Mailman 3 June 2014 - SciPy-User

How to display a RGB image (MFISH)?
by jp April 3, 2023

April 3, 2023

Hi, I have written a script to try to combine five images into a three channels RGB images<http://dip4fish.blogspot.com/2011/05/trying-to-combine-mfish-images-into-rg…> . The resulting RGB image looks like a grey scale image instead of a color one. I have three np.array Rnorm, Gnorm, Bnorm which are copied into a rgb array: *rgb = np.zeros((shape[0],shape[1],3),dtype=float) mxr=np.max(R) mxg=np.max(G) mxb=np.max(B) Rnorm=np.uint8((255*(R/mxr)))Gnorm=np.uint8((255*(R/mxg)))Bnorm=np.uint8((255*(R/mxb)))#copy each RGB component in an RGB array rgb[:,:,0]=Rnorm rgb[:,:,1]=Gnorm rgb[:,:,2]=Bnorm* *pylab.subplot(224, aspect='equal',frameon=False, xticks=[], yticks=[]) pylab.imshow(rgb) pylab.show()* Any advice? Thank you Jean-Patrick

3 4

Invitation to connect on LinkedIn
by Jelle Feringa via LinkedIn July 28, 2022

July 28, 2022

LinkedIn ------------ Jelle Feringa requested to add you as a connection on LinkedIn: ------------------------------------------ Jose, I'd like to add you to my professional network on LinkedIn. - Jelle Accept invitation from Jelle Feringa http://www.linkedin.com/e/-3wy1w2-gpkzwpgn-5n/Q6WKH0LACopGJkAw_6fSqajo6R7VM… View invitation from Jelle Feringa http://www.linkedin.com/e/-3wy1w2-gpkzwpgn-5n/Q6WKH0LACopGJkAw_6fSqajo6R7VM… ------------------------------------------ DID YOU KNOW your LinkedIn profile helps you control your public image when people search for you? Setting your profile as public means your LinkedIn profile will come up when people enter your name in leading search engines. Take control of your image! http://www.linkedin.com/e/-3wy1w2-gpkzwpgn-5n/ewp/inv-22/ -- (c) 2011, LinkedIn Corporation

2 1

Help required regding Eigenvectors
by Sudheer Phani March 14, 2022

March 14, 2022

Hello All, I have been encountering some strange problems with the calculation of eigen vectors using python. The problem is with the sign of the eigen vectors calculated. On Very few occasions the eigen vectors calculated using python have an opposite sign than the actual eigenvectors, though the magnitude is correct. Can any one of you plz help me in this regard and suggest me some solution if you have encoutered similar problems . Any help in this regd is appreciated Thanks in Advace regds Sudheer

5 5

Normalization for optimization in python
by Yuxiang Wang April 27, 2015

April 27, 2015

Dear all, During optimization, it is often helpful to normalize the input parameters to make them on the same order of magnitude, so the convergence can be much better. For example, if we want to minimize f(x), while a reasonable approximation is x0=[1e3, 1e-4], it might be helpful to normalize x0[0] and x0[1] to about the same order of magnitude (often O(1)). My question is, I have been using scipy.optimize and specifically, the L-BFGS-B algorithm. I was wondering that, do I need to normalize that manually by writing a function, or the algorithm already did it for me? Thank you! -Shawn -- Yuxiang "Shawn" Wang Gerling Research Lab University of Virginia yw5aj(a)virginia.edu +1 (434) 284-0836 https://sites.google.com/a/virginia.edu/yw5aj/

4 8

Optimizing odeint without a for loop
by Barrett B July 1, 2014

July 1, 2014

This is an initial attempt to model a neural network with two differential variables per cell--voltage, and the response variable n. I can go back and add more later as needed. I have a question about this code (obviously, all constants are given): -------- #ODE def f(X, t): N = len(X)/2 dV = np.zeros(N); dn = np.zeros(N) for i in range(N): E = X[i]; n = X[N+i] #dummy n_inf_E = 1/(1 + np.exp((E_half_n - E)/k_n)) m_inf_E = 1/(1 + np.exp((E_half_m - E)/k_m)) dV[i] = (I - g_L*(E - E_L) - g_Na*m_inf_E*(E - E_Na) - g_K*n*(E - E_K))/C for j in range(N): dV[i] += eps*connex[j, i]*X[j] dn[i] = (n_inf_E - n)/tau return np.concatenate((dV, dn)) connex = np.matrix([[-1,1,0], [1,-2,1], [0,1,-1]]) #connection matrix t = np.arange(0, stopTime, timeStep) V0 = np.array([0, -20, -50]) n0 = np.array([0.2, 0.4, 0.7]); N = len(n0) soln = odeint(f, np.concatenate((V0, n0)), t) ----------------- Is there a way to do this without the "for i in range (N)" loop; i.e., can I run through all the dV[i]'s in a more efficient method?

6 11

Re: [SciPy-User] SciPy-User Digest, Vol 130, Issue 23
by The Helmbolds June 27, 2014

June 27, 2014

If the matrix is not too large, the straightforward series expansion, for judiciously chosen n can be a remarkably good approximation: EXP = I # the identity matrix k = 1 for k in range(n): # or EXP = EXP*M / k >________________________________ > From: "scipy-user-request(a)scipy.org" <scipy-user-request(a)scipy.org> >To: scipy-user(a)scipy.org >Sent: Friday, June 27, 2014 10:00 AM >Subject: SciPy-User Digest, Vol 130, Issue 23 > > >Send SciPy-User mailing list submissions to > scipy-user(a)scipy.org > >To subscribe or unsubscribe via the World Wide Web, visit > http://mail.scipy.org/mailman/listinfo/scipy-user >or, via email, send a message with subject or body 'help' to > scipy-user-request(a)scipy.org > >You can reach the person managing the list at > scipy-user-owner(a)scipy.org > >When replying, please edit your Subject line so it is more specific >than "Re: Contents of SciPy-User digest..." > > >Today's Topics: > > 1. Re: Faster approach to calculate the matrix exponential > (Pauli Virtanen) > 2. Re: Faster approach to calculate the matrix exponential > (Stephan Hoyer) > 3. Re: Faster approach to calculate the matrix exponential > (Michael Kreim) > 4. Re: Faster approach to calculate the matrix exponential > (Troels Emtek?r Linnet) > > >---------------------------------------------------------------------- > >Message: 1 >Date: Thu, 26 Jun 2014 20:22:17 +0300 >From: Pauli Virtanen <pav(a)iki.fi> >Subject: Re: [SciPy-User] Faster approach to calculate the matrix > exponential >To: scipy-user(a)scipy.org >Message-ID: <lohks9$8rf$1(a)ger.gmane.org> >Content-Type: text/plain; charset=ISO-8859-1 > >26.06.2014 14:07, Troels Emtek?r Linnet kirjoitti: >> Using the scripy method does not sound viable. >> >> It was discussed here, post 2 >> >> http://thread.gmane.org/gmane.science.nmr.relax.devel/6446/match=eliminatin… > >That link doesn't load currently. > >Nevertheless, using the Scipy implementation inside for loops is faster >than using the eig() based approach, with a crossover (on my machine) >when the inner matrix size is bigger than 16x16. This comparison applies >to Scipy 0.14.0 --- the implementation in earlier Scipy versions is less >efficient. > >Benchmark: >""" >from scipy.linalg import expm > >def matrix_exponential_scipy(A, dtype=None): > B = np.empty(A.shape, dtype=A.dtype) > for i in range(A.shape[0]): > for j in range(A.shape[1]): > for k in range(A.shape[2]): > for l in range(A.shape[3]): > for m in range(A.shape[4]): > B[i,j,k,l,m] = expm(A[i,j,k,l,m]) > return B > >import numpy as np > >N = 18 >A = np.random.rand(3, 4, 5, 3, 2, N, N) > >start_1 = time.time() >B_1 = matrix_exponential_scipy(A) >end_1 = time.time() > >start_2 = time.time() >B_2 = matrix_exponential_rank_NE_NS_NM_NO_ND_x_x(A) >end_2 = time.time() > >print("Scipy:", end_1 - start_1) >print("Numpy:", end_2 - start_2) >""" >#-> ('Scipy:', 0.3015120029449463) >#-> ('Numpy:', 0.31020593643188477) > > >-- >Pauli Virtanen > > > >------------------------------ > >Message: 2 >Date: Thu, 26 Jun 2014 11:04:54 -0700 >From: Stephan Hoyer <shoyer(a)gmail.com> >Subject: Re: [SciPy-User] Faster approach to calculate the matrix > exponential >To: SciPy Users List <scipy-user(a)scipy.org> >Message-ID: > <CAEQ_TvdDYFZOZimHrCGbkPJwXGS=RCw5U-5h6-k0D9MA5Zgccw(a)mail.gmail.com> >Content-Type: text/plain; charset="utf-8" > >On Thu, Jun 26, 2014 at 5:32 AM, Charles R Harris <charlesr.harris(a)gmail.com >> wrote: >> >> Are you solving a differential equation? If so, an explicit solution may >> not be the best way to go. >> > >I agree. > >My experience with solving very similar quantum mechanics problems is that >using an ODE integrator like zofe (via scipy.integrate.ode) can be much >faster than solving the propagator exactly.. >

1 0

Faster approach to calculate the matrix exponential
by Troels Emtekær Linnet June 27, 2014

June 27, 2014

Dear NMR wizards. Do any of you know any faster approach to calculate the matrix exponential ? I currently do it via eigenvalue decomposition approach. But my profiling scripts tells me, that my bottleneck is the eig function. Stealing 86% of the time. The matrix exponential is essential calculated in all numerical models of NMR magnetisation evolution. My minimisation algorithm would calculate this at each step, with iterations between 1E4 to 1E7 iterations. I have looked through this paper, do you have any experience with some of the methods? They don't really come to a conclusion, but suggest method 3. (page 10) Moler, C. and Van Loan, C. (2003) Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later. SIAM Review, 45, 3-49. (http://dx.doi.org/10.1137/S00361445024180 or http://www.cs.cornell.edu/cv/researchpdf/19ways+.pdf). My multidimensional approach right now is: http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/matrix… The numerical models using it: http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/ns_cpm… http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/ns_cpm… http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/ns_mmq… http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/ns_mmq… http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/ns_r1r… http://svn.gna.org/svn/relax/branches/disp_spin_speed/lib/dispersion/ns_r1r… Troels Emtekær Linnet

7 9

scipy-0.14.0 build failure fedora 20
by Neal Becker June 24, 2014

June 24, 2014

I'm trying to build scipy-0.14.0 on fedora 20 (x86_64). I've tried different compile options, but most recently just using the scipy-0.14.0-2.fc21.src.rpm unchanged. It builds on py2 and py3, but when it tries to run self-test it explodes. I don't know which test it's running, but memory usage becomes much larger than my 16G, and I have to kill it. Any ideas?

2 2

lmfit: confidence intervals issue
by Antonino Ingargiola June 22, 2014

June 22, 2014

Hi, I'm not sure this is the right list for this kind of issues. I'm using lmfit to solve a non-linear least square problem to fit a "noisy" histogram to a model that is a single exponential decay convoluted with a "sharp" peak and a baseline noise floor. The problem is that unless I set constrains for the 'offset' parameter the confidence interval calculation fails in a strange way in scipy brentq function. You can see the full example notebook and the error message in this ipython notebook: http://nbviewer.ipython.org/urls/gist.githubusercontent.com/tritemio/901c2e… What can I do to fix the problem? If this is the normal behavior, a more informative error message would be helpful. Thanks in advance for any help, Antonio PS: Let me congratulate with the lmfit authors as this is a really amazing piece of software that should be included in scipy IMHO!

4 8

numpy.sort() failing
by Manolo Martínez June 21, 2014

June 21, 2014

Dear all, I have a couple of structured arrays that, as far as I can see, only differ in irrelevant ways; yet numpy.sort chokes on one, not the other: This structured array *cannot* be sorted with np.sort(array, order='weight')... array([([0, 0, 0], 0.0), ([0, 0, 1], 0.0), ([0, 0, 2], 0.0), ([0, 1, 0], 0.0), ([0, 1, 1], 0.0), ([0, 1, 2], 0.0), ([0, 2, 0], 0.0), ([0, 2, 1], 0.0), ([0, 2, 2], 0.0), ([1, 0, 0], 0.0), ([1, 0, 1], 0.0), ([1, 0, 2], 0.0), ([1, 1, 0], 0.0), ([1, 1, 1], 0.0), ([1, 1, 2], 0.0), ([1, 2, 0], 0.0), ([1, 2, 1], 0.0), ([1, 2, 2], 0.0), ([2, 0, 0], 4.08179211371555e-289), ([2, 0, 1], 0.0), ([2, 0, 2], 0.0), ([2, 1, 0], 1.0), ([2, 1, 1], 6.939504595227983e-225), ([2, 1, 2], 1.0626819127933375e-224), ([2, 2, 0], 1.1209583874093894e-260), ([2, 2, 1], 0.0), ([2, 2, 2], 0.0)], dtype=[('strat', 'O'), ('weight', '<f8')]) ... and this one can array([ ([[0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0]], -2.40235968796357e-21), ([[0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [1.0, 0.0, 0.0]], -3.3085614943210732e-21), ([[0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [2.0, 0.0, 0.0]], -1.2140117681585496e-23), ..., ([[2.0, 1.0, 0.0], [2.0, 1.0, 0.0], [2.0, 0.5, 0.5]], -1.154494002571172e-21), ([[2.0, 1.0, 0.0], [2.0, 1.0, 0.0], [2.0, 0.75, 0.25]], -1.3867815814699957e-22), ([[2.0, 1.0, 0.0], [2.0, 1.0, 0.0], [2.0, 1.0, 0.0]], 8.722432986511066e-20)], dtype=[('strat', 'O'), ('weight', '<f8')]) The first one leaves the following trace: In [111]: np.sort(aa, order=['weight', 'strat']) --------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-111-c3a3525a7694> in <module>() ----> 1 np.sort(aa, order=['weight', 'strat']) /usr/lib/python3.4/site-packages/numpy/core/fromnumeric.py in sort(a, axis, kind, order) 786 else: 787 a = asanyarray(a).copy() --> 788 a.sort(axis, kind, order) 789 return a 790 ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() I should say that I asked this very question some days ago over at #scipy, and a user there kindly run the offending array, and could not reproduce the problem. I'm a bit perplexed by this. python --version: 3.4.1 numpy.__version__: 1.8.1 ipython --version: 2.1.0 Thanks for any advice. Cheers, Manolo

1 0