Unit.
subs
(*args, **kwargs)¶Substitutes old for new in an expression after sympifying args.
two arguments, e.g. foo.subs(old, new)
replacements are processed in the order given with successive patterns possibly affecting replacements already made.
In this case the old/new pairs will be sorted by op count and in case of a tie, by number of args and the default_sort_key. The resulting sorted list is then processed as an iterable container (see previous).
If the keyword simultaneous
is True, the subexpressions will not be
evaluated until all the substitutions have been made.
See also
Examples
>>> from sympy import pi, exp, limit, oo
>>> from sympy.abc import x, y
>>> (1 + x*y).subs(x, pi)
pi*y + 1
>>> (1 + x*y).subs({x:pi, y:2})
1 + 2*pi
>>> (1 + x*y).subs([(x, pi), (y, 2)])
1 + 2*pi
>>> reps = [(y, x**2), (x, 2)]
>>> (x + y).subs(reps)
6
>>> (x + y).subs(reversed(reps))
x**2 + 2
>>> (x**2 + x**4).subs(x**2, y)
y**2 + y
To replace only the x**2 but not the x**4, use xreplace:
>>> (x**2 + x**4).xreplace({x**2: y})
x**4 + y
To delay evaluation until all substitutions have been made,
set the keyword simultaneous
to True:
>>> (x/y).subs([(x, 0), (y, 0)])
0
>>> (x/y).subs([(x, 0), (y, 0)], simultaneous=True)
nan
This has the added feature of not allowing subsequent substitutions to affect those already made:
>>> ((x + y)/y).subs({x + y: y, y: x + y})
1
>>> ((x + y)/y).subs({x + y: y, y: x + y}, simultaneous=True)
y/(x + y)
In order to obtain a canonical result, unordered iterables are sorted by count_op length, number of arguments and by the default_sort_key to break any ties. All other iterables are left unsorted.
>>> from sympy import sqrt, sin, cos
>>> from sympy.abc import a, b, c, d, e
>>> A = (sqrt(sin(2*x)), a)
>>> B = (sin(2*x), b)
>>> C = (cos(2*x), c)
>>> D = (x, d)
>>> E = (exp(x), e)
>>> expr = sqrt(sin(2*x))*sin(exp(x)*x)*cos(2*x) + sin(2*x)
>>> expr.subs(dict([A,B,C,D,E]))
a*c*sin(d*e) + b
The resulting expression represents a literal replacement of the old arguments with the new arguments. This may not reflect the limiting behavior of the expression:
>>> (x**3 - 3*x).subs({x: oo})
nan
>>> limit(x**3 - 3*x, x, oo)
oo
If the substitution will be followed by numerical evaluation, it is better to pass the substitution to evalf as
>>> (1/x).evalf(subs={x: 3.0}, n=21)
0.333333333333333333333
rather than
>>> (1/x).subs({x: 3.0}).evalf(21)
0.333333333333333314830
as the former will ensure that the desired level of precision is obtained.