title: How can I select Python 3.7 or Python 3.8 in iPython?
body: <p>My OS is <em>Windows 7</em></p>
  
  <p>I need to use <em>TensorFlow</em> with <em>Keras</em> in <em>Python</em>, but it doesn't work yet in <em>Python 3.8</em>. So I've installed again <em>Python 3.7</em> and his packages. </p>
  
  <p>Inside the IDE <em>PyCharm</em> was easy. It just add another interpreter in settings and it works. So when I create a project I need to select the right interpreter. </p>
  
  <p>I also use <em>iPython</em> to make small tests. I've searched in all <em>Stackoverflow</em> and I don't find any answer that works in 2020.  I use <em>iPython</em> and not the <em>Jupyter</em> environment. </p>
  
  <p>First, I've created the kernels, as iPython instructions said in its site. </p>
  
  <pre><code>python -m ipykernel install --user
  </code></pre>
  
  <p>It creates a <code>json</code> file in  <code>c:\Users\myname\AppData\Roaming\jupyter\kernels</code></p>
  
  <p>After I use <code>iPython -h</code> and he told me to use <em>kernel</em> subcommand:</p>
  
  <pre><code>Subcommands are launched as `ipython cmd [args]`. For information on using
  subcommand 'cmd', do: `ipython cmd -h`.
  </code></pre>
  
  <p>Feeling more motivated, I've tried the command <code>iPython kernel -h</code>. But I've got a very confusing text.</p>
  
  <p>So I've tried to use the <code>kernel</code> option,</p>
  
  <pre><code>C:&gt; iPython kernel
  </code></pre>
  
  <p>It gave me the following message:</p>
  
  <pre><code>To connect another client to this kernel, use:
      --existing kernel-3404.json
  </code></pre>
  
  <p>However, I'm not getting select the right kernel. </p>
  
  <p>I try to use use my kernel file, but with no success. </p>
  
  <pre><code>Invalid argument: '--kernel.json'
  </code></pre>
  
  <p>So I'm lost, because in no place I've found a detailed instructions for kernel subcommand in line command options for <em>iPython</em>.</p>