我想让用户输入特定的值，然后系统根据这些值计算出许多结果-我的程序仅具有几个功能就变得非常复杂。我提供了一个包含3个简单函数和6个具有以下关系的变量的示例：
我拥有的代码如下：
class MyCalculator:
def __init__(self):
self.a = None
self.b = None
self.c = None
self.d = None
self.e = None
self.f = None
def set(self, field, val):
if field == "a": self.a = val
if field == "b": self.b = val
if field == "c": self.c = val
if field == "d": self.d = val
if field == "e": self.e = val
for i in range(10): # circle round a few times to ensure everything has computed
if self.a and self.b:
self.c = self.a * self.b
if self.a and self.c:
self.b = self.c / self.a
if self.b and self.c:
self.a = self.c / self.b
if self.b and self.d:
self.e = self.b + self.d
if self.e and self.b:
self.d = self.e - self.b
if self.e and self.d:
self.b = self.e - self.d
if self.c and self.e:
self.f = self.c / self.e
if self.f and self.e:
self.e = self.f * self.e
if self.f and self.c:
self.e = self.c / self.f
def status(self):
print(f"a = {self.a} b = {self.b} c = {self.c} d = {self.d} e = {self.e} f = {self.f} ")
然后，如果我运行以下代码：
example1 = MyCalculator()
example1.set("a", 5)
example1.set("e", 7)
example1.set("c", 2)
example1.status()
这将打印出a = 5.0 b = 0.40000000000000036 c = 2.0000000000000018 d = 6.6 e = 7.0 f = 0.285714285714286
我想要一种更简单的方法，使用sympy和numpy之类的东西来达到相同的结果，但是到目前为止，我找不到任何能用的
参考方案
There's a live version of this solution online you can try for yourself
这是使用Sympy的完整解决方案。您需要做的就是在exprStr定义顶部的MyCalculator元组中输入所需的表达式，然后所有依赖项满足的东西都应该照顾好自己：
from sympy import S, solveset, Symbol
from sympy.parsing.sympy_parser import parse_expr
class MyCalculator:
# sympy assumes all expressions are set equal to zero
exprStr = (
'a*b - c',
'b + d - e',
'c/e - f'
)
# parse the expression strings into actual expressions
expr = tuple(parse_expr(es) for es in exprStr)
# create a dictionary to lookup expressions based on the symbols they depend on
exprDep = {}
for e in expr:
for s in e.free_symbols:
exprDep.setdefault(s, set()).add(e)
# create a set of the used symbols for input validation
validSymb = set(exprDep.keys())
def __init__(self, usefloat=False):
"""usefloat: if set, store values as standard Python floats (instead of the Sympy numeric types)
"""
self.vals = {}
self.numify = float if usefloat else lambda x: x
def set(self, symb, val, _exclude=None):
# ensure that symb is a sympy Symbol object
if isinstance(symb, str): symb = Symbol(symb)
if symb not in self.validSymb:
raise ValueError("Invalid input symbol.\n"
"symb: %s, validSymb: %s" % (symb, self.validSymb))
# initialize the set of excluded expressions, if needed
if _exclude is None: _exclude = set()
# record the updated value of symb
self.vals[symb] = self.numify(val)
# loop over all of the expressions that depend on symb
for e in self.exprDep[symb]:
if e in _exclude:
# we've already calculated an update for e in an earlier recursion, skip it
continue
# mark that e should be skipped in future recursions
_exclude.add(e)
# determine the symbol and value of the next update (if any)
nextsymbval = self.calc(symb, e)
if nextsymbval is not None:
# there is another symbol to update, recursively call self.set
self.set(*nextsymbval, _exclude)
def calc(self, symb, e):
# find knowns and unknowns of the expression
known = [s for s in e.free_symbols if s in self.vals]
unknown = [s for s in e.free_symbols if s not in known]
if len(unknown) > 1:
# too many unknowns, can't do anything with this expression right now
return None
elif len(unknown) == 1:
# solve for the single unknown
nextsymb = unknown[0]
else:
# solve for the first known that isn't the symbol that was just changed
nextsymb = known[0] if known[0] != symb else known[1]
# do the actual solving
sol = solveset(e, nextsymb, domain=S.Reals)
# evaluate the solution given the known values, then return a tuple of (next-symbol, result)
return nextsymb, sol.subs(self.vals).args[0]
def __str__(self):
return ' '.join(sorted('{} = {}'.format(k,v) for k,v in self.vals.items()))
测试出来：
mycalc = MyCalculator()
mycalc.set("a", 5)
mycalc.set("e", 7)
mycalc.set("c", 2)
print(mycalc)
输出：
a = 5 b = 2/5 c = 2 d = 33/5 e = 7 f = 2/7
关于Sympy的一件整洁的事情是它使用了理性的数学运算法则，从而避免了诸如2/7之类的任何奇怪的舍入错误。如果您希望将结果作为标准Python float值，则可以将usefloat标志传递给MyCalculator：
mycalc = MyCalculator(usefloat=True)
mycalc.set("a", 5)
mycalc.set("e", 7)
mycalc.set("c", 2)
print(mycalc)
输出：
a = 5.0 b = 0.4 c = 2.0 d = 6.6 e = 7.0 f = 0.2857142857142857
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