从嵌套数组中的字符串中删除字母并将其转换为整数

I have this:

我有这个：

[["a3"], ["b3"], ["c7", "c9"]]

I need to remove the letter from the strings and convert them into integers. I need:

我需要从字符串中删除字母并将它们转换为整数。我需要：

[[3], [3], [7, 9]]

I tried:

我试过了：

 [["a3"], ["b3"], ["c7", "c9"]].each do |a| 
   a.map do |string| 
     puts leave_num = string.slice!(0) 
     puts leave_num.to_i
   end  
 end

but I am sure there is a nicer way.

但我相信有更好的方法。

3 个解决方案

#1

a = [["a3"], ["b3"], ["c7", "c9"]]

a.map { |r| r.map { |e| e[/\d+/].to_i } }
# => [[3], [3], [7, 9]]

#2

One way to do this can be:

一种方法是：

a = [["a3"], ["b3"], ["c7", "c9"]]
a.map { |b| b.map { |c| c.scan(/\d+/)[0].to_i }}
# => [[3], [3], [7, 9]]

Basically I'm going through each element and returning integers using regex.

基本上我正在浏览每个元素并使用正则表达式返回整数。

#3

You can use in your code:

您可以在您的代码中使用：

string[0]  = ''
puts string  # instead puts leave_num = string.slice!(0)  
puts string.to_i  # instead puts leave_num.to_i

another option:

另外一个选择：

a.map{|e| e.map{|n| n[1..-1].to_i}}
 #=> [[3], [3], [7, 9]]

If your array elements has more letters like "d58as9a" so for such case try this although @Amadan and @Shivam's answers are correct:

如果您的数组元素有更多的字母，如“d58as9a”，那么对于这种情况尝试这个虽然@Amadan和@ Shivam的答案是正确的：

a = [["a3"], ["b3"], ["c7", "c9"], ["d58as9a", "d5d54d"] ]
 > a.map{|e| e.map{|n| n.scan(/\d/).join('').to_i}}
 => [[3], [3], [7, 9], [589, 554]]

Note: This will extract all digits from string.

注意：这将从字符串中提取所有数字。

#1

a = [["a3"], ["b3"], ["c7", "c9"]]

a.map { |r| r.map { |e| e[/\d+/].to_i } }
# => [[3], [3], [7, 9]]

#2