title: Matlab's imgaussfilt equivalent in Python
body: <p>I am trying to replicate Matlab's imgaussfilter behaviour in Python but I have been unable to reproduce the results. The docs don't help much since there is not explanation of what exactly is being done (for instance how is that function different from fspecial or how it differs from Octave's imsmooth (with Gaussian as an argument).</p>
  <p><strong>Matlab Code</strong></p>
  <pre><code>imgaussfilt(image,sigma)
  </code></pre>
  <p>with output</p>
  <pre><code> [[-0.02936392 -0.03168419 -0.0343706  ...  0.03136455  0.02864487
     0.02585145]
   [-0.03212093 -0.0347433  -0.03775943 ...  0.03507484  0.03209807
     0.02906134]
   [-0.03512981 -0.03808864 -0.04147075 ...  0.03873834  0.03549163
     0.03219311]
   ...
   [-0.07713804 -0.08337475 -0.08975262 ... -0.04314206 -0.03945251
    -0.03606256]
   [-0.07145457 -0.07714807 -0.08297654 ... -0.03986635 -0.03641579
    -0.03323718]
   [-0.06605107 -0.07122191 -0.07651892 ... -0.03684535 -0.03362917
    -0.03065128]]
  </code></pre>
  <p>The closest I have managed to get with Python has been:</p>
  <p><strong>Python</strong></p>
  <pre><code>skimage.filters.gaussian(image, sigma=s,mode = 'nearest',truncate=2.0)
  </code></pre>
  <p>with output</p>
  <pre><code>[[-0.02936397 -0.03168425 -0.03437067 ...  0.03136461  0.02864492
     0.0258515 ]
   [-0.03212099 -0.03474336 -0.03775951 ...  0.03507491  0.03209814
     0.0290614 ]
   [-0.03512988 -0.03808872 -0.04147083 ...  0.03873842  0.03549169
     0.03219317]
   ...
   [-0.07713819 -0.08337491 -0.08975279 ... -0.04314214 -0.03945259
    -0.03606263]
   [-0.07145471 -0.07714821 -0.0829767  ... -0.03986643 -0.03641586
    -0.03323725]
   [-0.06605119 -0.07122205 -0.07651907 ... -0.03684542 -0.03362923
    -0.03065134]]
  </code></pre>
  <p>which is similar but not exactly the same result. Is there a method that approximates it better? Do I need to change one of the parameters?</p>
  <p>EDIT:
  If you are looking for a close approximation using OpenCV @avisionx has provided a good starting point in the solutions.</p>