SympifyError：SympifyError：使用循环时索引

Question

在更改循环中的参数时，我在使用 simplify 时遇到问题。在添加循环之前，它工作得很好，所以我对出了什么问题有点困惑。这个想法是在参数变化时计算上述方程的不动点。我事先使用随机算法确定了参数。

使用的数据

index   c1  c2  c3  c4  c5
                
2   0.182984    2.016811    0.655393    1.581344    1000.0
3   0.481093    3.696431    0.174021    2.604066    1000.0
4   2.651888    0.665661    2.010521    1.004902    1000.0
5   4.356905    3.805205    0.169469    0.188154    1000.0
6   0.618898    1.205760    0.394822    0.624573    1000.0
7   1.628458    0.908339    0.117855    0.801636    1000.0
8   1.084346    0.251490    5.008077    4.606338    1000.0
9   0.314420    4.553279    0.279103    1.136288    1000.0
10  0.309323    3.447195    0.769426    1.058890    1000.0
11  1.353905    5.034620    3.025668    0.136687    1000.0
12  0.294230    0.590507    0.203964    0.105073    1000.0
13  0.433693    1.040195    0.197015    0.214636    1000.0
14  5.597691    2.734779    0.298786    6.869852    1000.0
15  0.106748    0.329506    1.642285    2.259433    1000.0
16  7.065243    0.138986    6.280275    0.265305    1000.0
17  0.676381    0.263757    6.540224    2.890927    1000.0
18  0.646750    2.573060    0.157341    1.779078    1000.0
19  2.829030    0.208247    0.102454    0.117786    1000.0
20  3.973703    0.134666    1.099034    4.255214    1000.0

df1 = df[df.columns[1]]
df2 = df[df.columns[2]]
df3 = df[df.columns[3]]
df4 = df[df.columns[4]]


EQ=[]
for i in df[:5]:
   a = df["c1"]
   b = df["c2"]
   c = df["c3"]
   d = df["c4"]
   Q = 1
   a1 = 0
   b1 = 0
   c1 = 0
   d1 = 0
   u,v  = sm.symbols('u,v', negative=False)

#  equations
   U = a * u -a1* v**2 - b*v+b1*v + Q
   V = c * u -c1*u*v- d*v + d1 + Q
# use sympy's way of setting equations to zero
   Uqual = sm.Eq(U, 0)
   Vqual = sm.Eq(V, 0)

# compute fixed points
   
  equilibria = sm.solve( (Uqual, Vqual), u,v)
  print('The fixed point(s) of this system are: %s' %equilibria)
  equilibria.append(equilibria)

SympifyError                              Traceback (most recent call last)
<ipython-input-81-7104e05ced6a> in <module>
     16     V = c * u -c1*u*v- d*v + d1 + Q
     17     # use sympy's way of setting equations to zero
---> 18     Uqual = sm.Eq(U, 0)
     19     Vqual = sm.Eq(V, 0)
     20 

~\anaconda3\lib\site-packages\sympy\core\relational.py in __new__(cls, lhs, rhs, **options)
    501             rhs = 0
    502         evaluate = options.pop('evaluate', global_parameters.evaluate)
--> 503         lhs = _sympify(lhs)
    504         rhs = _sympify(rhs)
    505         if evaluate:

~\anaconda3\lib\site-packages\sympy\core\sympify.py in _sympify(a)
    510 
    511     """
--> 512     return sympify(a, strict=True)
    513 
    514 

~\anaconda3\lib\site-packages\sympy\core\sympify.py in sympify(a, locals, convert_xor, strict, rational, evaluate)
    431 
    432     if strict:
--> 433         raise SympifyError(a)
    434 
    435     if iterable(a):

SympifyError: SympifyError: index
1       0.32539361355594*u - 0.153951771353544*v + 1
2      0.111286178007145*u - 0.211620881593914*v + 1
3      0.410704332996077*u - 0.338148622964363*v + 1
4       1.39126513227539*u - 0.715390758416011*v + 1
5       0.289981428632838*u - 3.76334113661812*v + 1
                           ...                      
96      0.450838908230239*u - 7.00849756407416*v + 1
97       4.59646738213032*u - 1.45107766000711*v + 1
98      6.28779804684458*u - 0.395831415205476*v + 1
99     0.196464087698782*u - 0.205057919337616*v + 1
100     1.69031014508742*u - 0.140571509904066*v + 1
Length: 100, dtype: object

Answer 1

在 isympy 会话中：

制作示例数据帧：

In [11]: import pandas as pd

In [12]: df = pd.DataFrame(np.arange(12).reshape(3,4))

In [13]: df
Out[13]: 
   0  1   2   3
0  0  1   2   3
1  4  5   6   7
2  8  9  10  11

设置非迭代案例：

In [15]: u,v  = symbols('u,v', negative=False)

In [16]: a,a1,b,b1 = 1,2,3,4

In [17]: U = a * u -a1* v**2 - b*v+b1*v

In [18]: U
Out[18]: 
       2    
u - 2⋅v  + v

与数据帧值对比：

In [19]: a,b = df[0],df[1]

In [20]: a,b
Out[20]: 
(0    0
 1    4
 2    8
 Name: 0, dtype: int64,
 0    1
 1    5
 2    9
 Name: 1, dtype: int64)

In [21]: U = a * u -a1* v**2 - b*v+b1*v

In [22]: U
Out[22]: 
0         -2*v**2 + 3*v
1      4*u - 2*v**2 - v
2    8*u - 2*v**2 - 5*v
dtype: object

此 U 是 pandas Series，带有对象元素（它们是sympy表达式）。但 U 本身并不是 sympy。

应用于简单表达式的 Eq：

In [23]: Eq(Out[18],0)
Out[23]: 
       2        
u - 2⋅v  + v = 0

您的错误 - 应用于系列的 Eq：

In [24]: Eq(Out[22],0)
---------------------------------------------------------------------------
SympifyError                              Traceback (most recent call last)
Input In [24], in <cell line: 1>()
----> 1 Eq(Out[22],0)

File /usr/local/lib/python3.8/dist-packages/sympy/core/relational.py:626, in Equality.__new__(cls, lhs, rhs, **options)
    624     rhs = 0
    625 evaluate = options.pop('evaluate', global_parameters.evaluate)
--> 626 lhs = _sympify(lhs)
    627 rhs = _sympify(rhs)
    628 if evaluate:

File /usr/local/lib/python3.8/dist-packages/sympy/core/sympify.py:528, in _sympify(a)
    502 def _sympify(a):
    503     """
    504     Short version of :func:`~.sympify` for internal usage for ``__add__`` and
    505     ``__eq__`` methods where it is ok to allow some things (like Python
   (...)
    526 
    527     """
--> 528     return sympify(a, strict=True)

File /usr/local/lib/python3.8/dist-packages/sympy/core/sympify.py:449, in sympify(a, locals, convert_xor, strict, rational, evaluate)
    446                 continue
    448 if strict:
--> 449     raise SympifyError(a)
    451 if iterable(a):
    452     try:

SympifyError: SympifyError: 0         -2*v**2 + 3*v
1      4*u - 2*v**2 - v
2    8*u - 2*v**2 - 5*v
dtype: object

Eq() 不具有“迭代系列”（甚至遍历列表）功能。

我们可以迭代（列表理解）并将 Eq 应用于系列的每个项：

In [25]: [Eq(U[i],0) for i in range(3)]
Out[25]: 
⎡     2                     2                   2          ⎤
⎣- 2⋅v  + 3⋅v = 0, 4⋅u - 2⋅v  - v = 0, 8⋅u - 2⋅v  - 5⋅v = 0⎦

作为一般规则，sympy 和 pandas/numpy 确实效果不好。

Answer 2

很难理解您在上面发布的代码中要实现的目标。因此，以下是我的猜测：

# NOTE: the following variables are of type
# pandas.core.series.Series. They are iterables
# (think of them as arrays)
a = df["c1"]
b = df["c2"]
c = df["c3"]
d = df["c4"]

# constants
Q = 1
a1 = 0
b1 = 0
c1 = 0
d1 = 0

# symbols
u, v  = symbols('u, v', negative=False)

#  equations
# NOTE: because a, b, c, d are iterables, then U, V 
# will be iterables too. Each element will be a SymPy
# expression because you used the symbols u and v.
U = a * u - a1 * v**2 - b * v + b1 * v + Q
V = c * u - c1 * u * v - d * v + d1 + Q

EQ = []
# loop over the equations and solve them
for u_eq, v_eq in zip(U, V):
    # here we are asking to solve u_eq=0 and v_eq=0
    equilibria = solve((u_eq, v_eq), (u, v))
    EQ.append(equilibria)
    print('The fixed point(s) of this system are: %s' % equilibria)