我无法通过Vision API的安全搜索/显式内容检测从URL运行图像。Python示例可在此处找到:
https://github.com/googleapis/python-vision/blob/HEAD/samples/snippets/detect/detect.py
如果我将下面的内容保存在python文件中,那么运行它的最佳方式是什么?我试过了!python detect.py安全搜索urihttp://www.photos-public-domain.com/wp-content/uploads/2011/01/old-vw-bug-and-van.jpg但它不起作用。也许我遗漏了一些代码或者运行方式不对
来自上面github的示例代码:
from google.cloud import vision
client = vision.ImageAnnotatorClient()
image = vision.Image()
image.source.image_uri = uri
response = client.safe_search_detection(image=image)
safe = response.safe_search_annotation
# Names of likelihood from google.cloud.vision.enums
likelihood_name = ('UNKNOWN', 'VERY_UNLIKELY', 'UNLIKELY', 'POSSIBLE',
'LIKELY', 'VERY_LIKELY')
print('Safe search:')
print('adult: {}'.format(likelihood_name[safe.adult]))
print('medical: {}'.format(likelihood_name[safe.medical]))
print('spoofed: {}'.format(likelihood_name[safe.spoof]))
print('violence: {}'.format(likelihood_name[safe.violence]))
print('racy: {}'.format(likelihood_name[safe.racy]))
if response.error.message:
raise Exception(
'{}\nFor more info on error messages, check: '
'https://cloud.google.com/apis/design/errors'.format(
response.error.message))
如果您只是执行了问题中包含的代码段,那么您没有正确地将值
uri
传递给代码。您需要解析在python命令上传递的参数。可以通过添加argparse来实现这一点使用命令
!python detect.py safe-search-uri http://www.photos-public-domain.com/wp-content/uploads/2011/01/old-vw-bug-and-van.jpg
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