title: bash, extract string before a colon
body: <p>If I have a file with rows like this</p>
  
  <pre><code>/some/random/file.csv:some string
  /some/random/file2.csv:some string2
  </code></pre>
  
  <p>Is there some way to get a file that only has the first part before the colon, e.g.</p>
  
  <pre><code>/some/random/file.csv
  /some/random/file2.csv
  </code></pre>
  
  <p>I would prefer to just use a bash one liner, but perl or python is also ok. </p>