My tools of the trade for python programming.
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# $Id: __init__.py,v 1.3 2010-09-10 21:23:59 wirawan Exp $
#
# wpylib.math main module
# Created: 20091204
# Wirawan Purwanto
#
pass
import numpy
ZERO_TOL = 5.0e-16
def ztol(val, tol=None, copy=True):
"""Rounds down values to zero if they are below tolerance."""
if tol == None: tol = ZERO_TOL
if "__iter__" not in dir(val):
if numpy.abs(val) < tol:
return 0
else:
return val
elif isinstance(val, numpy.ndarray):
if copy:
rslt = val.copy()
else:
rslt = val
numpy.putmask(rslt, numpy.abs(rslt) < tol, [0])
return rslt
else:
raise ValueError, "Unsupported datatype: %s" % str(type(val))
def epsilon(dtype):
"""A simple way to determine (at runtime) the precision of a given type
real number.
Precision is defined such that (1.0 + epsilon(dtype) > 1.0).
Below this number, the addition will not yield a different result.
"""
one = dtype(1.0)
small = one
small2 = small
while one + small > one:
small2 = small
small = dtype(small / 2)
return small2
def roundup(value, unit):
"""Rounds up a value to the next integer multiple of a unit."""
return numpy.ceil(float(value) / float(unit)) * unit
def choose(n,r):
"""Computes n! / {r! (n-r)!} . Note that the following condition must
always be fulfilled:
1 <= n
1 <= r <= n
Otherwise the result is not predictable!
Optimization: To minimize the # of multiplications and divisions, we
rewrite the expression as
n! n(n-1)...(n-r+1)
--------- = ----------------
r!(n-r)! r!
To avoid multiplication overflow as much as possible, we will evaluate
in the following STRICT order, from left to right:
n / 1 * (n-1) / 2 * (n-2) / 3 * ... * (n-r+1) / r
We can show that integer arithmatic operated in this order is exact
(i.e. no roundoff error).
Note: this implementation is based on my C++ cp.inc library.
For other implementations, see:
http://stackoverflow.com/questions/3025162/statistics-combinations-in-python
Published in stack overflow, see URL above.
"""
assert n >= 0
assert 0 <= r <= n
c = 1L
denom = 1
for (num,denom) in zip(xrange(n,n-r,-1), xrange(1,r+1,1)):
c = (c * num) // denom
return c