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A comment on Pythran: Crossing the Python Frontier [pdf]. Parent comment

Re:Pythran: Crossing the Python Frontier [pdf] by serge-ss-paille on Tuesday April 24, 2018 @08:32AM (#16911961)
(shameful author here)

One should read

  def rosen_explicit_loop(x): 
    s = 0. 
    n = x.shape[0] 
    for i in range(0, n - 1): 
      s += 100. * (x[i + 1] - x[i] ** 2.) ** 2. + (1 - x[i]) ** 2 
    return s
(edited)
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Re:Pythran: Crossing the Python Frontier [pdf] by chestervonwinch on Tuesday April 24, 2018 @09:00AM (#16912150)
...

    n = x.shape[0]
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Re:Pythran: Crossing the Python Frontier [pdf] by quietbritishjim on Tuesday April 24, 2018 @12:22PM (#16913987)
Or just len(x). This works perfectly well on numpy arrays and has the bonus that it works on regular lists/tuples of floats, so the first snippet doesn't rely on numpy.
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