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SymPy Expression Tree Manipulation

I’ve been working on some extensions to our special function computations in Prediction risk for global-local shrinkage regression and decided to employ SymPy as much as possible. Out of this came an implementation of a bivariate confluent hypergeometric function: the Humbert $$\Phi_1$$. This, and some numeric implementations, are available in a Python package and an R package.

In the course of this work there are expectations that appear as ratios of $$\Phi_1$$ functions, so it’s helpful to have a symbolic replacement routine to identify them. Pattern matching, finding, substitution and replacement are fairly standard in SymPy, so nothing special there; however, when you want something specific, it can get rather tricky.

Personally, I’ve found the approach offered by the sympy.strategies and sympy.unify frameworks the most appealing. See the original discussion here. The reason for their appeal is mostly due to their organization of the processes behind expression tree traversal and manipulation. It’s much easier to see how a very specific and non-trivial simplification or replacement could be accomplished and iteratively improved. These points are made very well in the posts here, so check them out.

Let’s say we want to write a function as_expectations that takes a sympy.Expr and replaces ratios of $$\Phi_1$$ functions according to the following pattern: $\begin{equation} E[X^n] = \frac{\Phi_1(\alpha, \beta, \gamma + n; x, y)}{\Phi_1(\alpha, \beta, \gamma; x, y)} \;. \label{eq:expectation} \end{equation}$

As an example, let’s set up a situation in which as_expectations would be used, and, from there, attempt to construct our function. Naturally, this will involve a test expression with terms that we know match Equation $$\eqref{eq:expectation}$$:

import sympy as sp

from hsplus.horn_symbolic import HornPhi1

a, b, g, z_1, z_2 = sp.symbols('a, b, g, z_1, z_2', real=True)
phi1_1 = HornPhi1((a, b), (g,), z_1, z_2)

n = sp.Dummy('n', integer=True, positive=True)
i = sp.Dummy('i', integer=True, nonnegative=True)

phi1_2 = HornPhi1((a, b), (g + n,), z_1, z_2)
phi1_3 = HornPhi1((a, b), (g + n - i,), z_1, z_2)

r_1 = phi1_2/phi1_1
r_2 = phi1_3/phi1_1

expr = a * r_1 - b * r_1 / g + sp.Sum(z_1/z_2 * r_2 - 3 * r_2, (i, 0,
n))

Our test expression expr looks like this

print(sp.latex(expr, mode='equation*', itex=True))

$\begin{equation*} \frac{a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g, \quad z_{1}, \quad z_{2}\right )\right)}}{\operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}} + \sum_{i=0}^{n} \left(\frac{z_{1} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad - i + n + g, \quad z_{1}, \quad z_{2}\right )\right)}}{z_{2} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}} - \frac{3 \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad - i + n + g, \quad z_{1}, \quad z_{2}\right )\right)}}{\operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}}\right) - \frac{b \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g, \quad z_{1}, \quad z_{2}\right )\right)}}{g \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}} \end{equation*}$

The ratios r_1 and r_2 should both be replaced by a symbol for $$E[X^m]$$, for $$m = n$$ and $$m = n - i$$ when $$i < n$$ respectively. We could allow $$E[X^0]$$, I suppose, but–for a more interesting discussion–let’s not.

We start by creating a SymPy pattern that expresses the mathematical form of $$E[X^m]$$ in Equation $$\eqref{eq:expectation}$$.

pnames = ('a', 'b', 'g', 'z_1', 'z_2')
phi1_wild_args_n = sp.symbols(','.join(n_ + '_w' for n_ in pnames),
cls=sp.Wild, real=True)

n_w = sp.Wild('n_w',
properties=(lambda x: x.is_integer and x.is_positive,),
exclude=(phi1_wild_args_n,))

phi1_wild_d = HornPhi1(phi1_wild_args_n[0:2],
phi1_wild_args_n[2:3],
*phi1_wild_args_n[3:5])

phi1_wild_n = HornPhi1(phi1_wild_args_n[0:2],
(phi1_wild_args_n + n_w,),
*phi1_wild_args_n[3:5])

C_w = sp.Wild('C_w', exclude=[sp.S.Zero])
E_pattern = phi1_wild_n / phi1_wild_d

E_fn = sp.Function("E", real=True)

When we find an $$E[X^m]$$ we’ll replace it with the symbolic function E_fn.

If we focus on only one of the terms (one we know matches E_pattern), r_1, we should find that our pattern suffices:

>>> r_1.match(E_pattern)
{n_w_: _n, z_2_w_: z_2, z_1_w_: z_1, a_w_: a, g_w_: g, b_w_: b}

However, building up to the complexity of expr, we see that a simple product doesn’t:

>>> (a * r_1).match(E_pattern)

Basically, the product has introduced some problems that arise from associativity. Here are the details for the root expression tree:

>>> (a * r_1).func
<class 'sympy.core.mul.Mul'>
>>> (a * r_1).args
(a, 1/HornPhi1(a, b, g, z_1, z_2), HornPhi1(a, b, _n + g, z_1, z_2))

The root operation is multiplication and the operation’s arguments are all terms in the product/division.

Any complete search for matches to E_pattern would have to consider all possible combinations of terms in (a * r_1).args, i.e. all possible groupings that arise due to associativity. The simple inclusion of another Wild term causes the match to succeed, since SymPy’s basic pattern matching does account for associativity in this case.

Here are a few explicit ways to make the match work:

>>> (a * r_1).match(C_w * E_pattern)
{a_w_: a, n_w_: _n, g_w_: g, z_2_w_: z_2, C_w_: a, b_w_: b, z_1_w_:
z_1}

or as a replacement:

res = (a * r_1).replace(C_w * E_pattern, C_w * E_fn(n_w,
*phi1_wild_args_n))
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} a E{\left (n,a,b,g,z_{1},z_{2} \right )} \end{equation*}$

and via rewriterule:

from sympy.unify.rewrite import rewriterule
rl = rewriterule(C_w * E_pattern,
C_w * E_fn(n_w, *phi1_wild_args_n),
phi1_wild_args_n + (n_w, C_w))
res = list(rl(a * r_1))
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} \left [ a E{\left (n,a,b,g,z_{1},z_{2} \right )}\right ] \end{equation*}$

The advantage in using rewriterule is that multiple matches will be returned. If we add another $$\Phi_1$$ in the numerator, so there are multiple possible $$E[X^m]$$, we get

phi1_4 = HornPhi1((a, b), (g + n + 1,), z_1, z_2)

res = list(rl(a * r_1 * phi1_4))
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} \left [ a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g, \quad z_{1}, \quad z_{2}\right )\right)} E{\left (n + 1,a,b,g,z_{1},z_{2} \right )}, \quad a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g, \quad z_{1}, \quad z_{2}\right )\right)} E{\left (n + 1,a,b,g,z_{1},z_{2} \right )}, \quad a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g + 1, \quad z_{1}, \quad z_{2}\right )\right)} E{\left (n,a,b,g,z_{1},z_{2} \right )}, \quad a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g, \quad z_{1}, \quad z_{2}\right )\right)} E{\left (n + 1,a,b,g,z_{1},z_{2} \right )}, \quad a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g, \quad z_{1}, \quad z_{2}\right )\right)} E{\left (n + 1,a,b,g,z_{1},z_{2} \right )}, \quad a \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad n + g + 1, \quad z_{1}, \quad z_{2}\right )\right)} E{\left (n,a,b,g,z_{1},z_{2} \right )}\right ] \end{equation*}$

FYI: the associativity of terms inside the function arguments is causing the seemingly duplicate results.

Naive use of Expr.replace doesn’t give all results; instead, it does something likely unexpected:

res = (a * r_1 * phi1_4).replace(C_w * E_pattern,
C_w * E_fn(n_w, *phi1_wild_args_n))
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} a E{\left (n,a,b,g,z_{1},z_{2} \right )} E{\left (n + 1,a,b,g,z_{1},z_{2} \right )} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)} \end{equation*}$

Returning to our more complicated expr…Just because we can match products doesn’t mean we’re finished, since we still need a good way to traverse the entire expression tree and match the sub-trees. More importantly, adding the multiplicative Wild term C_w is more of a hack than a direct solution, since we don’t want the matched contents of C_w.

Although Expr.replace/xreplace will match sub-expressions, we found above that it produces some odd results. Those results persist when applied to more complicated expressions:

res = expr.replace(C_w * E_pattern, C_w * E_fn(n_w,
*phi1_wild_args_n))
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} a E{\left (n,a,b,g,z_{1},z_{2} \right )} - \frac{b}{g} E{\left (n,a,b,g,z_{1},z_{2} \right )} + \sum_{i=0}^{n} \left(\frac{z_{1} E{\left (n,a,b,- i + g,z_{1},z_{2} \right )} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad - i + g, \quad z_{1}, \quad z_{2}\right )\right)}}{z_{2} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}} - \frac{3 E{\left (n,a,b,- i + g,z_{1},z_{2} \right )} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad - i + g, \quad z_{1}, \quad z_{2}\right )\right)}}{\operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}}\right) \end{equation*}$

Again, it looks like the matching was a little too liberal and introduced extra E and HornPhi1 terms. This is to be expected from the Wild matching in SymPy; it needs us to specify what not to match, as well. Our “fix” that introduced C_w is the exact source of the problem, but we can tell it not to match HornPhi1 terms and get better results:

C_w = sp.Wild('C_w', exclude=[sp.S.Zero, HornPhi1])
res = expr.replace(C_w * E_pattern, C_w * E_fn(n_w,
*phi1_wild_args_n))
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} a E{\left (n,a,b,g,z_{1},z_{2} \right )} - \frac{b}{g} E{\left (n,a,b,g,z_{1},z_{2} \right )} + \sum_{i=0}^{n} \left(\frac{z_{1} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad - i + n + g, \quad z_{1}, \quad z_{2}\right )\right)}}{z_{2} \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}} - \frac{3 \operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad - i + n + g, \quad z_{1}, \quad z_{2}\right )\right)}}{\operatorname{\Phi_1}{\left(\left ( a, \quad b, \quad g, \quad z_{1}, \quad z_{2}\right )\right)}}\right) \end{equation*}$

We’ve stopped it from introducing those superfluous E terms, but we’re still not getting replacements for the HornPhi1 ratios in the sums. Let’s single out those terms and see what’s going on:

res = r_2.find(C_w * E_pattern)
print(sp.latex(res, mode='equation*', itex=True))

$\begin{equation*} \left\{\right\} \end{equation*}$

The constrained integer Wild term, n_w, probably isn’t matching. Given the form of our pattern, n_w should match n - i, but n - i isn’t strictly positive, as required:

>>> (n - i).is_positive == True
False
>>> sp.ask(sp.Q.positive(n - i)) == True
False

Since $$n > 0$$ and $$i >= 0$$, the only missing piece is that $$n > i$$. The most relevant mechanism in SymPy to assess this information is the sympy.assumptions interface. We could add and retrieve the assumption sympy.Q.is_true(n > i) via sympy.assume.global_assumptions, or perform these operations inside of a Python with block, etc. This context management, via sympy.assumptions.assume.AssumptionsContext, would have to be performed manually, since I am not aware of any such mechanism offered by Sum and/or Basic.replace.

Unfortunately, these ideas sound good, but aren’t implemented:

>>> sp.ask(sp.Q.positive(n - i), sp.Q.is_true(n > i)) == True
False

See the documentation for sympy.assumptions.ask.ask; it explicitely states that inequalities aren’t handled, yet.

We could probably perform a manual reworking of sympy.Q.is_true(n > i) to sympy.Q.is_true(n - i > 0), which is of course equivalent to sympy.Q.positive(n - i): the result we want.

If one were to provide this functionality, there’s still the question of how the relevant AssumptionsContexts would be created and passed around/nested during the subexpression replacements. There is no apparent means of adding this sort of functionality through the Basic.replace interface, so this path looks less appealing. However, nesting with blocks from strategies in sympy.strategies does seem quite possible. For example, in sympy.strategies.traverse.sall, one could possibly wrap the return statement after the map(rule, ...) call in a with sympy.assuming(...): block that contains the assumptions for any variables arising as, say, the index of a Sum–like in our case. In this scenario, code in the subexpressions would be able to ask questions like sympy.Q.is_true(n > i) without altering the global assumptions context or the objects involved.

Anyway, that’s all I wanted to cover here. Perhaps later I’ll post a hack for the assumptions approach, but–at the very least–I’ll try to follow up with a more direct solution that uses sympy.strategies.

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