我目前正在上一门数学课程，其目的是理解概念和过程，而不是尽快解决问题。在求解方程式时，我想自己戳一下它们，而不是让它们为我求解。
假设我们有一个非常简单的等式z + 1 = 4-如果我自己解决这个问题，显然我会从两边都减去1，但是我不知道sympy是否提供了一种简单的方法来实现。目前，我能想到的最佳解决方案是：
from sympy import *
z = symbols('z')
eq1 = Eq(z + 1, 4)
Eq(eq1.lhs - 1, eq1.rhs - 1)
# Output:
# z == 3
更明显的表达式eq1 - 1仅从左侧减去。我该如何使用sympy这样一步一步地解决相等问题（即不让solve()方法只是给我答案）？任何有可能在对称相等的情况下进行操作的指针都将受到赞赏。
参考方案
https://github.com/sympy/sympy/issues/5031#issuecomment-36996878处有一个“ do”方法和讨论，使您可以对Equal的双方进行“ do”操作。它不被接受为SymPy的添加，但是它是您可以使用的简单附加组件。为方便起见，将其粘贴在此处：
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
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