The first input to lambdify should be either a SymPy expression or a list of SymPy expressions. I don't know enough about lambdify to know if we support list of lists of sympy expressions.
In general you won't be able to use lambdify with arbitrary n because the target expression (the second argument to lambdify) needs to already be evaluated. Building off of my previous example In [6]: result = simplify(E(prod(Xs))) In [7]: f = lambdify(mus+sigmas, result) In [8]: f(1,2,3,4,5,10,11,12,13,14) Out[8]: 120 Or, if you need your current interface In [9]: numeric_mus = [1,2,3,4,5] In [10]: numeric_sigmas = [10, 11, 12, 13, 14] In [11]: f(*(numeric_mus + numeric_sigmas)) Out[11]: 120 On Fri, Aug 30, 2013 at 3:09 PM, Janwillem van Dijk <jwe.van.d...@gmail.com>wrote: > T <https://groups.google.com/forum/#!topic/sympy/gRgNYef3oyc>hanks > Matthew, Is the following snippet of use? Yes and No. > The lines 5 and 6 are a useful tip; I missed the prod method, that is > nice. And yes for two N-distributions explicit solutions are to be found in > the litt and SymPy can find them as it can for longer products and also for > the uniform distribution which is also of use to me. However, I want both > the algebraic expression and numerical results. In the past lambdify has > helped me often and here it works also. > > My problem is that when I have PNs as the product of n distributions an mu > and sigma symbolic arrays of n means and stdevs then after: > MeanPns = E(PNs) > meanPNs = lambdify([mu, sigma], MeanPNs) > I cannot calculate the mean of the product of two (or more) Ns with > m=[1.0, 1.0] and s= [0.05, 0.15] by calling > average = meanPNs(m, s) > This gives an error saying that meanPNs expects 4 arguments an only 2 are > given. > So you need the intermediate step > ms = m + s > and than > average = meanPNs(*ms) > whereby meanPS gets the 4 Or rather 2n) arguments it wants. > This works but I do not consider it an elegant solution so I wondered > whether there is a more elegant solution. > > By the way, the purpose of it is to show in a course on uncertainty in > measurement that the Law of Propagation of Uncertainties (LPU) under > estimates the standard uncertainty and that the resulting distribution is > skewed in violation to the common assumption that the Central Limit Theorem > (CLT) would be valid. Showing this both through equations an numbers. > > Thanks again for paying attention, cheers, Janwillem > > > On Friday, 30 August 2013 06:00:21 UTC+2, Matthew wrote: > >> Is the following snippet of use? >> >> In [1]: from sympy.stats import * >> >> In [2]: n = 5 >> >> In [3]: mus = [Symbol('mu'+str(i), bounded=True) for i in range(n)] >> >> In [4]: sigmas = [Symbol('sigma'+str(i), positive=True) for i in range(n)] >> >> In [5]: Xs = [Normal('X'+str(i), mu, sigma) for i, (mu, sigma) in >> enumerate(zip(mus, sigmas))] >> >> In [6]: E(prod(Xs)) >> << Big Scary Thing >> >> >> In [7]: simplify(E(prod(Xs))) >> Out[7]: μ₀⋅μ₁⋅μ₂⋅μ₃⋅μ₄ >> >> Please note that you shouldn't have to switch down to numerics for this >> problem. There is a clean analytic form. >> >> >> >> On Thu, Aug 29, 2013 at 2:01 PM, Janwillem van Dijk >> <jwe.va...@gmail.com>wrote: >> >>> I have problems lambdifying a symbolic function that needs array >>> arguments. >>> The problem I want to solve is to find the statistics of the product of >>> Normal distributions. Attached is an example that works and even yields the >>> correct answers. However, I think that the part where the symbolic versions >>> are lambdified into Python callable functions can't win a beauty contest. >>> For example a part of the class : >>> >>> self.MeanNN = E(NN) >>> >>> self.meanNN = lambdify([mu, s], self.MeanNN) >>> >>> def meanOfNs(self, mu, s): >>> >>> mus = mu + s >>> >>> return self.meanNN(*mus) >>> >>> Than called as: >>> >>> Nn = ProductOfNormals(n) >>> >>> print('Mean: ', Nn.meanOfNs(mu, s)) >>> >>> Where mu and s are lists of length n >>> >>> >>> In case the arguments are spelled out it looks nicer but I want the >>> flexibility of variable number of distributions in the product. >>> >>> e.g.: >>> >>> self.MeanNN = E(NN) >>> >>> self.meanNN = lambdify([mu1, s1, mu2, s2], self.MeanNN) >>> >>> Than: >>> >>> NN = ProductNormalNormal() >>> >>> print('Mean: ', NN.meanNN(mu0, s0, mu1, s1)) >>> >>> Where all parms are scalar >>> >>> >>> Any tips on how to get a more elegant solution are welcome!! >>> >>> The complete example is attached. I am using SymPy 0.7.3 on Python 2.7.4 >>> on Linux. >>> >>> Cheers, Janwillem >>> >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "sympy" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to sympy+un...@**googlegroups.com. >>> To post to this group, send email to sy...@googlegroups.com. >>> >>> Visit this group at >>> http://groups.google.com/**group/sympy<http://groups.google.com/group/sympy> >>> . >>> For more options, visit >>> https://groups.google.com/**groups/opt_out<https://groups.google.com/groups/opt_out> >>> . >>> >> >> -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to sympy+unsubscr...@googlegroups.com. > To post to this group, send email to sympy@googlegroups.com. > Visit this group at http://groups.google.com/group/sympy. > For more options, visit https://groups.google.com/groups/opt_out. > -- You received this message because you are subscribed to the Google Groups "sympy" group. 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