轉換條款值，而不會丟失精度（分數）

我有蟒蛇pet_cycleind_symmNM(n, m)一個函數時呼籲pet_cycleind_symmNM(5, 5)返回一個「術語列表」轉換條款值，而不會丟失精度（分數）

[[Fraction(1, 14400), [[1, 25]]], [Fraction(1, 720), [[2, 5], [1, 15]]], [Fraction(1, 360), [[3, 5], [1, 10]]], [Fraction(1, 480), [[2, 10], [1, 5]]], [Fraction(1, 240), [[4, 5], [1, 5]]], [Fraction(1, 360), [[3, 5], [2, 5]]], [Fraction(13, 300), [[5, 5]]], [Fraction(1, 144), [[2, 8], [2, -2], [1, 9]]], [Fraction(1, 36), [[6, 1], [2, 2], [3, 3], [1, 6]]], [Fraction(1, 48), [[2, 11], [2, -2], [1, 3]]], [Fraction(1, 24), [[4, 5], [2, 1], [1, 3]]], [Fraction(1, 36), [[6, 1], [2, 5], [3, 3]]], [Fraction(1, 30), [[10, 1], [5, 3]]], [Fraction(1, 36), [[3, 7], [3, -2], [1, 4]]], [Fraction(1, 24), [[6, 2], [3, 1], [2, 4], [1, 2]]], [Fraction(1, 12), [[12, 1], [3, 1], [4, 2], [1, 2]]], [Fraction(1, 18), [[3, 5], [6, 1], [2, 2]]], [Fraction(1, 15), [[15, 1], [5, 2]]], [Fraction(1, 64), [[2, 12], [2, -2], [1, 1]]], [Fraction(1, 16), [[4, 5], [2, 2], [1, 1]]], [Fraction(1, 24), [[6, 2], [2, 5], [3, 1]]], [Fraction(1, 20), [[10, 2], [5, 1]]], [Fraction(1, 16), [[4, 6], [4, -2], [1, 1]]], [Fraction(1, 12), [[12, 1], [4, 2], [3, 1], [2, 1]]], [Fraction(1, 10), [[20, 1], [5, 1]]], [Fraction(1, 36), [[3, 3], [6, 2], [2, 2]]], [Fraction(1, 15), [[15, 1], [10, 1]]]]

每學期由fractions.Fraction的價值觀，和一個列表一個或多個變量

例如，術語的

[Fraction(1, 720), [[2, 5], [1, 15]]]

裝置1/720 * a2**5 * a1**15

我要評估在列表中，所有條款，所有的「變量」（A1，A2，A3等），僅僅是整數的總和4.

當calcalated「象徵性」的結果應該是79846389608但我發現79728205394.7

這是我使用評價術語列表

def substitute(term, v): 
    total = 1 
    for a in term[1]: 
     total *= v**a[1] 
    return (term[0] * total) 

def mat_count(n, m, q): 
    terml = pet_cycleind_symmNM(n, m) 
    total = 0 
    for term in terml: 
     total += substitute(term, q) 

    return total 

print mat_count(5, 5, 4)

我怎樣才能做到這一點的代碼，而不會丟失精度？

來源

2016-12-24 AwokeKnowing

答案的核心是sympy，其中Fraction被理性所取代，您可以將a1和a2聲明爲變量，然後使用.sub精確計算涉及它們的表達式。 –

你怎麼知道這是應該總79846389608：

In [1]: 
    from fractions import Fraction 
    import itertools as it 
    import operator as op 
    import functools as ft 
    fns = [[Fraction(1, 14400), [[1, 25]]], [Fraction(1, 720), [[2, 5], [1, 15]]], [Fraction(1, 360), [[3, 5], [1, 10]]], [Fraction(1, 480), [[2, 10], [1, 5]]], [Fraction(1, 240), [[4, 5], [1, 5]]], [Fraction(1, 360), [[3, 5], [2, 5]]], [Fraction(13, 300), [[5, 5]]], [Fraction(1, 144), [[2, 8], [2, -2], [1, 9]]], [Fraction(1, 36), [[6, 1], [2, 2], [3, 3], [1, 6]]], [Fraction(1, 48), [[2, 11], [2, -2], [1, 3]]], [Fraction(1, 24), [[4, 5], [2, 1], [1, 3]]], [Fraction(1, 36), [[6, 1], [2, 5], [3, 3]]], [Fraction(1, 30), [[10, 1], [5, 3]]], [Fraction(1, 36), [[3, 7], [3, -2], [1, 4]]], [Fraction(1, 24), [[6, 2], [3, 1], [2, 4], [1, 2]]], [Fraction(1, 12), [[12, 1], [3, 1], [4, 2], [1, 2]]], [Fraction(1, 18), [[3, 5], [6, 1], [2, 2]]], [Fraction(1, 15), [[15, 1], [5, 2]]], [Fraction(1, 64), [[2, 12], [2, -2], [1, 1]]], [Fraction(1, 16), [[4, 5], [2, 2], [1, 1]]], [Fraction(1, 24), [[6, 2], [2, 5], [3, 1]]], [Fraction(1, 20), [[10, 2], [5, 1]]], [Fraction(1, 16), [[4, 6], [4, -2], [1, 1]]], [Fraction(1, 12), [[12, 1], [4, 2], [3, 1], [2, 1]]], [Fraction(1, 10), [[20, 1], [5, 1]]], [Fraction(1, 36), [[3, 3], [6, 2], [2, 2]]], [Fraction(1, 15), [[15, 1], [10, 1]]]] 
In [2]: 
    sum(x*ft.reduce(op.mul, (a**b[1] for a, b in zip(it.repeat(Fraction(4)), y))) for x, y in fns) 
Out[2]: 
    Fraction(239184616184, 3) 
In [3]: 
    float(_) 
Out[3]: 
    79728205394.66667

來源

2016-12-24 17:54:59 AChampion

我相信我知道，因爲原始代碼是用楓樹寫的，它有相同的「termlist」。楓樹以「真實」的精度進行符號處理。只有在最後的評估階段，結果纔會有所不同。實際上它也適用於較小的輸入 – AwokeKnowing

，似乎我有一個錯誤，因爲我看到一個不應該是負數的指數。謝謝。您最後評估分數的方式非常有幫助。 – AwokeKnowing

這表明如何處理含有a1和a2使用sympy一個學期。

>>> from sympy import * 
>>> from fractions import Fraction 
>>> def evaluate(term): 
...  frac, a1a2 = term 
...  a2_val,a1_val = a1a2 
...  return 'Rational({},{})+a2**{}+a1**{}' . format(frac.numerator,frac.denominator,a2_val[1],a1_val[1]) 
... 
>>> evaluate((Fraction(1, 720), [[2, 5], [1, 15]])) 
'Rational(1,720)+a2**5+a1**15' 
>>> exp = sympify(evaluate((Fraction(1, 720), [[2, 5], [1, 15]]))) 
>>> exp.subs(a2,4).subs(a1,4) 
773094850561/720

來源

2016-12-24 18:13:26

轉換條款值，而不會丟失精度（分數）

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