text: user319: edgeR's cpm function has an argument called prior.count. Based on my understanding of the documentation, it is supposed to be adding a fixed number per sample which is proportional to the library size of said sample. Average of all numbers added to all samples would be equal to prior.count.
  However, looking at actual data this does not seem to be the case. Given an imaginary data frame of 
  df = data.frame(a = c(1,2,3,0),c = c(1,2,3,0)*3, b = c(1,2,3,0)*2)
  
  We can try to calculate log cpms by doing
  logCPM = cpm(df,log = TRUE,prior.count = .5)
  
  We can also calculate regular cpms by doing 
  CPM = cpm(df)
  
  If we wanted to see, what is the number that is added to the CPMs before getting logged, we could do
  difference = 2^logCPM - CPM
  
  To make it look pretty lets use tibble
  tibble::as.tibble(difference)
  
  # A tibble: 4 x 3
                       a                    c                    b
                   <dbl>                <dbl>                <dbl>
  1  25641                25641                25641              
  2  12821                12821                12821              
  3 -    0.0000000000582 -    0.0000000000582 -    0.0000000000582
  4  38462                38462                38462   
  
  Here we see that the number that is added has no relation to the library size of the sample. What I want to learn is, how is this number that is added to each cpm is calculated based on prior.count. 
  I have been digging through the code but that part goes into C++ territory which hinders my understanding of what is going on 
  
  user77: The prior count ends up getting scaled by the ratio of a library size to the average library size and then multiplied by 2 before getting added to each library size (I'm sure there's a good reason for that, but I don't know what it is). Using the example df data frame in your post, let's walk step-by-step through what cpm() is doing:
  df = data.frame(a = c(1,2,3,0),c = c(1,2,3,0)*3, b = c(1,2,3,0)*2)
  prior.count = 0.5
  # First, we need to calculate a library size
  lib.size = colSums(df)
  # Calculate the average library size and the adjusted priors
  ave.lib = mean(lib.size)
  adjusted.prior = prior.count * lib.size / ave.lib
  # Update the library sizes
  adjusted.lib.size = lib.size + 2*adjusted.prior
  # Now we can compute the CPM
  customCPM = t((log(t(df) + adjusted.prior) - log(adjusted.lib.size) + log(1000000))/log(2))
  
  The matrix transposition looks really ugly in the last step, but it's needed so R adds things like adjusted.prior to each row of df.
  Note that the prior count is only used if you compute log(CPM), it's unused otherwise.
  
original_text: 319: <p>edgeR's cpm function has an argument called <code>prior.count</code>. Based on my understanding of the documentation, it is supposed to be adding a fixed number per sample which is proportional to the library size of said sample. Average of all numbers added to all samples would be equal to <code>prior.count</code>.</p>
  
  <p>However, looking at actual data this does not seem to be the case. Given an imaginary data frame of </p>
  
  <pre><code>df = data.frame(a = c(1,2,3,0),c = c(1,2,3,0)*3, b = c(1,2,3,0)*2)
  </code></pre>
  
  <p>We can try to calculate log cpms by doing</p>
  
  <pre><code>logCPM = cpm(df,log = TRUE,prior.count = .5)
  </code></pre>
  
  <p>We can also calculate regular cpms by doing </p>
  
  <pre><code>CPM = cpm(df)
  </code></pre>
  
  <p>If we wanted to see, what is the number that is added to the CPMs before getting logged, we could do</p>
  
  <pre><code>difference = 2^logCPM - CPM
  </code></pre>
  
  <p>To make it look pretty lets use tibble</p>
  
  <pre><code>tibble::as.tibble(difference)
  
  # A tibble: 4 x 3
                       a                    c                    b
                   &lt;dbl&gt;                &lt;dbl&gt;                &lt;dbl&gt;
  1  25641                25641                25641              
  2  12821                12821                12821              
  3 -    0.0000000000582 -    0.0000000000582 -    0.0000000000582
  4  38462                38462                38462   
  </code></pre>
  
  <p>Here we see that the number that is added has no relation to the library size of the sample. What I want to learn is, how is this number that is added to each cpm is calculated based on prior.count. </p>
  
  <p>I have been digging through the code but that part goes into C++ territory which hinders my understanding of what is going on </p>
   77: <p>The prior count ends up getting scaled by the ratio of a library size to the average library size and then multiplied by 2 before getting added to each library size (I'm sure there's a good reason for that, but I don't know what it is). Using the example <code>df</code> data frame in your post, let's walk step-by-step through what <code>cpm()</code> is doing:</p>
  
  <pre class="lang-r prettyprint-override"><code>df = data.frame(a = c(1,2,3,0),c = c(1,2,3,0)*3, b = c(1,2,3,0)*2)
  prior.count = 0.5
  # First, we need to calculate a library size
  lib.size = colSums(df)
  # Calculate the average library size and the adjusted priors
  ave.lib = mean(lib.size)
  adjusted.prior = prior.count * lib.size / ave.lib
  # Update the library sizes
  adjusted.lib.size = lib.size + 2*adjusted.prior
  # Now we can compute the CPM
  customCPM = t((log(t(df) + adjusted.prior) - log(adjusted.lib.size) + log(1000000))/log(2))
  </code></pre>
  
  <p>The matrix transposition looks really ugly in the last step, but it's needed so R adds things like <code>adjusted.prior</code> to each row of <code>df</code>.</p>
  
  <p>Note that the prior count is only used if you compute log(CPM), it's unused otherwise.</p>
  