Is it possible to limit SFSpeechRecognizer's vocabulary?

Question

I'd like to use the ios SFSpeechRecognizer to recognize a closed set of words and phrases. Is there a way to specify these and remove all other possibilities? I cannot seem to find a way to do this.

score 3 · Answer 1 · answered Oct 13 '21 at 14:49

You can try to use contextualStrings property of SFSpeechAudioBufferRecognitionRequest to set high priority or to set some words that are not in vocabulary. For more info look here: https://developer.apple.com/documentation/speech/sfspeechrecognitionrequest/1649391-contextualstrings?language=objc

  SFSpeechAudioBufferRecognitionRequest *recognitionRequest = [[SFSpeechAudioBufferRecognitionRequest alloc] init];
    recognitionRequest.contextualStrings = @[@"Your specific word", @"short custom phrases that are unique to your app"];

score 1 · Answer 2 · answered Oct 08 '20 at 11:07

1

I think you cannot limit SFSpeechRecognizer's vocabulary, but you can fire action only if the words you want are recognized.

When you get your output, you can try something like this:

self.recognitionTask = speechRecognizer?.recognitionTask(with: recognitionRequest, resultHandler: { (result, error) in
    if result == "Desired word one" {
        // Do something
    } else if result == "Desired word two" {
        // Do something else
    } else {
        // Don't do anything
    }
})

answered Oct 08 '20 at 11:07

Giuseppe Garassino

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That’s too bad. My intention was to make certain words likelier to be candidates, not to ignore transcriptions containing out-of-vocabulary words. – synchronizer Oct 08 '20 at 23:21
Do you mean you want a personal vocabulary that contains "special" words like "expecto patronum"? – Giuseppe Garassino Oct 09 '20 at 07:07
No, and SFSpeechRecognizer is terrible at doing that even if it has that feature supposedly. I mean that if I had control over what algorithm they used to recognize words, supposing they had several candidates for a result, I’d want to make some of the candidates non-existent. For example, the recognizer is bad at math terms. “N numbers” yet erroneously recognizes words as company names I don’t care about without any way of knowing what other choices there were. I think there’s no way to guide the algorithm internally though. It is what it is. – synchronizer Oct 10 '20 at 15:42
I see... it would be very useful, but I don't think it is possible. Anyway, have you tried SiriKit? https://developer.apple.com/documentation/sirikit I never used it, but I remember I read somewhere you could have personal vocabulary. – Giuseppe Garassino Oct 11 '20 at 03:56
It won’t work for my purposes unfortunately. Thanks anyway. – synchronizer Oct 11 '20 at 15:23

score 0 · Answer 3 · answered Feb 27 '23 at 11:06

After some trials and error, the best solution is a mix between Mrsantateam and Giuseppe Garassino's answers.

you can first specify all commands that should be "prioritized" by setting contextualStrings...

var contextualString = ["hello world", "goodbye world"]
recognitionRequest.contextualStrings = contextualString

... And later filter those results in recognitionTask :

recognitionTask = speechRecognizer.recognitionTask(with: recognitionRequest) { result, error in
    var isFinal = false
    
    if let result = result {                
        for transcription in result.transcriptions {
            print("Pondering \(result.bestTranscription.formattedString)")
            if contextualString.contains(transcription.formattedString.lowercased()) {
                print("Found \(transcription.formattedString)")
            }
        }
    }
    
    //...
}

Sample code can be found at https://developer.apple.com/documentation/speech/recognizing_speech_in_live_audio if you want to play with it and test its limits

Is it possible to limit SFSpeechRecognizer's vocabulary?

3 Answers3