title: Repeat a scipy csr sparse matrix along axis 0
body: <p>I wanted to repeat the rows of a scipy csr sparse matrix, but when I tried to call numpy's repeat method, it simply treats the sparse matrix like an object, and would only repeat it as an object in an ndarray. I looked through the documentation, but I couldn't find any utility to repeats the rows of a scipy csr sparse matrix.</p>
  
  <p>I wrote the following code that operates on the internal data, which seems to work</p>
  
  <pre class="lang-python prettyprint-override"><code>def csr_repeat(csr, repeats):
      if isinstance(repeats, int):
          repeats = np.repeat(repeats, csr.shape[0])
      repeats = np.asarray(repeats)
      rnnz = np.diff(csr.indptr)
      ndata = rnnz.dot(repeats)
      if ndata == 0:
          return sparse.csr_matrix((np.sum(repeats), csr.shape[1]),
                                   dtype=csr.dtype)
      indmap = np.ones(ndata, dtype=np.int)
      indmap[0] = 0
      rnnz_ = np.repeat(rnnz, repeats)
      indptr_ = rnnz_.cumsum()
      mask = indptr_ &lt; ndata
      indmap -= np.int_(np.bincount(indptr_[mask],
                                    weights=rnnz_[mask],
                                    minlength=ndata))
      jumps = (rnnz * repeats).cumsum()
      mask = jumps &lt; ndata
      indmap += np.int_(np.bincount(jumps[mask],
                                    weights=rnnz[mask],
                                    minlength=ndata))
      indmap = indmap.cumsum()
      return sparse.csr_matrix((csr.data[indmap],
                                csr.indices[indmap],
                                np.r_[0, indptr_]),
                               shape=(np.sum(repeats), csr.shape[1]))
  </code></pre>
  
  <p>and be reasonably efficient, but I'd rather not monkey patch the class. Is there a better way to do this?</p>
  
  <h2>Edit</h2>
  
  <p>As I revisit this question, I wonder why I posted it in the first place. Almost everything I could think to do with the repeated matrix would be easier to do with the original matrix, and then apply the repetition afterwards. My assumption is that post repetition will always be the better way to approach this problem than any of the potential answers.</p>