是的,它是 Is there a natural_sort_by method for Ruby?
的副本,但我认为 dawg 和 Eric 在这里说得更清楚,至少对我来说这些答案更详尽..
我有一个这样的数组:
arr = ["file1.txt", "file11.txt", "file12.txt", "file2.txt", "file3.txt"]
我希望它像这样排序:
arr = ["file1.txt", "file2.txt", "file3.txt", "file11.txt", "file12.txt"]
我该怎么做?我试过 sort
,但我不是很清楚..
我这样按大小对文件进行排序:
files=Dir.entries("./").sort { |f| File.size(f) }.select { |f| File.file?(f) }
Eric 的回答是仅对文件名中的数字执行 Natural Sort Order 的好方法。如果所有文件名都具有相同的前缀,则有效。
如果您想添加第二个元素(例如,其中没有数字的文件名),您可以使用多元素 sort_by 创建一个列表:
filenames = ["file1.txt", "file11.txt", "file12.txt", "file2.txt", "file3.txt","file.txt", "File.txt"]
filenames.sort_by{ |name| [name[/\d+/].to_i, name] }
=> ["File.txt", "file.txt", "file1.txt", "file2.txt", "file3.txt", "file11.txt", "file12.txt"]
sort_by
实现的两个元素:
- 如果用正则表达式找到数字的整数值
name[/\d+/].to_i
then
- 姓名,如果没有数字或相同的数字
name
。
更稳健的是,您可以按数字拆分整个字符串并将每个字符串转换为一个 int:
> "abc123def456gh".split(/(\d+)/).map{ |e| Integer(e) rescue e}
=> ["abc", 123, "def", 456, "gh"]
所以你的自然排序变成:
arr.sort_by{ |s| s.split(/(\d+)/).map{ |e| Integer(e) rescue e}}
现在名字和数字(甚至是名字和数字的倍数)都被正确处理了:
> arr = ["file1.txt", "file11.txt", "file12.txt", "file2.txt", "file3.txt", "gfile10.txt", "gfile1.txt", "gfile.txt", "file.txt", "afile.txt","afile10.txt","afile2.txt" ]
> arr.sort_by{ |s| s.split(/(\d+)/).map{ |e| Integer(e) rescue e}}
=> ["afile2.txt", "afile10.txt", "afile.txt", "file1.txt", "file2.txt", "file3.txt", "file11.txt", "file12.txt", "file.txt", "gfile1.txt", "gfile10.txt", "gfile.txt"]