""" Created on 20. jan. 2011 @author: pab """ import numpy as np from numpy import exp, meshgrid __all__ = ['peaks', 'humps', 'magic'] def _magic_odd_order(n): ix = np.arange(n) + 1 J, I = np.meshgrid(ix, ix) A = np.mod(I + J - (n + 3) / 2, n) B = np.mod(I + 2 * J - 2, n) M = n * A + B + 1 return M def _magic_doubly_even_order(n): M = np.arange(1, n * n + 1).reshape(n, n) ix = np.mod(np.arange(n) + 1, 4) // 2 J, I = np.meshgrid(ix, ix) iz = np.flatnonzero(I == J) M.put(iz, n * n + 1 - M.flat[iz]) return M def _magic_even_order(n): p = n // 2 M0 = magic(p) M = np.hstack((np.vstack((M0, M0 + 3 * p * p)), np.vstack((M0 + 2 * p * p, M0 + p * p)))) if n > 2: k = (n - 2) // 4 jvec = np.hstack((np.arange(k), np.arange(n - k + 1, n))) for i in range(p): for j in jvec: temp = M[i][j] M[i][j] = M[i + p][j] M[i + p][j] = temp i = k j = 0 temp = M[i][j] M[i][j] = M[i + p][j] M[i + p][j] = temp j = i temp = M[i + p][j] M[i + p][j] = M[i][j] M[i][j] = temp return M def magic(n): """ Return magic square for n of any orders > 2. A magic square has the property that the sum of every row and column, as well as both diagonals, is the same number. Examples -------- >>> np.allclose(magic(3), ... [[8, 1, 6], ... [3, 5, 7], ... [4, 9, 2]]) True >>> np.allclose(magic(4), ... [[16, 2, 3, 13], ... [ 5, 11, 10, 8], ... [ 9, 7, 6, 12], ... [ 4, 14, 15, 1]]) True >>> np.allclose(magic(6), ... [[35, 1, 6, 26, 19, 24], ... [ 3, 32, 7, 21, 23, 25], ... [31, 9, 2, 22, 27, 20], ... [ 8, 28, 33, 17, 10, 15], ... [30, 5, 34, 12, 14, 16], ... [ 4, 36, 29, 13, 18, 11]]) True """ if (n < 3): raise ValueError('n must be greater than 2.') if np.mod(n, 2) == 1: return _magic_odd_order(n) elif np.mod(n, 4) == 0: return _magic_doubly_even_order(n) return _magic_even_order(n) def peaks(x=None, y=None, n=51): """ Return the "well" known MatLab (R) peaks function evaluated in the [-3,3] x,y range Example ------- >>> import matplotlib.pyplot as plt >>> x,y,z = peaks() h = plt.contourf(x,y,z) """ if x is None: x = np.linspace(-3, 3, n) if y is None: y = np.linspace(-3, 3, n) [x1, y1] = meshgrid(x, y) z = (3 * (1 - x1) ** 2 * exp(-(x1 ** 2) - (y1 + 1) ** 2) - 10 * (x1 / 5 - x1 ** 3 - y1 ** 5) * exp(-x1 ** 2 - y1 ** 2) - 1. / 3 * exp(-(x1 + 1) ** 2 - y1 ** 2)) return x1, y1, z def humps(x=None): """ Computes a function that has three roots, and some humps. Example ------- >>> import matplotlib.pyplot as plt >>> x = np.linspace(0,1) >>> y = humps(x) h = plt.plot(x,y) """ if x is None: y = np.linspace(0, 1) else: y = np.asarray(x) return 1.0 / ((y - 0.3) ** 2 + 0.01) + 1.0 / ((y - 0.9) ** 2 + 0.04) + \ 2 * y - 5.2 if __name__ == '__main__': from wafo.testing import test_docstrings test_docstrings(__file__)