Sympy：手动处理等式

Question

我现在正在学习一门数学课程，目标是理解概念和过程，而不是尽快完成题目。当我解决方程时，我希望自己能动手尝试，而不是让别人给我算出答案。

比如说，我们有一个很简单的方程 z + 1 = 4——如果让我自己来解这个方程，我肯定会从两边都减去1，但我不太清楚 sympy 是否提供了简单的方法来做到这一点。目前我想到的最好的解决方案是：

from sympy import *
z = symbols('z')
eq1 = Eq(z + 1, 4)
Eq(eq1.lhs - 1, eq1.rhs - 1)
# Output:
# z == 3

但是更明显的表达式 eq1 - 1 只是在左边减去1。那我该如何使用 sympy 一步一步地处理等式呢（也就是说，不用 solve() 方法直接给我答案）？如果能给我一些关于 sympy 等式操作的建议，我会非常感激。

Answer 1

这里有一个“do”方法和讨论，地址是 https://github.com/sympy/sympy/issues/5031#issuecomment-36996878，这个方法可以让你对等式的两边进行操作。虽然这个方法还没有被正式加入到SymPy中，但它是一个简单的附加功能，你可以使用。为了方便起见，这里把它贴出来：

def do(self, e, i=None, doit=False):
    """Return a new Eq using function given or a model
    model expression in which a variable represents each
    side of the expression.

    Examples
    ========

    >>> from sympy import Eq
    >>> from sympy.abc import i, x, y, z
    >>> eq = Eq(x, y)

    When the argument passed is an expression with one
    free symbol that symbol is used to indicate a "side"
    in the Eq and an Eq will be returned with the sides
    from self replaced in that expression. For example, to
    add 2 to both sides:

    >>> eq.do(i + 2)
    Eq(x + 2, y + 2)

    To add x to both sides:

    >>> eq.do(i + x)
    Eq(2*x, x + y)

    In the preceding it was actually ambiguous whether x or i
    was to be added but the rule is that any symbol that are
    already in the expression are not to be interpreted as the
    dummy variable. If we try to add z to each side, however, an 
    error is raised because now it is unclear whether i or z is being
    added:

    >>> eq.do(i + z)
    Traceback (most recent call last):
    ...
    ValueError: not sure what symbol is being used to represent a side

    The ambiguity must be resolved by indicating with another parameter 
    which is the dummy variable representing a side:

    >>> eq.do(i + z, i)
    Eq(x + z, y + z)

    Alternatively, if only one Dummy symbol appears in the expression then
    it will be automatically used to represent a side of the Eq.

    >>> eq.do(2*Dummy() + z)
    Eq(2*x + z, 2*y + z)

    Operations like differentiation must be passed as a
    lambda:

    >>> Eq(x, y).do(lambda i: i.diff(x))
    Eq(1, 0)

    Because doit=False by default, the result is not evaluated. to
    evaluate it, either use the doit method or pass doit=True.

    >>> _.doit == Eq(x, y).do(lambda i: i.diff(x), doit=True)
    True
    """
    if not isinstance(e, (FunctionClass, Lambda, type(lambda:1))):
      e = S(e)
      imaybe = e.free_symbols - self.free_symbols
      if not imaybe:
          raise ValueError('expecting a symbol')
      if imaybe and i and i not in imaybe:
          raise ValueError('indicated i not in given expression')
      if len(imaybe) != 1 and not i:
          d = [i for i in imaybe if isinstance(i, Dummy)]
          if len(d) != 1:
              raise ValueError(
                  'not sure what symbol is being used to represent a side')
          i = set(d)
      else:
          i = imaybe
      i = i.pop()
      f = lambda side: e.subs(i, side)
    else:
      f = e
    return self.func(*[f(side) for side in self.args], evaluate=doit)

from sympy.core.relational import Equality
Equality.do = do