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python dblquad invalid callable

I am trying to approximate the Gauss Linking integral for two straight lines in R^3 using dblquad. I've created this pair of lines as an object.
I have a form for the integrand in parametrisation variables s and t generated by a function gaussint(self,s,t) and this is working. I'm then just trying to define a function which returns the double integral over the two intervals [0,1].
Edit - the code for the function looks like this:
def gaussint(self,s,t):
formnum = self.newlens()[0]*self.newlens()[1]*np.sin(test.angle())*np.cos(test.angle())
formdenone = (np.cos(test.angle())**2)*(t*(self.newlens()[0]) - s*(self.newlens()[1]) + self.adists()[0] - self.adists()[1])**2
formdentwo = (np.sin(test.angle())**2)*(t*(self.newlens()[0]) + s*(self.newlens()[1]) + self.adists()[0] + self.adists()[1])**2
fullden = (4 + formdenone + formdentwo)**(3/2)
fullform = formnum/fullden
return fullform
The various other function calls here are just bits of linear algebra - lengths of lines, angle between them and so forth. s and t have been defined as symbols upstream, if they need to be.
The code for the integration then just looks like this (I've separated it out just to try and work out what was going on:
def approxint(self, s, t):
from scipy.integrate import dblquad
return dblquad(self.gaussint(s,t),0,1, lambda t:0,lambda t:1)
Running it gets me a lengthy bit of somewhat impenetrable process messages, followed by
ValueError: invalid callable given
Any idea where I'm going wrong?
Cheers.

How can I dynamically generate a python symbol statement?

I am trying to write a routine that normalizes (rewrites) a mathematical equation that may have more than one symbol on the LHS so that it only has one.
The following code illustrates what I want to do
Assume I have an equation
ln(x)-ln(x1)= -(a+by)
I want to solve for x or return
x=x1*exp(-a+by)
Using sympy I can do the following
from sympy import *
formula=' log(x)-log(x1) =-(a+b*y)'
lhs,rhs=formula.split('=',1)
x,x_1,y,a,b,y=symbols('x x_1 y a b y')
r=sympy.solve(eval(lhs)-eval(rhs),x)
r
==>
Output: [x1*exp(-a - b*y)]
I am trying to automate this for a range of input lines as follows
from sympy import *
import re
# eventually to be read ina loop from a file
formula="DLOG(SAUMMCREDBISCN/SAUNECONPRVTXN) =-0.142368233181-0.22796245228*(LOG(SAUMMCREDBISCN(-1)/SAUNECONPRVTXN(-1))+0.2*((SAUMMLOANINTRCN(-1)-SAUINTR(-1))/100)-LOG(SAUNYGDPMKTPKN(-1)))+0.576050997065*SAUNYGDPGAP_/100"
#try to convert formula into a string containing just the synbols
sym1=formula.replace("*"," ")
sym1=sym1.replace("DLOG"," ")
sym1=sym1.replace("LOG"," ")
sym1=sym1.replace("EXP"," ")
sym1=sym1.replace("RECODE"," ")
sym1=re.sub('[()/+-\=]',' ',sym1)
sym1=re.sub(' +',' ',sym1)
#This logic works for this particular formula
sym1
#now generate a string that has, instead of spaces between symbols
ss2=sym1.replace(' ',',')
#This is the part that does not work I want to generate a command that effectively says
#symbol,symbol2,..,symboln=symbols('symbol1 symbol2 ... symboln')
#tried this but it fails
eval(ss2)=symbols(sym1)
Generates the result
eval(ss2)=symbols(sym1)
^
SyntaxError: can't assign to function call
Any help for this py noob, would be greatly appreciated.

var('a b c') will inject symbol name 'a', 'b', 'c' into the namespace but perhaps #Blorgbeard is asking about lists or dicts because instead of creating many symbols you could put the symbols in a dictionary and then access them by name:
>>> formula=' log(x)-log(x1) =-(a+b*y)'
>>> eq = Eq(*map(S, formula.split('=', 1)))
>>> v = dict([(i.name, i) for i in eq.free_symbols]); v
{'y': y, 'b': b, 'x': x, 'x1': x1, 'a': a}
>>> solve(eq, v['x'])
[x1*exp(-a - b*y)]
So it is not actually necessary to use eval or to have a variable matching a symbol name: S can convert a string to an expression and free_symbols can identify which symbols are present. Putting them in a dictionary with keys being the Symbol name allows them to be retrieved from the dictionary with a string.

I think what you want to generate a string from these two statements and then you can eval that string
str_eq = f'{ss2} = {symbols(sym1)}'
print(str_eq)
>>' ,SAUMMCREDBISCN,SAUNECONPRVTXN,SAUMMCREDBISCN,SAUNECONPRVTXN,SAUMMLOANINTRCN,SAUINTR,SAUNYGDPMKTPKN,SAUNYGDPGAP_, = (SAUMMCREDBISCN, SAUNECONPRVTXN, SAUMMCREDBISCN, SAUNECONPRVTXN, SAUMMLOANINTRCN, SAUINTR, SAUNYGDPMKTPKN, SAUNYGDPGAP_)'
The first line says - give me a string but run the python code between the {} before returning it. For instance
print(f'Adding two numbers: (2+3) = {2 + 3}')
>> Adding two numbers: (2+3) = 5

First, maybe this could help make your code a bit denser.
If I understand correctly, you're trying to assign a variable using a list of names. You could use vars()[x]=:
import re
from sympy import symbols
symbol_names = re.findall("SAU[A-Z_]+",formula)
symbol_objs = symbols(symbol_names)
for sym_name,sym_obj in zip(symbol_names,symbol_objs):
vars()[sym] = sym_obj

A colleague of mine (Ib Hansen) gave me a very nice and elegant solution to the original problem (how to solve for the complicated expression) that by-passed the string manipulation solution that my original question was struggling with and which most answers addressed.
His solution is to use sympify and solve from sympy
from sympy import sympify, solve
import re
def normalize(var,eq,simplify=False,manual=False):
'''normalize an equation with respect to var using sympy'''
lhs, rhs = eq.split('=')
kat = sympify(f'Eq({lhs},{rhs})')
var_sym = sympify(var)
out = str(solve(kat, var_sym,simplify=simplify,manual=manual))[1:-1]
return f'{var} = {out}'
This works perfectly
from sympy import sympify, solve
import re
def norm(var,eq,simplify=False,manual=False):
'''normalize an equation with respect to var using sympy'''
lhs, rhs = eq.split('=')
kat = sympify(f'Eq({lhs},{rhs})')
var_sym = sympify(var)
out = str(solve(kat, var_sym,simplify=simplify,manual=manual))[1:-1] # simplify takes for ever
return f'{var} = {out}'
``` Re write equation spelling out the dlog (delta log) function that python does no know, and putting log into lowe case '''
test8=norm('saummcredbiscn','log(saummcredbiscn/sauneconprvtxn) -log(saummcredbiscn(-1)/sauneconprvtxn(-1)) =-0.142368233181-0.22796245228*(log(saummcredbiscn(-1)/sauneconprvtxn(-1))+0.2*((saummloanintrcn(-1)-sauintr(-1))/100)-log(saunygdpmktpkn(-1)))+0.576050997065*saunygdpgap_/100')
print (test8)
with result
saummcredbiscn = 0.867301828366361*sauneconprvtxn*(saummcredbiscn(-1.0)/sauneconprvtxn(-1.0))**(19300938693/25000000000)*saunygdpmktpkn(-1.0)**(5699061307/25000000000)*exp(0.00576050997065*saunygdpgap_ + 0.00045592490456*sauintr(-1.0) - 0.00045592490456*saummloanintrcn(-1.0)

converting a Fanuc g-code including Macro-B to basic syntax

I have a g-code written in Fanuc g-code format including Macro-B (more info here), for example
#101 = 2.0 (first variable)
#102 = 0.1 (second variable)
#103 = [#101 + #102 * 3] (third variable using simple arithmetic)
G01 X#101 Y#103 F0.1
which should be converted to:
G01 X1.0 Y2.3 F0.1
more elaborate examples here and here.
things to be changed:
all instances of a variable slot should be replace with its value:
(#\d+)\s*=\s*(-?\d*\.\d+|\d+\.\d*)
arithmetic +, -, * and / inside the [...] need to be calculated:
(#\d+)\s*=\s*\[(#\d+|(-?\d*\.\d+|\d+\.\d*))(\s*[+\-*/]\s*(#\d+|(-?\d*\.\d+|\d+\.\d*|\d+)))*\]
comments (...) could be ignored or removed.
I would appreciate if you could help me know how i can do this in Python and if the regex I have above is correct. Thanks for your support in advance.
P.S.1. Unfortunately I can't find the syntax highlighting for fenced code blocks for g-code
P.S.2. when changing floats to strings one should consider the issue with Python floating point handeling. I made this function to solve that:
def f32str(inputFloat):
"""
This function converts a Python float to a string with 3 decimals
"""
return str(f"{inputFloat:.3f}")

OK I found a piece of code which does the job. Assuming gcode is a multiline string read from a Fanuc G-code:
import re
import os
def f32str(inputFloat):
return str(f"{inputFloat:.3f}")
gcode = re.sub(r"\(.*?\)", "", gcode)
flag = len(re.findall(r"#\d+", gcode))
while 0 < flag:
cases = re.findall(r"((#\d+)\s*=\s*([-+]?\s*(\d*\.\d+|\d+\.?\d*)))", gcode)
for case in cases:
gcode = gcode.replace(case[0], "")
gcode = gcode.replace(case[0], case[1])
cases = re.findall(r"(\[(\s*[+-]?\s*(\d+(\.\d*)?|\d*\.\d+)(\s*[-+*\/]\s*[+-]?\s*(\d+(\.\d*)?|\d*\.\d+)\s*)*)\])", gcode)
for case in cases:
gcode = gcode.replace(case[0], f32str(eval(case[1])))
flag = len(re.findall(r"#\d+", gcode))
gcode = os.linesep.join([s for s in gcode.splitlines() if s.strip()])
this is probably the worst way to do this and there should be more efficient implementations. I will leave the rest to the Python experts.

concatenation in IPython.display in a notebook using Latex+symbols

Say I have an expression which I would like to display in LateX form and one which is a result of an analytical calculation where variables like theta appear and are pretty printed in the end. I would like to print both in one line. Here an example:
from IPython.display import display, Math, Latex
import numpy as np
from sympy import *
init_printing()
# In[1]:
name='\Gamma^'+str(1)+'_{'+str(2)+str(3)+'}'+'='
# In[2]:
theta = symbols('theta')
# In[3]:
display(Math(name),theta)
The last command prints name in a pretty form (LateX) as well as theta. However, a line-break is added which I would like to omit. How can this be achieved?

You need to form latex string first. Use sympy.latex() to make string from any printable sympy variable:
display(Math(name + latex(theta)))
I also wrote simple function with variable number of arguments for such output. No need to wrap variables with latex() here:
def prlat(*args):
str = ''
for arg in args:
str += latex(arg)
display(Math(str))
prlat(name, theta, '= 0')

SymPy/SciPy: solving a system of ordinary differential equations with different variables

I am new to SymPy and Python in general, and I am currently working with Python 2.7 and SymPy 0.7.5 with the objective to:
a) read a system of differential equations from a text file
b) solve the system
I already read this question and this other question, and they are almost what I am looking for, but I have an additional issue: I do not know in advance the form of the system of equations, so I cannot create the corresponding function using def inside the script, as in this example. The whole thing has to be managed at run-time.
So, here are some snippets of my code. Suppose I have a text file system.txt containing the following:
dx/dt = 0.0387*x - 0.0005*x*y
dy/dt = 0.0036*x*y - 0.1898*y
What I do is:
# imports
import sympy
import scipy
import re as regex
# define all symbols I am going to use
x = sympy.Symbol('x')
y = sympy.Symbol('y')
t = sympy.Symbol('t')
# read the file
systemOfEquations = []
with open("system.txt", "r") as fp :
for line in fp :
pattern = regex.compile(r'.+?\s+=\s+(.+?)$')
expressionString = regex.search(pattern, line) # first match ends in group(1)
systemOfEquations.append( sympy.sympify( expressionString.group(1) ) )
At this point, I am stuck with the two symbolic expressions inside the systemOfEquation list. Provided that I can read the initial conditions for the ODE system from another file, in order to use scipy.integrate.odeint, I would have to convert the system into a Python-readable function, something like:
def dX_dt(X, t=0):
return array([ 0.0387*X[0] - 0.0005*X[0]*X[1] ,
-0.1898*X[1] + 0.0036*X[0]*X[1] ])
Is there a nice way to create this at run-time? For example, write the function to another file and then import the newly created file as a function? (maybe I am being stupid here, but remember that I am relatively new to Python :-D)
I've seen that with sympy.utilities.lambdify.lambdify it's possible to convert a symbolic expression into a lambda function, but I wonder if this can help me...lambdify seems to work with one expression at the time, not with systems.
Thank you in advance for any advice :-)
EDIT:
With minimal modifications, Warren's answer worked flawlessly. I have a list of all symbols inside listOfSymbols; moreover, they appear in the same order as the columns of data X that will be used by odeint. So, the function I used is
def dX_dt(X, t):
vals = dict()
for index, s in enumerate(listOfSymbols) :
if s != time :
vals[s] = X[index]
vals[time] = t
return [eq.evalf(subs=vals) for eq in systemOfEquations]
I just make an exception for the variable 'time' in my specific problem. Thanks again! :-)

If you are going to solve the system in the same script that reads the file (so systemOfEquations is available as a global variable), and if the only variables used in systemOfEquations are x, y and possibly t, you could define dX_dt in the same file like this:
def dX_dt(X, t):
vals = dict(x=X[0], y=X[1], t=t)
return [eq.evalf(subs=vals) for eq in systemOfEquations]
dX_dt can be used in odeint. In the following ipython session, I have already run the script that creates systemOfEquations and defines dX_dt:
In [31]: odeint(dX_dt, [1,2], np.linspace(0, 1, 5))
Out[31]:
array([[ 1. , 2. ],
[ 1.00947534, 1.90904183],
[ 1.01905178, 1.82223595],
[ 1.02872997, 1.73939226],
[ 1.03851059, 1.66032942]]