title: Regex separate based on specific space - R (dplyr::separate)
body: <p>I want to separate a column based on a specific space.</p>
  <p>For example, I do not want to separate in every space, but only if the space have met certain condition.</p>
  <p>I've tried to separate every space that met the condition, but this remove the space and the condition. For example, if I try to remove every space that is preceded by a letter and after there is a number this remove the letter, the space and the number.</p>
  <p>In the code I tried using separate from dplyr, but if there is another optimal solution I would take!</p>
  <p>Thanks in advance!</p>
  <p>Code</p>
  <pre><code>library(tidyverse)
  
  df &lt;- tibble(
  column = c(&quot;Current Assets 3a 10.001&quot;, &quot;Cash and Equivalents 2b 1.009&quot;, &quot;Debt 2.050&quot; )
  )
  
  
  df %&gt;% 
    dplyr::separate(value,
             into = c(&quot;column1&quot;,&quot;column2&quot;,&quot;column3&quot;),
             sep = 'insert regex pattern here')
  
  #Ideally i would want something like that
  
  tibble(
    column1 = c(&quot;Current Assets&quot;, &quot;Cash and Equivalents&quot;, &quot;Debt&quot;),
    column2 = c(&quot;3a&quot;, &quot;2b&quot;, NA),
    column3 = c(10.001, 1.009, 2.050)
    
  )
  
  </code></pre>