Basic Operations

From

Here we discuss some of the most basic operations needed for expression manipulation in SymPy. Some more advanced operations will be discussed later in the :ref:advanced expression manipulation <tutorial-manipulation> section.

    >>> from sympy import *
    >>> x, y, z = symbols("x y z")
In Julia:
julia> using SymPy

julia> x, y, z = symbols("x y z")
(x, y, z)
  • We didn't replicate from sympy import *, though this is mostly done through the command import_from(sympy). By default, SymPy only makes available a priviledged collection of the functions available through the sympy object. The import_from imports most all of the rest.

  • If a function is not imported, it may be referenced through qualification, asin sympy.expand_trig, as will be seen in the following.

  • the use of symbols to construct symbolic values is more easily facilitated with the macro @syms, used as follows

julia> @syms x, y, z
(x, y, z)

Substitution

One of the most common things you might want to do with a mathematical expression is substitution. Substitution replaces all instances of something in an expression with something else. It is done using the subs method. For example

    >>> expr = cos(x) + 1
    >>> expr.subs(x, y)
    cos(y) + 1
In Julia:
julia> expr = cos(x) + 1
cos(x) + 1

julia> expr.subs(x, y)
cos(y) + 1

Julia also allows "call" notation using a pairs to indicate the substitution:

julia> expr(x => y)
cos(y) + 1

Substitution is usually done for one of two reasons:

  1. Evaluating an expression at a point. For example, if our expression is cos(x) + 1 and we want to evaluate it at the point x = 0, so that we get cos(0) + 1, which is 2.
   >>> expr.subs(x, 0)
   2
In Julia:
julia> expr(x => 0)
2

  1. Replacing a subexpression with another subexpression. There are two reasons we might want to do this. The first is if we are trying to build an expression that has some symmetry, such as x^{x^{x^x}}. To build this, we might start with x**y, and replace y with x**y. We would then get x**(x**y). If we replaced y in this new expression with x**x, we would get x**(x**(x**x)), the desired expression.
   >>> expr = x**y
   >>> expr
   x**y
   >>> expr = expr.subs(y, x**y)
   >>> expr
   x**(x**y)
   >>> expr = expr.subs(y, x**x)
   >>> expr
   x**(x**(x**x))
In Julia:
julia> expr = x^y
 y
x

julia> expr = expr(y => x^y)
 ⎛ y⎞
 ⎝x ⎠
x

julia> expr = expr(y => x^x)
 ⎛ ⎛ x⎞⎞
 ⎜ ⎝x ⎠⎟
 ⎝x    ⎠
x

The second is if we want to perform a very controlled simplification, or perhaps a simplification that SymPy is otherwise unable to do. For example, say we have \sin(2x) + \cos(2x), and we want to replace \sin(2x) with 2\sin(x)\cos(x). As we will learn later, the function expand_trig does this. However, this function will also expand \cos(2x), which we may not want. While there are ways to perform such precise simplification, and we will learn some of them in the :ref:advanced expression manipulation <tutorial-manipulation> section, an easy way is to just replace \sin(2x) with 2\sin(x)\cos(x).

   >>> expr = sin(2*x) + cos(2*x)
   >>> expand_trig(expr)
   2*sin(x)*cos(x) + 2*cos(x)**2 - 1
   >>> expr.subs(sin(2*x), 2*sin(x)*cos(x))
   2*sin(x)*cos(x) + cos(2*x)
In Julia:
  • expand_trig is not exported, so we qualify it:
julia> expr = sin(2*x) + cos(2*x)
sin(2⋅x) + cos(2⋅x)

julia> sympy.expand_trig(expr) |> string
"2*sin(x)*cos(x) + 2*cos(x)^2 - 1"

julia> expr(sin(2*x) => 2*sin(x)*cos(x))
2⋅sin(x)⋅cos(x) + cos(2⋅x)

There are two important things to note about subs. First, it returns a new expression. SymPy objects are immutable. That means that subs does not modify it in-place. For example

   >>> expr = cos(x)
   >>> expr.subs(x, 0)
   1
   >>> expr
   cos(x)
   >>> x
   x
In Julia:
julia> expr = cos(x)
cos(x)

julia> expr(x => 0)
1

julia> expr
cos(x)

julia> x
x

Quick Tip

SymPy expressions are immutable. No function will change them in-place.

Here, we see that performing expr.subs(x, 0) leaves expr unchanged. In fact, since SymPy expressions are immutable, no function will change them in-place. All functions will return new expressions.

To perform multiple substitutions at once, pass a list of (old, new) pairs to subs.

    >>> expr = x**3 + 4*x*y - z
    >>> expr.subs([(x, 2), (y, 4), (z, 0)])
    40
In Julia:
julia> expr = x^3 + 4*x*y - z;  string(expr)
"x^3 + 4*x*y - z"

julia> expr.subs([(x, 2), (y, 4), (z, 0)])
40

Or, using pairs:

julia> expr(x=>2, y=>4, z=>0)
40

It is often useful to combine this with a list comprehension to do a large set of similar replacements all at once. For example, say we had x^4 - 4x^3 + 4x^2 - 2x + 3 and we wanted to replace all instances of x that have an even power with y, to get y^4 - 4x^3 + 4y^2 - 2x + 3.

    >>> expr = x**4 - 4*x**3 + 4*x**2 - 2*x + 3
    >>> replacements = [(x**i, y**i) for i in range(5) if i % 2 == 0]
    >>> expr.subs(replacements)
    -4*x**3 - 2*x + y**4 + 4*y**2 + 3
In Julia:
julia> expr = x^4 - 4*x^3 + 4*x^2 - 2*x + 3
 4      3      2
x  - 4⋅x  + 4⋅x  - 2⋅x + 3

julia> replacements = [(x^i, y^i) for i in 1:5 if iseven(i)]
2-element Vector{Tuple{Sym, Sym}}:
 (x^2, y^2)
 (x^4, y^4)

julia> expr.subs(replacements)
     3          4      2
- 4⋅x  - 2⋅x + y  + 4⋅y  + 3

Converting Strings to SymPy Expressions

The sympify function (that's sympify, not to be confused with simplify) can be used to convert strings into SymPy expressions.

For example

    >>> str_expr = "x**2 + 3*x - 1/2"
    >>> expr = sympify(str_expr)
    >>> expr
    x**2 + 3*x - 1/2
    >>> expr.subs(x, 2)
    19/2
In Julia:

As sympify is not passed a symbolic value, it is qualified:

julia> str_expr = "x^2 + 3*x - 1/2"
"x^2 + 3*x - 1/2"

julia> expr = sympy.sympify(str_expr)
 2         1
x  + 3⋅x - ─
           2
julia> expr.subs(x, 2)
19/2

Alert:

sympify uses eval. Don't use it on unsanitized input.

evalf

To evaluate a numerical expression into a floating point number, use evalf.

    >>> expr = sqrt(8)
    >>> expr.evalf()
    2.82842712474619
In Julia:
  • We must use a symbolic value for 8:
julia> expr = sqrt(Sym(8))
2⋅√2

julia> expr.evalf()
2.82842712474619
N is different in SymPy.jl

More importantly, SymPy.jl treats N differently from evalf. N is used to convert a SymPy numeric (or Boolean) value to a Julian counterpart. The main difference between N(x) and convert(T, x), is that rather than specify the Julia type as T, N works to guess the appropriate type for the SymPy object.

julia> N(sqrt(8))   # brings back as BigFloat
2.8284271247461903
julia> N(sqrt(9))   # an Int
3.0

SymPy can evaluate floating point expressions to arbitrary precision. By default, 15 digits of precision are used, but you can pass any number as the argument to evalf. Let's compute the first 100 digits of \pi.

    >>> pi.evalf(100)
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068
In Julia:
julia> PI.evalf(100)
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068

To numerically evaluate an expression with a Symbol at a point, we might use subs followed by evalf, but it is more efficient and numerically stable to pass the substitution to evalf using the subs flag, which takes a dictionary of Symbol: point pairs.

    >>> expr = cos(2*x)
    >>> expr.evalf(subs={x: 2.4})
    0.0874989834394464
In Julia:

A Dict can be used:

julia> expr = cos(2*x)
cos(2⋅x)

julia> expr.evalf(subs=Dict(x => 2.4))
0.0874989834394464

Sometimes there are roundoff errors smaller than the desired precision that remain after an expression is evaluated. Such numbers can be removed at the user's discretion by setting the chop flag to True.

    >>> one = cos(1)**2 + sin(1)**2
    >>> (one - 1).evalf()
    -0.e-124
    >>> (one - 1).evalf(chop=True)
    0
In Julia:
  • we need to use symbolic values for 1 in defining _one:
julia> _one = cos(Sym(1))^2 + sin(Sym(1))^2
   2         2
cos (1) + sin (1)

julia> (_one - 1).evalf()
-0.e-124
julia> (_one - 1).evalf(chop=true)
0

N with Julia

The N function is used to convert a symbolic number or boolean into a Julia counterpart.

julia> two = Sym(2)
2

julia> a,b,c,d = two, sqrt(two), two^20, two^100
(2, sqrt(2), 1048576, 1267650600228229401496703205376)

julia> N.((a,b,c,d))
(2, 1.414213562373095048801688724209698078569671875376948073176679737990732478462102, 1048576, 1267650600228229401496703205376)

lambdify

subs and evalf are good if you want to do simple evaluation, but if you intend to evaluate an expression at many points, there are more efficient ways. For example, if you wanted to evaluate an expression at a thousand points, using SymPy would be far slower than it needs to be, especially if you only care about machine precision. Instead, you should use libraries like NumPy <http://www.numpy.org/>_ and SciPy <http://www.scipy.org/>_.

The easiest way to convert a SymPy expression to an expression that can be numerically evaluated is to use the lambdify function. lambdify acts like a lambda function, except it converts the SymPy names to the names of the given numerical library, usually NumPy. For example

    >>> import numpy # doctest:+SKIP
    >>> a = numpy.arange(10) # doctest:+SKIP
    >>> expr = sin(x)
    >>> f = lambdify(x, expr, "numpy") # doctest:+SKIP
    >>> f(a) # doctest:+SKIP
    [ 0.          0.84147098  0.90929743  0.14112001 -0.7568025  -0.95892427
     -0.2794155   0.6569866   0.98935825  0.41211849]
Alert

lambdify uses eval. Don't use it on unsanitized input.

In Julia:
  • lambdify is defined seperately and with a different argument order: lambdify(ex, vars=free_symbols(ex)).
julia> a = 0:10
0:10

julia> @syms x
(x,)

julia> expr = sin(x)
sin(x)

julia> fn = lambdify(expr);

julia> fn.(a)
11-element Vector{Float64}:
  0.0
  0.8414709848078965
  0.9092974268256817
  0.1411200080598672
 -0.7568024953079282
 -0.9589242746631385
 -0.27941549819892586
  0.6569865987187891
  0.9893582466233818
  0.4121184852417566
 -0.5440211108893698
Technical note

The lambdify function converts a symbolic expression into a Julia expression, and then creates a function using invokelatest to avoid world age issues.

More performant functions can be produced using the following pattern:

julia> ex = sin(x)^2 + x^2
 2      2
x  + sin (x)

julia> body = convert(Expr, ex)
:(SymPy.__POW__(x, 2) + SymPy.__POW__(sin(x), 2))

julia> syms = Symbol.(free_symbols(ex))
1-element Vector{Symbol}:
 :x

julia> fn = eval(Expr(:function, Expr(:call, gensym(), syms...), body));

julia> fn(pi)
9.869604401089358

You can use other libraries than NumPy. For example, to use the standard library math module, use "math".

    >>> f = lambdify(x, expr, "math")
    >>> f(0.1)
    0.0998334166468
In Julia:
  • this doesn't apply, so is not implemented.

To use lambdify with numerical libraries that it does not know about, pass a dictionary of sympy_name:numerical_function pairs. For example

    >>> def mysin(x):
    ...     """
    ...     My sine. Note that this is only accurate for small x.
    ...     """
    ...     return x
    >>> f = lambdify(x, expr, {"sin":mysin})
    >>> f(0.1)
    0.1
In Julia:
  • The fns dictionary coud be used to do this, though due to the call of eval, we must do this in the proper module:
julia> mysin(x) = cos(x)
mysin (generic function with 1 method)

julia> ex = sin(x)
sin(x)

julia> body = SymPy.walk_expression(ex, fns=Dict("sin" => :mysin))
:(mysin(x))

julia> syms = (:x,)
(:x,)

julia> fn = eval(Expr(:function, Expr(:call, gensym(), syms...), body));

julia> fn(0)
1.0

TODO

Write an advanced numerics section