Sunday, April 5, 2009

Succinctly Groovy

I was reading a post on displaying Google Visualization charts in Grails on mrhaki's blog and noticed something interesting. In his code, mrhaki wanted to get a handle on the newest (most recently modified) file in a directory. His approach was to list the files in the directory, sort them (in ascending order), reverse that, and then get the zeroth item:
def xmlFile = dir.listFiles().sort{ file -> file.lastModified() }.reverse()[0]
It's a nice example of how you can express a somewhat complex flow in a single line of Groovy. Still, it seemed like overkill to me and another reader who failed to identify him or herself. That reader linked to the Groovy Collections page and suggested using a Comparator that sorted the list in descending order to start with, but didn't include a code snippet. I liked the fact that this approach avoids the extra work to reverse the sorted list, but I wasn't sure it would be possible to express it so compactly and it still seemed more complicated than necessary. I looked over the Collections page and was reminded of the min(Closure) and max(Closure) methods available on Groovy's enhanced collections. So, rather than sorting the list of files, you can just ask for the max as determined by last modified date:
def xmlFile = dir.listFiles().max{ it.lastModified() }
Nice and simple, eh? Ignoring the overhead of imports and class/method declaration, neither of which are necessary with the Groovy snippet, here's the most compact way I know of expressing this using the Java standard libraries:
File xmlFile = Collections.max(Arrays.asList(dir.listFiles()), new Comparator<File>() {
  public int compare(File file1, File file2) {
    return (int)(file1.lastModified() - file2.lastModified());
I'm glad Groovy's out there!


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