title: awk select rows that has column 1 partially matching column 2
body: <p>I have an &quot;example_file&quot; like below,</p>
  <pre><code>a2022.1 a2022.1 80
  a2022.1 a2022.2 90
  a2022.1 a2023.1 80
  a2022.2 a2022.2 90
  a2022.2 a2023.1 40
  a2022.3 a2022.1 50
  b20225.1 a2022.1 80
  </code></pre>
  <p>and I would like to select lines which has the third column &gt; 80 and also has the 1st column (before the dot) different from the 2nd column (before the dot).</p>
  <p>So the desired result would be</p>
  <pre><code>a2022.1 a2023.1 80
  b20225.1 a2022.1 80
  </code></pre>
  <p>So I already managed to exclude the lines with 3rd column &lt;80 and with the same 1st and 2nd column by using the line below</p>
  <pre><code>awk '($3 &gt; 80 &amp;&amp; $1!=$2)' example_file
  </code></pre>
  <p>and I tried <code>awk '($3 &gt; 80 &amp;&amp; $1!~$2)' example_file</code>, but it does not exclude the line</p>
  <pre><code>a2022.1 a2022.2 90
  </code></pre>
  <p>I thought it could be easy, but I can't figure it out. Could somebody help? Thanks!</p>