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"""
Functions which are common and require SciPy Base and Level 1 SciPy
(special, linalg)
"""
from __future__ import division, print_function, absolute_import
from scipy.lib.six.moves import xrange
from numpy import exp, log, asarray, arange, newaxis, hstack, product, array, \
where, zeros, extract, place, pi, sqrt, eye, poly1d, dot, \
r_, rollaxis, sum, fromstring
__all__ = ['logsumexp', 'factorial','factorial2','factorialk','comb',
'central_diff_weights', 'derivative', 'pade', 'lena', 'ascent', 'face']
# XXX: the factorial functions could move to scipy.special, and the others
# to numpy perhaps?
def logsumexp(a, axis=None, b=None):
"""Compute the log of the sum of exponentials of input elements.
Parameters
----------
a : array_like
Input array.
axis : int, optional
Axis over which the sum is taken. By default `axis` is None,
and all elements are summed.
.. versionadded:: 0.11.0
b : array-like, optional
Scaling factor for exp(`a`) must be of the same shape as `a` or
broadcastable to `a`.
.. versionadded:: 0.12.0
Returns
-------
res : ndarray
The result, ``np.log(np.sum(np.exp(a)))`` calculated in a numerically
more stable way. If `b` is given then ``np.log(np.sum(b*np.exp(a)))``
is returned.
See Also
--------
numpy.logaddexp, numpy.logaddexp2
Notes
-----
Numpy has a logaddexp function which is very similar to `logsumexp`, but
only handles two arguments. `logaddexp.reduce` is similar to this
function, but may be less stable.
Examples
--------
>>> from scipy.misc import logsumexp
>>> a = np.arange(10)
>>> np.log(np.sum(np.exp(a)))
9.4586297444267107
>>> logsumexp(a)
9.4586297444267107
With weights
>>> a = np.arange(10)
>>> b = np.arange(10, 0, -1)
>>> logsumexp(a, b=b)
9.9170178533034665
>>> np.log(np.sum(b*np.exp(a)))
9.9170178533034647
"""
a = asarray(a)
if axis is None:
a = a.ravel()
else:
a = rollaxis(a, axis)
a_max = a.max(axis=0)
if b is not None:
b = asarray(b)
if axis is None:
b = b.ravel()
else:
b = rollaxis(b, axis)
out = log(sum(b * exp(a - a_max), axis=0))
else:
out = log(sum(exp(a - a_max), axis=0))
out += a_max
return out
def factorial(n,exact=0):
"""
The factorial function, n! = special.gamma(n+1).
If exact is 0, then floating point precision is used, otherwise
exact long integer is computed.
- Array argument accepted only for exact=0 case.
- If n<0, the return value is 0.
Parameters
----------
n : int or array_like of ints
Calculate ``n!``. Arrays are only supported with `exact` set
to False. If ``n < 0``, the return value is 0.
exact : bool, optional
The result can be approximated rapidly using the gamma-formula
above. If `exact` is set to True, calculate the
answer exactly using integer arithmetic. Default is False.
Returns
-------
nf : float or int
Factorial of `n`, as an integer or a float depending on `exact`.
Examples
--------
>>> arr = np.array([3,4,5])
>>> sc.factorial(arr, exact=False)
array([ 6., 24., 120.])
>>> sc.factorial(5, exact=True)
120L
"""
if exact:
if n < 0:
return 0
val = 1
for k in xrange(1,n+1):
val *= k
return val
else:
from scipy import special
n = asarray(n)
sv = special.errprint(0)
vals = special.gamma(n+1)
sv = special.errprint(sv)
return where(n >= 0,vals,0)
def factorial2(n, exact=False):
"""
Double factorial.
This is the factorial with every second value skipped, i.e.,
``7!! = 7 * 5 * 3 * 1``. It can be approximated numerically as::
n!! = special.gamma(n/2+1)*2**((m+1)/2)/sqrt(pi) n odd
= 2**(n/2) * (n/2)! n even
Parameters
----------
n : int or array_like
Calculate ``n!!``. Arrays are only supported with `exact` set
to False. If ``n < 0``, the return value is 0.
exact : bool, optional
The result can be approximated rapidly using the gamma-formula
above (default). If `exact` is set to True, calculate the
answer exactly using integer arithmetic.
Returns
-------
nff : float or int
Double factorial of `n`, as an int or a float depending on
`exact`.
Examples
--------
>>> factorial2(7, exact=False)
array(105.00000000000001)
>>> factorial2(7, exact=True)
105L
"""
if exact:
if n < -1:
return 0
if n <= 0:
return 1
val = 1
for k in xrange(n,0,-2):
val *= k
return val
else:
from scipy import special
n = asarray(n)
vals = zeros(n.shape,'d')
cond1 = (n % 2) & (n >= -1)
cond2 = (1-(n % 2)) & (n >= -1)
oddn = extract(cond1,n)
evenn = extract(cond2,n)
nd2o = oddn / 2.0
nd2e = evenn / 2.0
place(vals,cond1,special.gamma(nd2o+1)/sqrt(pi)*pow(2.0,nd2o+0.5))
place(vals,cond2,special.gamma(nd2e+1) * pow(2.0,nd2e))
return vals
def factorialk(n,k,exact=1):
"""
n(!!...!) = multifactorial of order k
k times
Parameters
----------
n : int, array_like
Calculate multifactorial. Arrays are only supported with exact
set to False. If `n` < 0, the return value is 0.
exact : bool, optional
If exact is set to True, calculate the answer exactly using
integer arithmetic.
Returns
-------
val : int
Multi factorial of `n`.
Raises
------
NotImplementedError
Raises when exact is False
Examples
--------
>>> sc.factorialk(5, 1, exact=True)
120L
>>> sc.factorialk(5, 3, exact=True)
10L
"""
if exact:
if n < 1-k:
return 0
if n <= 0:
return 1
val = 1
for j in xrange(n,0,-k):
val = val*j
return val
else:
raise NotImplementedError
def comb(N,k,exact=0):
"""
The number of combinations of N things taken k at a time.
This is often expressed as "N choose k".
Parameters
----------
N : int, ndarray
Number of things.
k : int, ndarray
Number of elements taken.
exact : int, optional
If `exact` is 0, then floating point precision is used, otherwise
exact long integer is computed.
Returns
-------
val : int, ndarray
The total number of combinations.
Notes
-----
- Array arguments accepted only for exact=0 case.
- If k > N, N < 0, or k < 0, then a 0 is returned.
Examples
--------
>>> k = np.array([3, 4])
>>> n = np.array([10, 10])
>>> sc.comb(n, k, exact=False)
array([ 120., 210.])
>>> sc.comb(10, 3, exact=True)
120L
"""
if exact:
if (k > N) or (N < 0) or (k < 0):
return 0
val = 1
for j in xrange(min(k, N-k)):
val = (val*(N-j))//(j+1)
return val
else:
from scipy import special
k,N = asarray(k), asarray(N)
lgam = special.gammaln
cond = (k <= N) & (N >= 0) & (k >= 0)
sv = special.errprint(0)
vals = exp(lgam(N+1) - lgam(N-k+1) - lgam(k+1))
sv = special.errprint(sv)
return where(cond, vals, 0.0)
def central_diff_weights(Np, ndiv=1):
"""
Return weights for an Np-point central derivative.
Assumes equally-spaced function points.
If weights are in the vector w, then
derivative is w[0] * f(x-ho*dx) + ... + w[-1] * f(x+h0*dx)
Parameters
----------
Np : int
Number of points for the central derivative.
ndiv : int, optional
Number of divisions. Default is 1.
Notes
-----
Can be inaccurate for large number of points.
"""
if Np < ndiv + 1:
raise ValueError("Number of points must be at least the derivative order + 1.")
if Np % 2 == 0:
raise ValueError("The number of points must be odd.")
from scipy import linalg
ho = Np >> 1
x = arange(-ho,ho+1.0)
x = x[:,newaxis]
X = x**0.0
for k in range(1,Np):
X = hstack([X,x**k])
w = product(arange(1,ndiv+1),axis=0)*linalg.inv(X)[ndiv]
return w
def derivative(func, x0, dx=1.0, n=1, args=(), order=3):
"""
Find the n-th derivative of a function at a point.
Given a function, use a central difference formula with spacing `dx` to
compute the `n`-th derivative at `x0`.
Parameters
----------
func : function
Input function.
x0 : float
The point at which `n`-th derivative is found.
dx : int, optional
Spacing.
n : int, optional
Order of the derivative. Default is 1.
args : tuple, optional
Arguments
order : int, optional
Number of points to use, must be odd.
Notes
-----
Decreasing the step size too small can result in round-off error.
Examples
--------
>>> def f(x):
... return x**3 + x**2
...
>>> derivative(f, 1.0, dx=1e-6)
4.9999999999217337
"""
if order < n + 1:
raise ValueError("'order' (the number of points used to compute the derivative), "
"must be at least the derivative order 'n' + 1.")
if order % 2 == 0:
raise ValueError("'order' (the number of points used to compute the derivative) "
"must be odd.")
# pre-computed for n=1 and 2 and low-order for speed.
if n == 1:
if order == 3:
weights = array([-1,0,1])/2.0
elif order == 5:
weights = array([1,-8,0,8,-1])/12.0
elif order == 7:
weights = array([-1,9,-45,0,45,-9,1])/60.0
elif order == 9:
weights = array([3,-32,168,-672,0,672,-168,32,-3])/840.0
else:
weights = central_diff_weights(order,1)
elif n == 2:
if order == 3:
weights = array([1,-2.0,1])
elif order == 5:
weights = array([-1,16,-30,16,-1])/12.0
elif order == 7:
weights = array([2,-27,270,-490,270,-27,2])/180.0
elif order == 9:
weights = array([-9,128,-1008,8064,-14350,8064,-1008,128,-9])/5040.0
else:
weights = central_diff_weights(order,2)
else:
weights = central_diff_weights(order, n)
val = 0.0
ho = order >> 1
for k in range(order):
val += weights[k]*func(x0+(k-ho)*dx,*args)
return val / product((dx,)*n,axis=0)
def pade(an, m):
"""
Return Pade approximation to a polynomial as the ratio of two polynomials.
Parameters
----------
an : (N,) array_like
Taylor series coefficients.
m : int
The order of the returned approximating polynomials.
Returns
-------
p, q : Polynomial class
The pade approximation of the polynomial defined by `an` is
`p(x)/q(x)`.
Examples
--------
>>> from scipy import misc
>>> e_exp = [1.0, 1.0, 1.0/2.0, 1.0/6.0, 1.0/24.0, 1.0/120.0]
>>> p, q = misc.pade(e_exp, 2)
>>> e_exp.reverse()
>>> e_poly = np.poly1d(e_exp)
Compare ``e_poly(x)`` and the pade approximation ``p(x)/q(x)``
>>> e_poly(1)
2.7166666666666668
>>> p(1)/q(1)
2.7179487179487181
"""
from scipy import linalg
an = asarray(an)
N = len(an) - 1
n = N - m
if n < 0:
raise ValueError("Order of q <m> must be smaller than len(an)-1.")
Akj = eye(N+1, n+1)
Bkj = zeros((N+1, m), 'd')
for row in range(1, m+1):
Bkj[row,:row] = -(an[:row])[::-1]
for row in range(m+1, N+1):
Bkj[row,:] = -(an[row-m:row])[::-1]
C = hstack((Akj, Bkj))
pq = linalg.solve(C, an)
p = pq[:n+1]
q = r_[1.0, pq[n+1:]]
return poly1d(p[::-1]), poly1d(q[::-1])
def lena():
"""
Get classic image processing example image, Lena, at 8-bit grayscale
bit-depth, 512 x 512 size.
Parameters
----------
None
Returns
-------
lena : ndarray
Lena image
Examples
--------
>>> import scipy.misc
>>> lena = scipy.misc.lena()
>>> lena.shape
(512, 512)
>>> lena.max()
245
>>> lena.dtype
dtype('int32')
>>> import matplotlib.pyplot as plt
>>> plt.gray()
>>> plt.imshow(lena)
>>> plt.show()
"""
import pickle
import os
fname = os.path.join(os.path.dirname(__file__),'lena.dat')
f = open(fname,'rb')
lena = array(pickle.load(f))
f.close()
return lena
def ascent():
"""
Get an 8-bit grayscale bit-depth, 512 x 512 derived image for easy use in demos
The image is derived from accent-to-the-top.jpg at
http://www.public-domain-image.com/people-public-domain-images-pictures/
Parameters
----------
None
Returns
-------
ascent : ndarray
convenient image to use for testing and demonstration
Examples
--------
>>> import scipy.misc
>>> ascent = scipy.misc.ascent()
>>> ascent.shape
(512, 512)
>>> ascent.max()
255
>>> import matplotlib.pyplot as plt
>>> plt.gray()
>>> plt.imshow(ascent)
>>> plt.show()
"""
import pickle
import os
fname = os.path.join(os.path.dirname(__file__),'ascent.dat')
with open(fname, 'rb') as f:
ascent = array(pickle.load(f))
return ascent
def face(gray=False):
"""
Get a 1024 x 768, color image of a raccoon face.
raccoon-procyon-lotor.jpg at http://www.public-domain-image.com
Parameters
----------
gray : bool, optional
If True then return color image, otherwise return an 8-bit gray-scale
Returns
-------
face : ndarray
image of a racoon face
Examples
--------
>>> import scipy.misc
>>> face = scipy.misc.face()
>>> face.shape
(768, 1024, 3)
>>> face.max()
230
>>> face.dtype
dtype('uint8')
>>> import matplotlib.pyplot as plt
>>> plt.gray()
>>> plt.imshow(face)
>>> plt.show()
"""
import bz2
import os
with open(os.path.join(os.path.dirname(__file__), 'face.dat'), 'rb') as f:
rawdata = f.read()
data = bz2.decompress(rawdata)
face = fromstring(data, dtype='uint8')
face.shape = (768, 1024, 3)
if gray is True:
face = (0.21 * face[:,:,0] + 0.71 * face[:,:,1] + 0.07 * face[:,:,2]).astype('uint8')
return face
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