Mailman 3 November 2011 - SciPy-User

zscore axis functionality is borked
by Alacast Dec. 17, 2011

Dec. 17, 2011

axis=0 (the default) works fine. axis=1, etc, is clearly wrong. Am I misunderstanding how to use this, or is this a bug? In [16]: i = rand(4,4) In [17]: i Out[17]: array([[ 0.85367762, 0.25348857, 0.23572615, 0.50403358], [ 0.70199066, 0.81872151, 0.47357357, 0.20425537], [ 0.31042673, 0.25837984, 0.73550134, 0.57970176], [ 0.42828877, 0.60988596, 0.04059321, 0.73944219]]) In [18]: zscore(i, axis=0) Out[18]: array([[ 1.30128758, -0.96195723, -0.52119142, -0.01453907], [ 0.59653471, 1.38544585, 0.39284654, -1.55756529], [-1.22271057, -0.94164388, 1.39942427, 0.37494213], [-0.67511172, 0.51815526, -1.27107939, 1.19716222]]) In [19]: zscore(i[:,0]) Out[19]: array([ 1.30128758, 0.59653471, -1.22271057, -0.67511172]) In [20]: zscore(i[:,0])==zscore(i,axis=0)[:,0] Out[20]: array([ True, True, True, True], dtype=bool) In [21]: zscore(i, axis=1) Out[21]: array([[-0.99378502, -1.59397407, -1.61173649, -1.34342906], [-1.6379836 , -1.52125275, -1.86640069, -2.13571889], [-2.09968257, -2.15172946, -1.67460796, -1.83040754], [-1.29796925, -1.11637205, -1.68566481, -0.98681582]]) #The above is obviously wrong, as everything has a negative z score In [22]: zscore(i[0,:]) Out[22]: array([ 1.56824016, -0.83321371, -0.90428403, 0.16925757]) In [23]: zscore(i[0,:])==zscore(i,axis=1)[0,:] Out[23]: array([False, False, False, False], dtype=bool) #Using axis=1 produces different results from taking a row directly. In [24]: zscore(i, axis=-1) Out[24]: array([[-0.99378502, -1.59397407, -1.61173649, -1.34342906], [-1.6379836 , -1.52125275, -1.86640069, -2.13571889], [-2.09968257, -2.15172946, -1.67460796, -1.83040754], [-1.29796925, -1.11637205, -1.68566481, -0.98681582]]) #Getting rows by using axis=-1 is no better (this is the same result as axis=1

3 9

No Scipy 0.10 reference manual
by Miha Marolt Dec. 3, 2011

Dec. 3, 2011

There is no link to the Scipy 0.10 reference guide on the documentation page (http://docs.scipy.org/doc/).

3 5

Subclassing scipy sparse matrix class
by Per Nielsen Dec. 3, 2011

Dec. 3, 2011

Hi all I am trying to create a subclass of the sparse matrix class in scipy, to add some extra methods I need. I have tried to follow the guide on: http://www.scipy.org/Subclasses but without much luck, the view method does not exist for the sparse matrix class. Below is a script I have created ----------------- #!/usr/bin/env python from scipy.sparse.csr import csr_matrix as spmatrix class sparsematrix_addons(spmatrix): """ subclass for standard scipy sparse class to add missing functionality """ def __new__(cls, matrix): obj = spmatrix.__init__(cls, matrix) return obj def square_spmat(self, M): return M ** 2 def ravel_spmat(self, M): pass if __name__ == '__main__': import numpy as np x = (np.random.rand(10, 10) * 2).astype(int).astype(float) xsp = sparsematrix_addons(x) -------------------- However, this generates the following error: TypeError: unbound method __init__() must be called with csr_matrix instance as first argument (got type instance instead) I am not strong in python OOP, or OOP in general, so I am sure this is a rather trivial problem to solve. Anyone got any solutions or ideas to point me in the right direction? Thanks in advance, Per

3 4

creating sparse indicator arrays
by josef.pktd＠gmail.com Dec. 1, 2011

Dec. 1, 2011

Is there a simple or fast way to create a sparse indicator array, `a` below, without going through the dense matrix first? >>> from scipy import sparse >>> g = np.array([0, 0, 1, 1]) #categories, integers, >>> u = np.arange(2) #unique's, range(number_categories) >>> g[:,None] == u array([[ True, False], [ True, False], [False, True], [False, True]], dtype=bool) this is the one I want: >>> a = sparse.csc_matrix((g[:,None] == u)) >>> a <4x2 sparse matrix of type '<type 'numpy.int8'>' with 4 stored elements in Compressed Sparse Column format> >>> a.todense() matrix([[1, 0], [1, 0], [0, 1], [0, 1]], dtype=int8) Thanks, Josef

2 4

fastest linear interpolation ?
by josef.pktd＠gmail.com Nov. 30, 2011

Nov. 30, 2011

Does anyone know which linear interpolation function in numpy or scipy is the fastest for evaluating a large number of points? for mainly or only 1d cases I'd like to approximate a non-linear function on a fine grid by linear interpolation. Number of points/segments should be large. The setup can be expensive, and includes calculating all the points, but I want fast evaluation. Thanks, Josef

1 0

Help installing scipy
by Barris, Wes Nov. 30, 2011

Nov. 30, 2011

I have a new CentOS 6 (64-bit) system. It is running the latest rpm version of python: python-2.6.5-3.el6_0.2.x86_64 I have also installed numpy: rpm -qa | fgrep numpy numpy-1.3.0-6.2.el6.x86_64 I have downloaded the latest software source of scipy and am trying to install it. When I run the first command I get an error: python setup.py build Traceback (most recent call last): File "setup.py", line 196, in <module> setup_package() File "setup.py", line 147, in setup_package from numpy.distutils.core import setup ImportError: No module named distutils.core I don't know anything about python so I am not sure what is supposed to provide this module or where to look for it. python -c 'from numpy.f2py.diagnose import run; run()' Traceback (most recent call last): File "<string>", line 1, in <module> ImportError: No module named f2py.diagnose python -c 'import os,sys;print os.name,sys.platform' posix linux2 uname -a Linux cvirdlux01.tyson.com 2.6.32-71.29.1.el6.x86_64 #1 SMP Mon Jun 27 19:49:27 BST 2011 x86_64 x86_64 x86_64 GNU/Linux -- Wes Barris This email and any files transmitted with it are confidential and intended solely for the use of the addressee. If you are not the intended addressee, then you have received this email in error and any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. Please notify us immediately of your unintended receipt by reply and then delete this email and your reply. Cobb-Vantress, Inc. and its subsidiaries and affiliates will not be held liable to any person resulting from the unintended or unauthorized use of any information contained in this email or as a result of any additions or deletions of information originally contained in this email.

3 2

how can I create B-splines of mutidimensional values?
by Kay F. Jahnke Nov. 28, 2011

Nov. 28, 2011

Hi group! I have the following problem: I have multidimensional values which I have sampled over a 2D grid. Now I want to interpolate the values (using B-spline interpolation) for arbirary (x,y) locations. Obviously this can be done by creating a separate B-spline for each dimension of of the values, interpolating at (x,y) and putting the results of the interpolations together, forming the multidimensional result. For performance reasons, this approach isn't optimal, though. Since the splines are evaluated at the same location, a fair deal of the calculations would be identical. But I haven't found a way to create a B-spline of the multidimensional values to exploit the fact that I'm evaluating several splines at the same position. I suppose my problem would be quite common - one typical case would be RGB image data: it would be silly to have separate splines for the R,G and B channels and triplicate the part of the calculation which only depends on the position where a value is interpolated. What I'm missing is a mechanism to calculate the spline with n-dimensional coefficients and interpolation routines yielding multidimensional values when used with these n-dimensional spline coefficients. Am I missing something? I'm not very proficient with numpy/scipy, so maybe I just can't see the obvious... Helpful hints welcome. Kay

2 9

meshgrid for 3D
by Bala subramanian Nov. 28, 2011

Nov. 28, 2011

Friends, I have a data file containing three vectors x,y,z and want to create a coordinate matrices from three vectors. While i know that the numpy's meshgrid function can be used for two vectors, i dnt any tool which i can use for three dimension. Kindly suggest me some solution. Thanks, Bala

2 1

Confusion about lognormal distribution functions
by tazz_ben Nov. 27, 2011

Nov. 27, 2011

Hi Group - So, what I'm trying to do is draw a firm size from a lognormal distribution in a simulation (I'm using a fortuna RNG outside of the scope of this question -- why instead of twister deals with my research question, for this purposes it is just important to say using the built in random draw from a specific distribution wouldn't work). But when I do something like this: from scipy.stats import lognorm lognorm.ppf(.5,1,50,50) The numbers that come out make no sense (I'm right in believing "loc" = "mean" and "scale" = "standard deviation"?). I've tried logging the numbers, un-logging the numbers, etc. I'm very confused on what it is doing.

3 2

Re: [SciPy-User] Problem with ODEINT
by Lofgren, Eric Nov. 27, 2011

Nov. 27, 2011

> Eric, > You have given odeint an initial condition of length 5, but the function > that defines your system is returning a vector of only length 3. Don't do > that. > ... > Warren Warren- This does indeed seem to solve the problem, I haven't hit any errors in 1,000 or so runs, and it does indeed make the code run considerably faster. Thank you for the advice and help. Eric

1 0