本文整理汇总了Python中sympy.count_ops函数的典型用法代码示例。如果您正苦于以下问题:Python count_ops函数的具体用法?Python count_ops怎么用?Python count_ops使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了count_ops函数的15个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: test_simplify_ratio
def test_simplify_ratio():
# roots of x**3-3*x+5
roots = ['(5/2 + 21**(1/2)/2)**(1/3)*(1/2 - I*3**(1/2)/2)'
' + 1/((1/2 - I*3**(1/2)/2)*(5/2 + 21**(1/2)/2)**(1/3))',
'(5/2 + 21**(1/2)/2)**(1/3)*(1/2 + I*3**(1/2)/2)'
' + 1/((1/2 + I*3**(1/2)/2)*(5/2 + 21**(1/2)/2)**(1/3))',
'-1/(5/2 + 21**(1/2)/2)**(1/3) - (5/2 + 21**(1/2)/2)**(1/3)']
for r in roots:
r = S(r)
assert count_ops(simplify(r, ratio=1)) <= count_ops(r)
# If ratio=oo, simplify() is always applied:
assert simplify(r, ratio=oo) is not r
开发者ID:haz,项目名称:sympy,代码行数:12,代码来源:test_simplify.py
示例2: expression_complexity
def expression_complexity(expr, complexity=None):
'''
Returns the complexity of an expression (either string or sympy)
The complexity is defined as 1 for each arithmetic operation except divide which is 2,
and all other operations are 20. This can be overridden using the complexity
argument.
Note: calling this on a statement rather than an expression is likely to lead to errors.
Parameters
----------
expr: `sympy.Expr` or str
The expression.
complexity: None or dict (optional)
A dictionary mapping expression names to their complexity, to overwrite default behaviour.
Returns
-------
complexity: int
The complexity of the expression.
'''
subs = {'ADD':1, 'DIV':2, 'MUL':1, 'SUB':1}
if complexity is not None:
subs.update(complexity)
ops = sympy.count_ops(expr, visual=True)
for atom in ops.atoms():
if hasattr(atom, 'name'):
subs[atom.name] = 20 # unknown operations assumed to have a large cost
return ops.evalf(subs=subs)
开发者ID:ZeitgeberH,项目名称:brian2,代码行数:30,代码来源:sympytools.py
示例3: subs_matrix_verbose
def subs_matrix_verbose(A_expr,subs,simultaneous=False):
print "flashlight.sympy: subs_matrix_verbose(...) begin..."
A_subs_expr = sympy.Matrix.zeros(A_expr.rows,A_expr.cols)
for r in range(A_expr.rows):
for c in range(A_expr.cols):
print " ",r,c,len(subs),sympy.count_ops(A_expr[r,c])
A_subs_expr[r,c] = A_expr[r,c].subs(subs,simultaneous=simultaneous)
print "flashlight.sympy: subs_matrix_verbose(...) end."
return A_subs_expr
开发者ID:mikeroberts3000,项目名称:flashlight,代码行数:10,代码来源:sympy_utils.py
示例4: test_issue_2827_trigsimp_methods
def test_issue_2827_trigsimp_methods():
measure1 = lambda expr: len(str(expr))
measure2 = lambda expr: -count_ops(expr)
# Return the most complicated result
expr = (x + 1)/(x + sin(x)**2 + cos(x)**2)
ans = Matrix([1])
M = Matrix([expr])
assert trigsimp(M, method='fu', measure=measure1) == ans
assert trigsimp(M, method='fu', measure=measure2) != ans
# all methods should work with Basic expressions even if they
# aren't Expr
M = Matrix.eye(1)
assert all(trigsimp(M, method=m) == M for m in
'fu matching groebner old'.split())
# watch for E in exptrigsimp, not only exp()
eq = 1/sqrt(E) + E
assert exptrigsimp(eq) == eq
开发者ID:baoqchau,项目名称:sympy,代码行数:17,代码来源:test_trigsimp.py
示例5: TestTraverseExpr
def TestTraverseExpr(tmpExp):
# Inspired from sympy-master/sympy/core/expr.py:
assert isinstance(tmpExp, sympy.Expr) #fraction) #Equality)
assert tmpExp.is_Mul #fraction) #Equality)
print("Number of operators on the rhs exp: %s" % \
sympy.count_ops(tmpExp, visual=True))
args = tmpExp.as_ordered_factors() #order=order)
print("args = %s" % args)
print("args[1] = %s" % args[1])
assert args[1].is_Pow
args1 = args[1].args
print args1
print args1[0]
assert args1[1] == -1
#print("args1 (the denominator factors) = %s" % str(args1))
#print type(args[1])
assert args[1].is_Mul
开发者ID:alexsusu,项目名称:video-diff,代码行数:18,代码来源:parse.py
示例6: render_node
def render_node(self, node):
expr = NodeRenderer(use_vectorisation_idx=False).render_node(node)
if is_scalar_expression(expr, self.variables) and not has_non_float(expr,
self.variables):
if expr in self.optimisations:
name = self.optimisations[expr]
else:
# Do not pull out very simple expressions (including constants
# and numbers)
sympy_expr = str_to_sympy(expr)
if sympy.count_ops(sympy_expr, visual=False) < 2:
return expr
self.n += 1
name = '_lio_const_'+str(self.n)
self.optimisations[expr] = name
return name
else:
return NodeRenderer.render_node(self, node)
开发者ID:SudShekhar,项目名称:brian2,代码行数:19,代码来源:translation.py
示例7: expression_complexity
def expression_complexity(expr, complexity=None):
'''
Returns the complexity of an expression (either string or sympy)
The complexity is defined as 1 for each arithmetic operation except divide which is 2,
and all other operations are 20. This can be overridden using the complexity
argument.
Note: calling this on a statement rather than an expression is likely to lead to errors.
Parameters
----------
expr: `sympy.Expr` or str
The expression.
complexity: None or dict (optional)
A dictionary mapping expression names to their complexity, to overwrite default behaviour.
Returns
-------
complexity: int
The complexity of the expression.
'''
if isinstance(expr, str):
# we do this because sympy.count_ops doesn't handle inequalities (TODO: handle sympy as well str)
for op in ['<=', '>=', '==', '<', '>']:
expr = expr.replace(op, '+')
# work around bug with rand() and randn() (TODO: improve this)
expr = expr.replace('rand()', 'rand(0)')
expr = expr.replace('randn()', 'randn(0)')
subs = {'ADD':1, 'DIV':2, 'MUL':1, 'SUB':1}
if complexity is not None:
subs.update(complexity)
ops = sympy.count_ops(expr, visual=True)
for atom in ops.atoms():
if hasattr(atom, 'name'):
subs[atom.name] = 20 # unknown operations assumed to have a large cost
return ops.evalf(subs=subs)
开发者ID:brian-team,项目名称:brian2numba,代码行数:37,代码来源:sympytools.py
示例8: Read
def Read(fileName):
fin = open(fileName)
while (True):
str = fin.readline()
if str:
#pass
str = str.rstrip(rstripStr)
else:
break
if str.startswith("#"):
continue
for index, v in enumerate(defVars):
cmpStr = v + " = "
#print("cmpStr = %s" % cmpStr)
#" - v"
if str.startswith(cmpStr):
#TestTraverseExpr(tmpExp)
tmpExp = Parse(str[len(cmpStr) : ])
lhs = sympy.Symbol(v)
print (tmpExp - lhs)
lhs, rhs = PreprocessEq(lhs, tmpExp)
#assert False
#eqList[index] = Parse(str[len(cmpStr) : ] + " - " + v)
eqList[index] = rhs - lhs
#print("eqList[index] = %s" % str(eqList[index]))
print("eqList[%d] = %s" % (index, eqList[index]))
#print(separatevars(expr=eqList[index], symbols=[xdst, xsrc]))
print("Number of operators: %s" % \
sympy.count_ops(eqList[index], visual=True))
GenCode(eqList[index])
开发者ID:alexsusu,项目名称:video-diff,代码行数:37,代码来源:parse.py
示例9: count
def count(val):
return count_ops(val, visual=True)
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:2,代码来源:test_count_ops.py
示例10: count_ops
def count_ops(self, visual=None):
"""wrapper for count_ops that returns the operation count."""
from sympy import count_ops
return count_ops(self, visual)
return sum(a.count_ops(visual) for a in self.args)
开发者ID:wxgeo,项目名称:sympy,代码行数:5,代码来源:basic.py
示例11: _mellin_transform
def _mellin_transform(f, x, s_, integrator=_default_integrator, simplify=True):
""" Backend function to compute mellin transforms. """
from sympy import re, Max, Min, count_ops
# We use a fresh dummy, because assumptions on s might drop conditions on
# convergence of the integral.
s = _dummy('s', 'mellin-transform', f)
F = integrator(x**(s-1) * f, x)
if not F.has(Integral):
return _simplify(F.subs(s, s_), simplify), (-oo, oo), True
if not F.is_Piecewise:
raise IntegralTransformError('Mellin', f, 'could not compute integral')
F, cond = F.args[0]
if F.has(Integral):
raise IntegralTransformError('Mellin', f, 'integral in unexpected form')
def process_conds(cond):
"""
Turn ``cond`` into a strip (a, b), and auxiliary conditions.
"""
a = -oo
b = oo
aux = True
conds = conjuncts(to_cnf(cond))
t = Dummy('t', real=True)
for c in conds:
a_ = oo
b_ = -oo
aux_ = []
for d in disjuncts(c):
d_ = d.replace(re, lambda x: x.as_real_imag()[0]).subs(re(s), t)
if not d.is_Relational or (d.rel_op != '<' and d.rel_op != '<=') \
or d_.has(s) or not d_.has(t):
aux_ += [d]
continue
soln = _solve_inequality(d_, t)
if not soln.is_Relational or \
(soln.rel_op != '<' and soln.rel_op != '<='):
aux_ += [d]
continue
if soln.lhs == t:
b_ = Max(soln.rhs, b_)
else:
a_ = Min(soln.lhs, a_)
if a_ != oo and a_ != b:
a = Max(a_, a)
elif b_ != -oo and b_ != a:
b = Min(b_, b)
else:
aux = And(aux, Or(*aux_))
return a, b, aux
conds = [process_conds(c) for c in disjuncts(cond)]
conds = filter(lambda x: x[2] is not False, conds)
conds.sort(key=lambda x: (x[0]-x[1], count_ops(x[2])))
if not conds:
raise IntegralTransformError('Mellin', f, 'no convergence found')
a, b, aux = conds[0]
return _simplify(F.subs(s, s_), simplify), (a, b), aux
开发者ID:ALGHeArT,项目名称:sympy,代码行数:63,代码来源:transforms.py
示例12: simplify
from sympy import simplify, cos, sin, trigsimp, cancel
from sympy import sqrt, count_ops, oo, symbols, log
from sympy.abc import x, y
expr = (2*x + 3*x**2)/(4*x*sin(y)**2 + 2*x*cos(y)**2)
expr
simplify(expr)
trigsimp(expr)
cancel(_)
root = 4/(sqrt(2)+3)
simplify(root, ratio=1) == root
count_ops(simplify(root, ratio=oo)) > count_ops(root)
x, y = symbols('x y', positive=True)
expr2 = log(x) + log(y) + log(x)*log(1/y)
expr3 = simplify(expr2)
expr3
count_ops(expr2)
count_ops(expr3)
print(count_ops(expr2, visual=True))
print(count_ops(expr3, visual=True))
开发者ID:xenron,项目名称:sandbox-da-python,代码行数:23,代码来源:8823_06_new11.py
示例13: intersect
def intersect(self, other, **flags):
if not isinstance(other, Circle):
raise BaryException()
if sympy.count_ops(self.eqn()) > sympy.count_ops(other.eqn()):
raise BaryException()
return [Point(*sol) for sol in hsolve([self.eqn(), ((self._eqn - other._eqn) / (x + y + z)).simplify()], x, y, z, **flags)]
开发者ID:Delfad0r,项目名称:python-bary,代码行数:6,代码来源:bary.py
示例14: count_ops
def count_ops(self, visual=None):
"""wrapper for count_ops that returns the operation count."""
from sympy import count_ops
return count_ops(self, visual)
开发者ID:bladewang,项目名称:sympy,代码行数:4,代码来源:basic.py
示例15: cgen_ncomp
#.........这里部分代码省略.........
tpf = (xFj - xPj)/(xTj - xPj)
xP = [(((xF[i]/ppf)*(beta[i]**(NT+1) - 1))/(beta[i]**(NT+1) - beta[i]**(-NP))) \
for i in r]
xT = [(((xF[i]/tpf)*(1 - beta[i]**(-NP)))/(beta[i]**(NT+1) - beta[i]**(-NP))) \
for i in r]
rfeed = xFj / xF[k]
rprod = xPj / xP[k]
rtail = xTj / xT[k]
# setup constraint equations
numer = [ppf*xP[i]*log(rprod) + tpf*xT[i]*log(rtail) - xF[i]*log(rfeed) for i in r]
denom = [log(beta[j]) * ((beta[i] - 1.0)/(beta[i] + 1.0)) for i in r]
LoverF = sum([n/d for n, d in zip(numer, denom)])
SWUoverF = -1.0 * sum(numer)
SWUoverP = SWUoverF / ppf
prod_constraint = (xPj/xFj)*ppf - (beta[j]**(NT+1) - 1)/\
(beta[j]**(NT+1) - beta[j]**(-NP))
tail_constraint = (xTj/xFj)*(sum(xT)) - (1 - beta[j]**(-NP))/\
(beta[j]**(NT+1) - beta[j]**(-NP))
#xp_constraint = 1.0 - sum(xP)
#xf_constraint = 1.0 - sum(xF)
#xt_constraint = 1.0 - sum(xT)
# This is NT(NP,...) and is correct!
#nt_closed = solve(prod_constraint, NT)[0]
# However, this is NT(NP,...) rewritten (by hand) to minimize the number of NP
# and M* instances in the expression. Luckily this is only depends on the key
# component and remains general no matter the number of components.
nt_closed = (-MW[0]*log(alpha) + Mstar*log(alpha) + log(xTj) + log((-1.0 + xPj/\
xF[0])/(xPj - xTj)) - log(alpha**(NP*(MW[0] - Mstar))*(xF[0]*xPj - xPj*xTj)/\
(-xF[0]*xPj + xF[0]*xTj) + 1))/((MW[0] - Mstar)*log(alpha))
# new expression for normalized flow rate
# NOTE: not needed, solved below
#loverf = LoverF.xreplace({NT: nt_closed})
# Define the constraint equation with which to solve NP. This is chosen such to
# minimize the number of ops in the derivatives (and thus np_closed). Other,
# more verbose possibilities are commented out.
#np_constraint = (xP[j]/sum(xP) - xPj).xreplace({NT: nt_closed})
#np_constraint = (xP[j]- sum(xP)*xPj).xreplace({NT: nt_closed})
#np_constraint = (xT[j]/sum(xT) - xTj).xreplace({NT: nt_closed})
np_constraint = (xT[j] - sum(xT)*xTj).xreplace({NT: nt_closed})
# get closed form approximation of NP via symbolic derivatives
stat = _aggstatus(stat, " order-{0} NP approximation".format(nporder), aggstat)
d0NP = np_constraint.xreplace({NP: NP0})
d1NP = diff(np_constraint, NP, 1).xreplace({NP: NP0})
if 1 == nporder:
np_closed = NP0 - d1NP / d0NP
elif 2 == nporder:
d2NP = diff(np_constraint, NP, 2).xreplace({NP: NP0})/2.0
# taylor series polynomial coefficients, grouped by order
# f(x) = ax**2 + bx + c
a = d2NP
b = d1NP - 2*NP0*d2NP
c = d0NP - NP0*d1NP + NP0*NP0*d2NP
# quadratic eq. (minus only)
#np_closed = (-b - sqrt(b**2 - 4*a*c)) / (2*a)
# However, we need to break up this expr as follows to prevent
# a floating point arithmetic bug if b**2 - 4*a*c is very close
# to zero but happens to be negative. LAME!!!
np_2a = 2*a
np_sqrt_base = b**2 - 4*a*c
np_closed = (-NP_b - sqrt(NP_sqrt_base)) / (NP_2a)
else:
raise ValueError("nporder must be 1 or 2")
# generate cse for writing out
msg = " minimizing ops by eliminating common sub-expressions"
stat = _aggstatus(stat, msg, aggstat)
exprstages = [Eq(NP_b, b), Eq(NP_2a, np_2a),
# fix for floating point sqrt() error
Eq(NP_sqrt_base, np_sqrt_base), Eq(NP_sqrt_base, Abs(NP_sqrt_base)),
Eq(NP1, np_closed), Eq(NT1, nt_closed).xreplace({NP: NP1})]
cse_stages = cse(exprstages, numbered_symbols('n'))
exprothers = [Eq(LpF, LoverF), Eq(PpF, ppf), Eq(TpF, tpf),
Eq(SWUpF, SWUoverF), Eq(SWUpP, SWUoverP)] + \
[Eq(*z) for z in zip(xPi, xP)] + [Eq(*z) for z in zip(xTi, xT)]
exprothers = [e.xreplace({NP: NP1, NT: NT1}) for e in exprothers]
cse_others = cse(exprothers, numbered_symbols('g'))
exprops = count_ops(exprstages + exprothers)
cse_ops = count_ops(cse_stages + cse_others)
msg = " reduced {0} ops to {1}".format(exprops, cse_ops)
stat = _aggstatus(stat, msg, aggstat)
# create function body
ccode, repnames = cse_to_c(*cse_stages, indent=6, debug=debug)
ccode_others, repnames_others = cse_to_c(*cse_others, indent=6, debug=debug)
ccode += ccode_others
repnames |= repnames_others
msg = " completed in {0:.3G} s".format(time.time() - start_time)
stat = _aggstatus(stat, msg, aggstat)
if aggstat:
print(stat)
return ccode, repnames, stat
开发者ID:FlanFlanagan,项目名称:pyne,代码行数:101,代码来源:enrich_multi_sym.py
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