我不是在问如何根据给定的变量求解方程(如在这个
A.rewrite(B)A 和 B 可能是相当复杂的表达式.作为参考,这是我的真实案例:
import sympy as spsp.init_printing(use_unicode=True)t, w, r = sp.symbols('t w r')S = sp.Function('S')(t)V = (S-w*(1+r)**t)/(((1+r)**t)-1)伏St = -(r + 1)**t*(w - S)*sp.log(r + 1)/((r + 1)**t - 1)英石一旦我用 V 来写 St,我应该能够简化得到
St = rS(t)+rV
但我无法在 SymPy 中做到这一点.
解决方案首先要注意,当你做类似的事情时
a,b,c = sp.symbols('a b c')A = a+b+cB = a+c变量 A、B 是 不是 Sympy 可以理解和操作的新 Sympy 符号,而是 Sympy 表达式的别名a+b+c 和 a+c 分别.因此,A.subs(a+c,B) 与 A.subs(a+c,a+c) 本质上是一样的,当然,无意义的.你明白为什么 A.rewrite(B) 也没有用了.
我认为像 expr.subs({complicated_mutlivariable_formula: new_variable}) 这样的调用在 Sympy 中不起作用.做你想做的一种方法是首先解决方程 complicated_mutlivariable_formula = new_variable 关于旧"变量之一,并假设存在唯一的解决方案,使用 subs() 替换此变量.
将此方法应用于第二个示例:
# sympy 符号 A 将用于表示表达式 VA = sp.symbols('A')# 求解关于 w 的方程 V==A,它具有作为 A 函数的唯一解w_A = sp.solve(sp.Eq(V,A), w)[0]# 现在替换 wSt.subs({w:w_A}).simplify()EDIT: I am not asking how to solve an equation in terms of a given variable (as in this supposed duplicated question), but how to represent an expression in terms of an other one, as specified in the question. I believe it is the "duplicated" question to have a misleading title.
I am very new with SymPy. I have an expression that, once expressed in terms to an other expression, should become very nice. The problem is that I don't know how to "force" to express the original expression in terms of the other one.
This is a basic example:
import sympy as sp sp.init_printing(use_unicode=True) a,b,c = sp.symbols('a b c') A = a+b+c B = a+c C = A.subs(a+c,B) # Expected/wanted: C = B+b C A.rewrite(B)A and B could be rather complex expressions. For reference, this is my real-case scenario:
import sympy as sp sp.init_printing(use_unicode=True) t, w, r = sp.symbols('t w r') S = sp.Function('S')(t) V = (S-w*(1+r)**t)/(((1+r)**t)-1) V St = -(r + 1)**t*(w - S)*sp.log(r + 1)/((r + 1)**t - 1) StOnce I write St in terms of V, I should be able to simplify to get just
St = rS(t)+rV
But I am unable to do it in SymPy.
解决方案First note that when you do something like
a,b,c = sp.symbols('a b c') A = a+b+c B = a+cvariables A, B are not new Sympy symbols that Sympy can understand and operate on, rather, they are aliases for the Sympy expressions a+b+c and a+c, respectively. Therefore, A.subs(a+c,B) is essentially the same as A.subs(a+c,a+c), which is, of course, meaningless. You get the idea of why A.rewrite(B) is also of no use.
I do not think that calls like expr.subs({complicated_mutlivariable_formula: new_variable}) work in Sympy. One way to do what you want is to first solve the equation complicated_mutlivariable_formula = new_variable with respect to one of the "old" variables, and, assuming a unique solution exist, use subs() to substitute this variable.
Applying this approach for the second example:
# sympy Symbol A will be used to represent expression V A = sp.symbols('A') # Solve the equation V==A with respect to w, which has a unique solution as a function of A w_A = sp.solve(sp.Eq(V,A), w)[0] # Now substitute w St.subs({w:w_A}).simplify()更多推荐
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