[Numpy-tickets] [NumPy] #292: Make repmat work for arbitrary dimensions

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Sat Sep 23 11:12:21 CDT 2006

#292: Make repmat work for arbitrary dimensions
 Reporter:  baxissimo    |       Owner:  somebody
     Type:  enhancement  |      Status:  new     
 Priority:  normal       |   Milestone:          
Component:  numpy.lib    |     Version:          
 Severity:  normal       |    Keywords:          
 repmat() should work with inputs of any shape and take a tuple for the
 output number of repeats.

 If this is considered too big a change to replace repmat's 3-argument
 signature with an incompatible 2-arg one, then a new function 'reparray'
 should be added.

 An example implementation is below
 def reparray(A, tup):
     """Repeat an array the number of times given in the integer tuple,

     Similar to repmat, but works for arrays of any dimension.
     reparray(A,(m,n)) is equivalent to repmat(A,m,n)

     If tup has length d, the result will have dimension of max(d, A.ndim).
     If tup is scalar it is treated as a 1-tuple.

     If A.ndim < d, A is promoted to be d-dimensional by prepending new
     So a shape (3,) array is promoted to (1,3) for 2-D replication,
     or shape (1,1,3) for 3-D replication.
     If this is not the desired behavior, promote A to d-dimensions
     before calling this function.

     If d < A.ndim, tup is promoted to A.ndim by appending 1's to it.  Thus
     for an A.shape of (2,3,4,5), a tup of (2,2) is treated as (2,2,1,1)

     >>> a = array([0,1,2])
     >>> reparray(a,2)
     array([0, 1, 2, 0, 1, 2])
     >>> reparray(a,(1,2))
     array([[0, 1, 2, 0, 1, 2]])
     >>> reparray(a,(2,2))
     array([[0, 1, 2, 0, 1, 2],
            [0, 1, 2, 0, 1, 2]])
     >>> reparray(a,(2,1,2))
     array([[[0, 1, 2, 0, 1, 2]],

            [[0, 1, 2, 0, 1, 2]]])

     See Also:
        repmat, repeat
     if numpy.isscalar(tup):
         tup = (tup,)
     d = len(tup)
     c = numpy.array(A,copy=False,subok=True,ndmin=d)
     shape = list(c.shape)
     n = c.size
     for i, nrep in enumerate(tup):
         if nrep!=1:
             c = c.reshape(-1,n).repeat(nrep,0)
         dim_in = shape[i]
         dim_out = dim_in*nrep
         shape[i] = dim_out
         n /= dim_in
     return c.reshape(shape)
 A more complete version with two implementations and tests and timing code
 is attached.

Ticket URL: <http://projects.scipy.org/scipy/numpy/ticket/292>
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