0

我在我的应用程序中使用 OpenEars 框架来检测语音识别。这个想法是听几句话,然后根据所说的词执行一些动作。好吧,识别工作和单词被检测到,但问题是“SNAPPIE”这个词我遇到了一些问题——其他听起来不像的词也会被识别为“SNAPPIE”,有时即使我咳嗽。这是我用来设置监听器的一段代码:

private func setupOpenEars() {

    _openEarsEventsObserver = OEEventsObserver()
    _openEarsEventsObserver.delegate = self

    OEPocketsphinxController.sharedInstance().disablePreferredBufferSize = true

    do {
      try OEPocketsphinxController.sharedInstance().setActive(true)
    } catch let error {
      print("There was an error setting Pocketsphinx as active: \(error.localizedDescription)")
    }

    let grammarDictionary = ["ThisCanBeSaidOnce": [ "SNAPPIE", "TAKE SNAPSHOT" ]]
    let languageModelGenerator = OELanguageModelGenerator()
    let fileName = "FirstOpenEarsDynamicLanguageModel"
    let error = languageModelGenerator.generateGrammar(from: grammarDictionary, withFilesNamed: fileName, forAcousticModelAtPath: OEAcousticModel.path(toModel: "AcousticModelEnglish"))

    if let languageError = error {
      print("Dynamic language generator reported error: \(languageError.localizedDescription)")
    } else {
      _pathToGeneratedLanguageModel = languageModelGenerator.pathToSuccessfullyGeneratedLanguageModel(withRequestedName: fileName)
      _pathToGeneratedDictionary = languageModelGenerator.pathToSuccessfullyGeneratedDictionary(withRequestedName: fileName)
    }

    _correctPathToMyLanguageModelFile = "\(NSSearchPathForDirectoriesInDomains(FileManager.SearchPathDirectory.cachesDirectory, FileManager.SearchPathDomainMask.userDomainMask, true)[0])/\(fileName).gram"
    OEPocketsphinxController.sharedInstance().vadThreshold = 3.5
  }

是否有可能因为“SNAPPIE”实际上不是“有效”词而成为问题?

4

1 回答 1

0

您可以使用任何在线随机单词生成器工具创建一个随机单词列表,然后将您想要识别的特定关键字附加到词汇表中。这将提高您对特定关键字的识别度。您仍然需要将假设文本与您的关键字列表进行比较,但大多数情况下该解决方案已经足够好了。

于 2018-06-19T10:12:41.047 回答