code: # The remove_unused_levels defined here was copied based on the source code
  # defined in pandas.core.indexes.muli.py
  
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  import numpy as np
  
  
  # for pandas 0.19
  def remove_unused_levels(self):
      """
      create a new MultiIndex from the current that removing
      unused levels, meaning that they are not expressed in the labels
      The resulting MultiIndex will have the same outward
      appearance, meaning the same .values and ordering. It will also
      be .equals() to the original.
      .. versionadded:: 0.20.0
      Returns
      -------
      MultiIndex
      Examples
      --------
      >>> i = pd.MultiIndex.from_product([range(2), list('ab')])
      MultiIndex(levels=[[0, 1], ['a', 'b']],
                 labels=[[0, 0, 1, 1], [0, 1, 0, 1]])
      >>> i[2:]
      MultiIndex(levels=[[0, 1], ['a', 'b']],
                 labels=[[1, 1], [0, 1]])
      The 0 from the first level is not represented
      and can be removed
      >>> i[2:].remove_unused_levels()
      MultiIndex(levels=[[1], ['a', 'b']],
                 labels=[[0, 0], [0, 1]])
      """
      import pandas.core.algorithms as algos
  
      new_levels = []
      new_labels = []
  
      changed = False
      for lev, lab in zip(self.levels, self.labels):
  
          # Since few levels are typically unused, bincount() is more
          # efficient than unique() - however it only accepts positive values
          # (and drops order):
          uniques = np.where(np.bincount(lab + 1) > 0)[0] - 1
          has_na = int(len(uniques) and (uniques[0] == -1))
  
          if len(uniques) != len(lev) + has_na:
              # We have unused levels
              changed = True
  
              # Recalculate uniques, now preserving order.
              # Can easily be cythonized by exploiting the already existing
              # "uniques" and stop parsing "lab" when all items are found:
              uniques = algos.unique(lab)
              if has_na:
                  na_idx = np.where(uniques == -1)[0]
                  # Just ensure that -1 is in first position:
                  uniques[[0, na_idx[0]]] = uniques[[na_idx[0], 0]]
  
              # labels get mapped from uniques to 0:len(uniques)
              # -1 (if present) is mapped to last position
              label_mapping = np.zeros(len(lev) + has_na)
              # ... and reassigned value -1:
              label_mapping[uniques] = np.arange(len(uniques)) - has_na
  
              lab = label_mapping[lab]
  
              # new levels are simple
              lev = lev.take(uniques[has_na:])
  
          new_levels.append(lev)
          new_labels.append(lab)
  
      result = self._shallow_copy()
  
      if changed:
          result._reset_identity()
          result._set_levels(new_levels, validate=False)
          result._set_labels(new_labels, validate=False)
  
      return result
  
apis: numpy.where numpy.bincount pandas.core.algorithms.unique