text: user7056: I have a data.frame with results from a CRISPR screen.
  I want to calculate the % enrichment of a variable of interest var from baseline to treatment group.
  Here is the distribution of var in linear:
  
  I can do scale var with log transformation
  
  But many rows with var = 0 will be lost.
  Moreover, I cannot compare % enrichment by dividing treatmentgroup::var to Control/baseline::var because I will run into the problem of dividing by zero.
  I've seen people circumvent this problem by adding a small value ex)0.01 to the entire column of var.
  But I'm wondering if there's a better way of circumventing this problem. For example, a simple transformation that can deal with zeros and negative values.
  Appreciate any input!
  
  user8641: Cube root transformation is another way to deal with zeros without dropping them.
  var^(1/3)
  
  EDIT as Michael pointed out, square root transformation is another--perhaps more common--transformation that leaves zeros as zeros.
  sqrt(var)
  
  
original_text: 7056: <p>I have a <code>data.frame</code> with results from a CRISPR screen.</p>
  
  <p>I want to calculate the % enrichment of a variable of interest <code>var</code> from baseline to treatment group.</p>
  
  <p>Here is the distribution of <code>var</code> in linear:
  <a href="https://i.stack.imgur.com/cDOXS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cDOXS.png" alt="enter image description here"></a></p>
  
  <p>I can do scale <code>var</code> with log transformation
  <a href="https://i.stack.imgur.com/tzXXW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tzXXW.png" alt="enter image description here"></a></p>
  
  <p>But many rows with <code>var</code> = 0 will be lost.
  Moreover, I cannot compare % enrichment by dividing <code>treatmentgroup::var</code> to <code>Control/baseline::var</code> because I will run into the problem of dividing by zero.</p>
  
  <p>I've seen people circumvent this problem by adding a small value ex)0.01 to the entire column of <code>var</code>.
  But I'm wondering if there's a better way of circumventing this problem. For example, a simple transformation that can deal with zeros and negative values.</p>
  
  <p>Appreciate any input!</p>
   8641: <p>Cube root transformation is another way to deal with zeros without dropping them.</p>
  
  <pre><code>var^(1/3)
  </code></pre>
  
  <p><strong>EDIT</strong> as Michael pointed out, square root transformation is another--perhaps more common--transformation that leaves zeros as zeros.</p>
  
  <pre><code>sqrt(var)
  </code></pre>
  