python - 语音识别 Python 出现奇怪的请求错误

Question

使用以下代码进行语音识别根本不起作用

with sr.Microphone() as source:
# read the audio data from the default microphone
audio = r.record(source, duration=4)
print("Recognizing...")
# convert speech to text
# recognize speech using Google Speech Recognition
try:
    # for testing purposes, we're just using the default API key
    # to use another API key, use `r.recognize_google(audio, key="GOOGLE_SPEECH_RECOGNITION_API_KEY")`
    # instead of `r.recognize_google(audio)`
    print("Google Speech Recognition thinks you said in English: -  " + r.recognize_google(audio, language = "en-US"))
except sr.UnknownValueError:
    print("Google Speech Recognition could not understand audio")
except sr.RequestError as e:
    print("Could not request results from Google Speech Recognition service; {0}".format(e))

这是完整的错误，似乎请求只是失败了，但是如果我上传音频文件作为源，则相同的代码似乎可以正常工作。我已经检查过 sr.Microphone 并且默认选项也正确链接到我的实际麦克风......

    ---------------------------------------------------------------------------
HTTPError                                 Traceback (most recent call last)
~/anaconda3/lib/python3.7/site-packages/speech_recognition/__init__.py in recognize_google(self, audio_data, key, language, show_all)
    839         try:
--> 840             response = urlopen(request, timeout=self.operation_timeout)
    841         except HTTPError as e:

~/anaconda3/lib/python3.7/urllib/request.py in urlopen(url, data, timeout, cafile, capath, cadefault, context)
    221         opener = _opener
--> 222     return opener.open(url, data, timeout)
    223 

~/anaconda3/lib/python3.7/urllib/request.py in open(self, fullurl, data, timeout)
    530             meth = getattr(processor, meth_name)
--> 531             response = meth(req, response)
    532 

~/anaconda3/lib/python3.7/urllib/request.py in http_response(self, request, response)
    640             response = self.parent.error(
--> 641                 'http', request, response, code, msg, hdrs)
    642 

~/anaconda3/lib/python3.7/urllib/request.py in error(self, proto, *args)
    568             args = (dict, 'default', 'http_error_default') + orig_args
--> 569             return self._call_chain(*args)
    570 

~/anaconda3/lib/python3.7/urllib/request.py in _call_chain(self, chain, kind, meth_name, *args)
    502             func = getattr(handler, meth_name)
--> 503             result = func(*args)
    504             if result is not None:

~/anaconda3/lib/python3.7/urllib/request.py in http_error_default(self, req, fp, code, msg, hdrs)
    648     def http_error_default(self, req, fp, code, msg, hdrs):
--> 649         raise HTTPError(req.full_url, code, msg, hdrs, fp)
    650 

HTTPError: HTTP Error 400: Bad Request

During handling of the above exception, another exception occurred:

RequestError                              Traceback (most recent call last)
<ipython-input-109-50b94b08a896> in <module>
      3     audio = r.record(source, duration=4)
      4     print("Recognizing...")
----> 5     r.recognize_google(audio, language = "en-US")

~/anaconda3/lib/python3.7/site-packages/speech_recognition/__init__.py in recognize_google(self, audio_data, key, language, show_all)
    840             response = urlopen(request, timeout=self.operation_timeout)
    841         except HTTPError as e:
--> 842             raise RequestError("recognition request failed: {}".format(e.reason))
    843         except URLError as e:
    844             raise RequestError("recognition connection failed: {}".format(e.reason))

RequestError: recognition request failed: Bad Request

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Answer 1

这可能不是您编写的代码的解决方案。但会帮助许多其他尝试使用 Google 的人 SpeechRecognition并收到相同的模糊错误消息。
我通过将音频输入文件缩短为 less than 10 MB 解决了同样的问题.
目前，同步请求的配额约为 1 分钟 (10 MB)。
报价documentation :

There is a limit of 10 MB on all single requests sent to the APIusing local files. In the case of the Recognize andLongRunningRecognize methods, this limit applies to the size of therequest sent. In the case of the StreamingRecognize method, the 10 MB limit applies to both the initial StreamingRecognize request and the size of each individual message in the stream. Exceeding this limit will throw an error.