title: "SyntaxError: invalid syntax" New to coding and trying to learn, but here is what I tried
body: <p>This is the error I am getting:</p>
  <pre><code>  File &quot;&lt;ipython-input-30-56bb89b51cc1&gt;&quot;, line 4
      pass = 50 #@param {type:&quot;integer&quot;}
           ^
  SyntaxError: invalid syntax
  </code></pre>
  <p>I am using a Google Colab for VQGAN+CLIP AI Text to Image Generation: <a href="https://colab.research.google.com/drive/1go6YwMFe5MX6XM9tv-cnQiSTU50N9EeT?usp=sharing" rel="nofollow noreferrer">https://colab.research.google.com/drive/1go6YwMFe5MX6XM9tv-cnQiSTU50N9EeT?usp=sharing</a>
  and the code with the error is not included in the Colab, but is found here: <a href="https://tuscriaturas.miraheze.org/wiki/Ayuda:Generar_im%C3%A1genes_con_VQGAN%2BCLIP/English#Code_for_generating_the_zip" rel="nofollow noreferrer">https://tuscriaturas.miraheze.org/wiki/Ayuda:Generar_im%C3%A1genes_con_VQGAN%2BCLIP/English#Code_for_generating_the_zip</a>
  The default code is supposed to save every 50th image to a .zip automatically. Deleting the line that gives the error saves every single image, which in this particular case is 1,350 images. I tried looking up this issue, but my lack of knowledge doesn't help and I can't find an example of this exact issue. I tried changing 'pass' to 'intervals' thinking that would work for some reason and it just ignored that line entirely. I hope I gave enough detail and I apologize if this is a stupid issue but I am trying to learn and have hit a wall. I appreciate any help. Here is the code for the entire cell.</p>
  <pre><code># @title Create a zip with all of the images (or some)
  initial_image = 0 #@param {type:&quot;integer&quot;}
  final_image = 1350 #@param {type:&quot;integer&quot;}
  pass = 50 #@param {type:&quot;integer&quot;}
  filename = &quot;files.zip&quot; #@param {type:&quot;string&quot;}
  
  import zipfile
  
  if final_image == -1:
    final_image = i
  
  zipf = zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED)
  for i in tqdm(range(initial_image, final_image+1)):
    fname = f&quot;{i:04}.png&quot;
    zipf.write(f&quot;steps/{fname}&quot;, f&quot;{filename.split('.')[0]}-{fname}&quot;)
  zipf.close()
  print(filename, &quot;created. Downloading… A download dialog will be open when the download is ready.&quot;)
  from google.colab import files
  files.download(filename)
  </code></pre>