Tweaked version of code at http://mail.scipy.org/pipermail/scipy-user/attachments/20060803/ad246212/attachment.py with minor bug fix and improved documentation.

hough.py

## Copyright (C) 2006 Stefan van der Walt <stefan@sun.ac.za>
##
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## modification, are permitted provided that the following conditions are
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## THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR
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## DISCLAIMED. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT,
## INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES
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## Modified 2013-08-14 by Thomas A Caswell tcaswell@gmail.com

import numpy as np

itype = np.uint16 # See ticket 225


def houghtf(img, angles=None):
    """Perform the straight line Hough transform.

    Parameters
    ----------
      img : a boolean np.ndarray
          The image to be processed
          
      angles : np.ndarray
          The angles to process in degrees

          
    Returns
    -------
      H : np.ndarray
         the Hough transform coefficients, columns are fixed :math:`\theta`,
         rows are fixed :math:`\rho`
         
      angles : np.ndarray
         The :math:`\theta` values 
      distances : np.ndarray
         The :math:`\rho` values


    """
    if img.ndim != 2:
        raise ValueError("Input must be a two-dimensional array")

    img = img.astype(bool)

    # numpy did not like `if angles` -> truth value of array ambiguous 
    if angles is None:
        angles = np.linspace(-90,90,180)

    theta = angles / 180. * np.pi
    d = np.ceil(np.hypot(*img.shape))
    # can't just change this number due to implicit hashing below
    nr_bins = 2*d - 1
    bins = np.linspace(-d,d,nr_bins)
    out = np.zeros((nr_bins,len(theta)),dtype=itype)

    rows,cols = img.shape
    x,y = np.mgrid[:rows,:cols]

    for i,(cT,sT) in enumerate(zip(np.cos(theta),np.sin(theta))):
        # this line is tricksy implicit hashing
        rho = np.round_(cT*x[img] + sT*y[img]) - bins[0] + 1
        rho = rho.astype(itype)
        rho[(rho < 0) | (rho > nr_bins)] = 0
        bc = np.bincount(rho.flat)[1:]
        out[:len(bc),i] = bc

    return out,angles,bins