Mailman 3 October 2012 - SciPy-User

Re: [SciPy-User] Projected Area
by Lorenzo Isella Oct. 2, 2012

Oct. 2, 2012

On Mon, 01 Oct 2012 21:54:29 +0200, <scipy-user-request(a)scipy.org> wrote: > Date: Mon, 1 Oct 2012 12:39:48 -0600 > From: "jkhilmer(a)chemistry.montana.edu" > <jkhilmer(a)chemistry.montana.edu> > Subject: Re: [SciPy-User] Projected Area > To: SciPy Users List <scipy-user(a)scipy.org> > Message-ID: > <CA+S87B_=gAKA56wKdHLd48unBsASCXzcfQu+-iuS3KHe57B0LA(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > Lorenzo, > Were the previous suggestions not viable due to speed or precision? > http://thread.gmane.org/gmane.comp.python.scientific.user/30450/focus=30464 > Jonathan Hello, Thanks for the link. That was another question I asked some time ago. Here the situation is simpler and I think I do not want to over-engineer the solution: so far it looks like I can get some decent results with an old-fashoned Monte Carlo integration. Cheers Lorenzo

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interpolate interp1d and rbf
by Janwillem Oct. 2, 2012

Oct. 2, 2012

I am using interpolate.interp1d and interpolate.rbf and could simplify my scripts greatly if I could get the original axes values from the interpolating function like f = interpolate.interp1d(x, y) and than somewhere else original_x = f.get_original_x() Is there such a thing as "get_original_x" Or if not possible is there at least a way to find the range of the axes to prevent extrapolation errors. I had a look at f.nodes but with no success My scipy is 0.9.0 Many thanks, Janwillem

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Scaling clip-art image
by klo uo Oct. 2, 2012

Oct. 2, 2012

Hi, while looking for a way to produce quality resize of line drawings I was suggested this algorithm: https://secure.wikimedia.org/wikipedia/en/wiki/Hqx It produces by far best results on fixed resize ratios: 2x, 3x, 4x Color edges are crisp, lines almost sharp as wanted, and I was wondering does scipy has similar function, or different approach that may give good results on clip-art images resize?

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Re: [SciPy-User] Projected Area
by Lorenzo Isella Oct. 1, 2012

Oct. 1, 2012

Hello, And thanks for your reply. Unfortunately, the situation is not this easy. The dimer example was somehow misleading. It is not so straightforward to calculate the area of multiple overlapping circles (in particular when the intersection of 4-5 circles is not empty). I think I will have to resort to some Monte Carlo integration. Cheers Lorenzo On Mon, 01 Oct 2012 16:59:21 +0200, <scipy-user-request(a)scipy.org> wrote: > On Mon, Oct 1, 2012 at 10:34 AM, Lorenzo Isella > <lorenzo.isella(a)gmail.com> wrote: >> Dear All, >> I hope this is not too off-topic. >> I need to know if there is already some ready-to-use SciPy algorithm >> (or at least if this is easy to implement or not). >> Consider a dimer, i.e. 2 spheres with a single contact point. This >> dimer can have any orientation in the 3D and I have the (x,y,z) >> coordinates of the centre of the 2 spheres. >> For a given orientation, I want to project the dimer on, let's say, >> the xy plane and evaluate the area of the surface of its projection. >> I spoke about a dimer since it is easy to start discussing a simple >> case, but in general I will deal with objects consisting of several >> non-overlapping spheres such that any sphere has at least a contact >> point with another sphere. > There is nothing implemented in scipy for this. For the case of > spheres projected (orthographically?) onto a plane, the shadows are > probably-overlapping circles (the contact point is irrelevant). It > looks like there is an analytical solution to the area of the > intersection for circles: > http://mathworld.wolfram.com/Circle-CircleIntersection.html > You can probably just add up the areas of each circle, then subtract > out one copy of each area of intersection to get the area of the > union.

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Request help on fsolve
by The Helmbolds Oct. 1, 2012

Oct. 1, 2012

Please help me out here. I’m trying to rewrite the docstring for the `fsolve.py` routine located on my machine in: C:/users/owner/scipy/scipy/optimize/minpack.py The specific issue I’m having difficulty with is understanding the outputs described in fsolve’s docstring as: 'fjac': the orthogonal matrix, q, produced by the QR factorization of the final approximate Jacobian matrix, stored column wise 'r': upper triangular matrix produced by QR factorization of same matrix These are described in SciPy’s minpack/hybrd.f file as: ‘fjac’ is an output n by n array which contains the orthogonal matrix q produced by the qr factorization of the final approximate jacobian. ‘r’ is an output array of length lr which contains the upper triangular matrix produced by the qr factorization of the final approximate jacobian, stored rowwise. For ease in writing, in what follows let’s use the symbols ‘Jend’ for the final approximate Jacobian matrix, and use ‘Q’ and ‘R’ for its QR decomposition matrices. Now consider the problem of finding the solution to the following three nonlinear equations in three unknowns (u, v, w), which we will refer to as ‘E’: 2 * a * u + b * v + d - w * v = 0 b * u + 2 * c * v + e - w * u = 0 u * v - f = 0 where (a, b, c, d, e, f ) = (2, 3, 7, 8, 9, 2). For inputs to fsolve, we identify (u, v, w) = (x[0], x[1], x[2]). Now fsolve gives the solution array: [uend vend wend] = [ 1.79838825 1.11210691 16.66195357]. With these values, the above three equations E are satisfied to an accuracy of about 9 significant figures. The Jacobian matrix for the three LHS functions in E is: J = np.matrix([[2*a, b-w, -v], [b-w, 2*c, -u], [v, u, 0.]]) Note that it’s symmetric, and if we compute its value using the above fsolve’s ‘end’ solution values we get: Jend = [[ 4. 19.66195357 1.11210691], [ 19.66195357 14. 1.79838825], [ 1.11210691 1.79838825 0. ]] Using SciPy’s linalg package, this Jend has the QR decomposition: Qend = [[-0.28013447 -0.91516674 -0.28981807] [ 0.95679602 -0.24168763 -0.16164302] [ 0.07788487 -0.32257856 0.94333293]] Rend = [[-14.278857 17.08226116 -1.40915124] [ -0. 9.69946027 1.45241144] [ -0. 0. 0.61300558]] and Qend * Rend = Jend to within about 15 significant figures. However, fsolve gives the QR decomposition: qretm = [[-0.64093238 0.75748326 0.1241966 ] [-0.62403598 -0.60841098 0.4903215 ] [-0.44697291 -0.23675978 -0.8626471 ]] rret = [ -7.77806716 30.02199802 -0.819055 -10.74878184 2.00090268 1.02706198] and converting rret to a NumPy matrix gives: rretm = [[ -7.77806716 30.02199802 -0.819055 ] [ 0. -10.74878184 2.00090268] [ 0. 0. 1.02706198]] Now qret and rretm bear no obvious relation to Qend and Rend. Although qretm is orthogonal to about 16 significant figures, we find the product: qretm * rretm = [[ 4.98521509 -27.38409295 2.16816676] [ 4.85379376 -12.19513008 -0.2026608 ] [ 3.47658529 -10.87414051 -0.99362993]] which bears no obvious relationship to Jend. The hybrdj.f routine in minpack refers to a permutation matrix, p, such that we should have in our notation: p*Jend = qretm*rretm, but fsolve apparently does not return the matrix p, and I don’t see any permutation of Jend that would equal qretm*rretm. If we reinterpret rret as meaning the matrix: rretaltm = [[ -7.77806716 30.02199802 -10.74878184] [ 0. -0.819055 2.00090268] [ 0. 0. 1.02706198]] then we get the product: qretm * rretaltm = [[ 4.98521509 -19.86249109 8.53245022] [ 4.85379376 -18.2364849 5.99384603] [ 3.47658529 -13.22510045 3.44468895]] which again bears no obvious relationship to Jend. Using the transpose of qretm in the above product is no help. So please help me out here. What are the fjac and r values that fsolve returns? How are they related to the above Qend, Rend, and Jend? How is the user supposed to use them? Bob H

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Projected Area
by Lorenzo Isella Oct. 1, 2012

Oct. 1, 2012

Dear All, I hope this is not too off-topic. I need to know if there is already some ready-to-use SciPy algorithm (or at least if this is easy to implement or not). Consider a dimer, i.e. 2 spheres with a single contact point. This dimer can have any orientation in the 3D and I have the (x,y,z) coordinates of the centre of the 2 spheres. For a given orientation, I want to project the dimer on, let's say, the xy plane and evaluate the area of the surface of its projection. I spoke about a dimer since it is easy to start discussing a simple case, but in general I will deal with objects consisting of several non-overlapping spheres such that any sphere has at least a contact point with another sphere. Any suggestion is appreciated. Cheers Lorenzo

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