title: grep -v pattern and also remove 1 line before and 4 lines after
body: <p>I would like to grep a pattern and remove the line of the matching pattern and also 1 line before and 4 lines after the context. I tried:</p>
  
  <pre><code>grep -v -A 4 -B 1
  </code></pre>
  
  <p>Thanks in advance!</p>
  
  <p>Example:</p>
  
  <pre><code>Rule: r1
  Owner: Process explorer.exe Pid 1544
  0x01ec350f  8b 45 a8 0f b6 00 8d 4d a8 ff 14 85 c8 7f ed 01   .E.....M........
  0x01ec351f  84 c0 75 ec 8b 4d fc e8 ba f5 fe ff f7 85 b0 fd   ..u..M..........
  0x01ec352f  ff ff 00 00 01 00 75 13 33 c0 50 50 50 68 48 28   ......u.3.PPPhH(
  0x01ec353f  eb 01 33 d2 8b cb e8 b0 57 ff ff f7 05 8c 9b ed   ..3.....W.......
  </code></pre>
  
  <p>I would like to grep "explorer.exe" and remove the line and also 1 line before and 4 lines after.</p>