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Python

'''
Created on 20. jan. 2011
@author: pab
'''
import numpy as np
from numpy import exp, meshgrid
__all__ = ['peaks', 'humps', 'magic']
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
--------
9 years ago
>>> 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: # odd order
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
elif np.mod(n, 4) == 0: # doubly even order
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])
else: # singly even order
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 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__':
8 years ago
from wafo.testing import test_docstrings
test_docstrings(__file__)