title: provide a url to s3.upload_file()
body: <p>Currently I have the process of uploading files to s3 broken down to two parts:</p>
  <ol>
  <li>download the file locally :</li>
  </ol>
  <pre><code>#downloads file from a url to local
  def download_url(url, save_path):  
      r = requests.get(url, stream=True)
      with open(save_path, 'wb') as fd:
          for chunk in r.iter_content(chunk_size=1024 * 1024):
              fd.write(chunk)
  </code></pre>
  <ol start="2">
  <li>upload to s3 :</li>
  </ol>
  <pre><code>def create_aws_resource(aws_access_key_id, aws_secret_access_key, aws_default_region):
      s3 = boto3.client('s3', aws_access_key_id=aws_access_key_id,aws_secret_access_key=aws_secret_access_key)
      return s3
  </code></pre>
  <pre><code>def save_to_s3(bucket,path,save_path,link_save_path, aws_access_key_id, aws_secret_access_key, aws_default_region):
     s3 = create_aws_resource(aws_access_key_id, aws_secret_access_key, aws_default_region)
     os.chdir(save_path)
     filename = link_save_path.split('/')[-1]
     s3_path = path+'/'+filename
     s3.upload_file(link_save_path, bucket, s3_path)
  </code></pre>
  <p>This process needs to be run on a server and hence I want to eliminate the first step of downloading locally.
  Is there a way to directly store files (any files - .xlsx, .pdf, .ods, .xls, .html) to s3?
  example : want to upload the file on <a href="http://seriestoriche.istat.it/fileadmin/documenti/Tavola_17.1.xls" rel="nofollow noreferrer">http://seriestoriche.istat.it/fileadmin/documenti/Tavola_17.1.xls</a> to s3 directly.
  Any help would be highly appreciated !!</p>