panel (top/bottom)", "1", "22\"", "21 1/2\"", "1/2\""],
  ["bnb >Top (front/back)", "bnb >Top (front/back)", "2", "22\"", "3\"", "1/2\""],
  ["bnb >Top (sides)", "bnb >Top (sides)", "2", "15 1/2\"", "3\"", "1/2\""]
]

target_array = [
  ["bnb", "<b>(BASE CABINET)</b>", "1", "22\"", "22\"", "22\""],
  ["bnb", "Edgebanding (sides)", "2", "22\"", "1/2\"", "1/2\""],
  ["bnb", "Edgebanding (top/bottom)", "2", "21\"", "1/2\"", "1/2\""],
  ["bnb", "Panel (back)", "1", "21\"", "21\"", "1/2\""],
  ["bnb", "Panel (sides)", "2", "21 1/2\"", "21\"", "1/2\""],
  ["bnb", "Panel (top/bottom)", "1", "22\"", "21 1/2\"", "1/2\""],
  ["bnb", "Top (front/back)", "2", "22\"", "3\"", "1/2\""],
  ["bnb", "Top (sides)", "2", "15 1/2\"", "3\"", "1/2\""]
]

arr = source_array.map do |a|
  [
   a[0].split('>').first.strip,
   a[1].split('>', -1)[1..-1].join('>')
  ] + a[2..-1]
end

arr == target_array # => true
于 2012-04-14T02:08:14.220 回答
0

Array.each_with_index可能是你想要的。通过索引检查，我们可以确定是第一个还是第二个

arr=[
["bnb ><b>(BASE CABINET)</b>", "bnb ><b>(BASE CABINET)</b>", "1", "22\"", "22\"", "22\""],
["bnb >Edgebanding (sides)", "bnb >Edgebanding (sides)", "2", "22\"", "1/2\"", "1/2\""],
["bnb >Edgebanding (top/bottom)", "bnb >Edgebanding (top/bottom)", "2", "21\"", "1/2\"", "1/2\""],
["bnb >Panel (back)", "bnb >Panel (back)", "1", "21\"", "21\"", "1/2\""],
["bnb >Panel (sides)", "bnb >Panel (sides)", "2", "21 1/2\"", "21\"", "1/2\""],
["bnb >Panel (top/bottom)", "bnb >Panel (top/bottom)", "1", "22\"", "21 1/2\"", "1/2\""],
["bnb >Top (front/back)", "bnb >Top (front/back)", "2", "22\"", "3\"", "1/2\""],
["bnb >Top (sides)", "bnb >Top (sides)", "2", "15 1/2\"", "3\"", "1/2\""]
]

new_arr = 
arr.map do |sub_arr|
    sub_arr.each_with_index.map do |str,i|
        if i == 0
            str.gsub(/\s\>.*/,'')
        elsif i == 1
            str.gsub('bnb >','')
        else
            str
        end
    end
end

p new_arr

这是一个有效的初步解决方案。你总是可以把它变成你自己的风格，例如，将 do end 更改为大括号，使用 map！，将 if 语句更改为同一行等。

于 2012-04-14T02:15:57.550 回答