Mailman 3 March 2010 - NumPy-Discussion

matrix operation
by gerardob March 17, 2010

March 17, 2010

Let A and B be two n x n matrices. I would like to have another n x n matrix C such that C_ij = min {A_ij, B_ij} Example: A = numpy.array([[2,3],[10,12]]) B = numpy.array([[1,4],[9,13]]) Output C = [[1,3],[9,12]] The function min(axis) of numpy seems to be only unary. Thanks. -- View this message in context: http://old.nabble.com/matrix-operation-tp27936387p27936387.html Sent from the Numpy-discussion mailing list archive at Nabble.com.

6 6

Problem migrating PDL's index() into NumPy
by Miroslav Sedivy March 17, 2010

March 17, 2010

Hello, being quite new to NumPy and having used previously PDL in Perl, I am currently migrating one of my PDL projects into NumPy. Most of the functions can be migrated without problems and there are functions in NumPy that allow me to do things in much clearer way than in PDL. However, I have a problem with the following operation: There are two 2D arrays with dimensions: A[10000,1000] and B[10000,100]. The first dimension of both arrays corresponds to a list of 10000 objects. The array A contains for each of 10000 objects 1000 integer values between 0 and 99, so that for each of 10000 objects a corresponding value can be found in the array B. I need a new array C[10000,1000] with values from B the following way: for x in range(10000): for y in range(1000): C[x,y] = B[x,A[x,y]] In Perl's PDL, this can be done with $C = $B->index($A) If in NumPy I do C = B[A], then I do not get a [10000,1000] 2D array, but rather a [10000,1000,1000] 3D array, in which I can find the correct values on the following positions: for x in range(10000): for y in range(1000): C[x,y,y] which may seem nice, but it needs 1000 times more memory and very probably 1000 times more time to calculate... Impossible with such large arrays... :-( Could anyone help me, please? Regards, Miroslav Sedivy

2 4

size of a specific dimension of a numpy array
by gerardo.berbeglia March 17, 2010

March 17, 2010

I would like to know a simple way to know the size of a given dimension of a numpy array. Example A = numpy.zeros((10,20,30),float) The size of the second dimension of the array A is 20. Thanks. -- View this message in context: http://old.nabble.com/size-of-a-specific-dimension-of-a-numpy-array-tp27933… Sent from the Numpy-discussion mailing list archive at Nabble.com.

2 1

Subclassing ma.MaskedArray
by David Carmean March 17, 2010

March 17, 2010

I understand that ma.MaskedArray is a subclass of ndarray; in addition to the requirements for subclassing the latter, what does ma.MaskedArray add to the list? I.e. what do I have to watch out for? Basically I need a version of Luke Campagnola's MetaArray ( http://www.scipy.org/Cookbook/MetaArray ) that works with masked arrays. My first attempts have failed. Thanks.

2 1

Some help on matlab to numpy translation
by Nicolas Rougier March 16, 2010

March 16, 2010

Hello, I'm trying to translate a small matlab program for the simulation in a 2D flow in a channel past a cylinder and since I do not have matlab access, I would like to know if someone can help me, especially on array indexing. The matlab source code is available at: http://www.lbmethod.org/openlb/lb.examples.html and below is what I've done so far in my translation effort. In the matlab code, there is a "ux" array of shape (1,lx,ly) and I do not understand syntax: "ux(:,1,col)" with "col = [2:(ly-1)]". If someone knows, that would help me a lot... Nicolas ''' Channel flow past a cylindrical obstacle, using a LB method Copyright (C) 2006 Jonas Latt Address: Rue General Dufour 24, 1211 Geneva 4, Switzerland E-mail: Jonas.Latt(a)cui.unige.ch ''' import numpy # GENERAL FLOW CONSTANTS lx = 250 ly = 51 obst_x = lx/5.+1 # position of the cylinder; (exact obst_y = ly/2.+1 # y-symmetry is avoided) obst_r = ly/10.+1 # radius of the cylinder uMax = 0.02 # maximum velocity of Poiseuille inflow Re = 100 # Reynolds number nu = uMax * 2.*obst_r / Re # kinematic viscosity omega = 1. / (3*nu+1./2.) # relaxation parameter maxT = 4000 # total number of iterations tPlot = 5 # cycles # D2Q9 LATTICE CONSTANTS # matlab: t = [4/9, 1/9,1/9,1/9,1/9, 1/36,1/36,1/36,1/36]; # matlab: cx = [ 0, 1, 0, -1, 0, 1, -1, -1, 1]; # matlab: cy = [ 0, 0, 1, 0, -1, 1, 1, -1, -1]; # matlab: opp = [ 1, 4, 5, 2, 3, 8, 9, 6, 7]; # matlab: col = [2:(ly-1)]; t = numpy.array([4/9., 1/9.,1/9.,1/9.,1/9., 1/36.,1/36.,1/36.,1/36.]) cx = numpy.array([ 0, 1, 0, -1, 0, 1, -1, -1, 1]) cy = numpy.array([ 0, 0, 1, 0, -1, 1, 1, -1, -1]) opp = numpy.array([ 1, 4, 5, 2, 3, 8, 9, 6, 7]) col = numpy.arange(2,(ly-1)) # matlab: [y,x] = meshgrid(1:ly,1:lx); # matlab: obst = (x-obst_x).^2 + (y-obst_y).^2 <= obst_r.^2; # matlab: obst(:,[1,ly]) = 1; # matlab: bbRegion = find(obst); y,x = numpy.meshgrid(numpy.arange(ly),numpy.arange(lx)) obst = (x-obst_x)**2 + (y-obst_y)**2 <= obst_r**2 obst[:,0] = obst[:,ly-1] = 1 bbRegion = numpy.nonzero(obst) # INITIAL CONDITION: (rho=0, u=0) ==> fIn(i) = t(i) # matlab: fIn = reshape( t' * ones(1,lx*ly), 9, lx, ly); fIn = numpy.ones((lx,ly,9)) fIn [:,:] = t # MAIN LOOP (TIME CYCLES) # matlab: for cycle = 1:maxT for cycle in range(maxT): # MACROSCOPIC VARIABLES # matlab: rho = sum(fIn); # matlab: ux = reshape ( ... # matlab: (cx * reshape(fIn,9,lx*ly)), 1,lx,ly) ./rho; # matlab: uy = reshape ( ... # matlab: (cy * reshape(fIn,9,lx*ly)), 1,lx,ly) ./rho; rho = fIn.sum(-1) ux = (cx*fIn).sum(-1)/rho uy = (cx*fIn).sum(-1)/rho # MACROSCOPIC (DIRICHLET) BOUNDARY CONDITIONS # Inlet: Poiseuille profile # matlab: L = ly-2; y = col-1.5; # matlab: ux(:,1,col) = 4 * uMax / (L*L) * (y.*L-y.*y); # matlab: uy(:,1,col) = 0; # matlab: rho(:,1,col) = 1 ./ (1-ux(:,1,col)) .* ( ... # matlab: sum(fIn([1,3,5],1,col)) + ... # matlab: 2*sum(fIn([4,7,8],1,col)) ); L = ly-2.0 y = col-1.5 # Is that right ? ux[0,1:-1] = 4 * uMax / (L*L) * (y.*L-y.*y) # Is that right ? uy[0,1:-1] = 0 rho[0,1:-1] = 1./(1-ux[1,1:-1])*(sum(fIn[ ([1,3,5],1,col)) + 2*sum(fIn([4,7,8],1,col))) # Outlet: Zero gradient on rho/ux # matlab: rho(:,lx,col) = rho(:,lx-1,col); # matlab: uy(:,lx,col) = 0; # matlab: ux(:,lx,col) = ux(:,lx-1,col); # % Outlet: Zero gradient on rho/ux # rho(:,lx,col) = rho(:,lx-1,col); # uy(:,lx,col) = 0; # ux(:,lx,col) = ux(:,lx-1,col); # % COLLISION STEP # for i=1:9 # cu = 3*(cx(i)*ux+cy(i)*uy); # fEq(i,:,:) = rho .* t(i) .* ( 1 + cu + 1/2*(cu.*cu) - 3/2*(ux.^2+uy.^2) ); # fOut(i,:,:) = fIn(i,:,:) - omega .* (fIn(i,:,:)-fEq(i,:,:)); # end # % MICROSCOPIC BOUNDARY CONDITIONS # for i=1:9 # % Left boundary # fOut(i,1,col) = fEq(i,1,col) + 18*t(i)*cx(i)*cy(i)* ( fIn(8,1,col) - fIn(7,1,col)-fEq(8,1,col)+fEq(7,1,col) ); # % Right boundary # fOut(i,lx,col) = fEq(i,lx,col) + 18*t(i)*cx(i)*cy(i)* ( fIn(6,lx,col) - fIn(9,lx,col)-fEq(6,lx,col)+fEq(9,lx,col) ); # % Bounce back region # fOut(i,bbRegion) = fIn(opp(i),bbRegion); # % STREAMING STEP # for i=1:9 # fIn(i,:,:) = circshift(fOut(i,:,:), [0,cx(i),cy(i)]);

6 11

How to fix the diagonal values of a matrix
by gerardo.berbeglia March 16, 2010

March 16, 2010

How can i take out the diagonal values of a matrix and fix them to zero? Example: input: [[2,3,4],[3,4,5],[4,5,6]] output: [[0,3,4],[3,0,5],[4,5,0]] Thanks. -- View this message in context: http://old.nabble.com/How-to-fix-the-diagonal-values-of-a-matrix-tp27917991… Sent from the Numpy-discussion mailing list archive at Nabble.com.

3 4

Documentation for dtypes with named fields
by Sam Tygier March 16, 2010

March 16, 2010

Hello I can't find much documentation for using arrays where the dtype has named fields (is there a term for this sort of array). there is some in http://docs.scipy.org/doc/numpy/reference/arrays.dtypes.html http://docs.scipy.org/doc/numpy/reference/generated/numpy.dtype.html but that's most creation. for example if i have a = zeros(5, dtype=[('a','f'), ('b','f'), ('c','i')]) and fill it with values. if i want to fill a row, i can do a[0] = (0.1,0.2,3) but if i use a list it wont work a[0] = [0.1,0.2,3] is there an explanation of why? now if i wanted to do some sort of conversion, eg multiply columns 'a' and 'b' by 10, is there a way to do a *= [10,10,1] like i would do with a if were a zeros((5,3)) array is a['a'] *= 10 a['b'] *= 10 the best method? also whats the best way to take a slice of the columns? if i want just the 'a' and 'b' columns. if it were a 2d array i could do a[0:2] but a['a':'b'] does not work. is there a way to do this? thanks Sam

4 14

Array mapping question
by Jorge Scandaliaris March 16, 2010

March 16, 2010

Hi, I have a 1D array containing indexes to specific measurements. As this array is a slice of a bigger one, the indexes don't necessarily start at 0 nor they are sequential. For example, I can have an array A where In [34]: A.shape Out[34]: (4764,) In [35]: ctab = np.unique(A) In [36]: ctab Out[36]: array([48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62], dtype=int64) I would like to map these indexes to a sequence starting from zero. The usual look up table approach doesn't work here. I can solve this using a dictionary, but then I am forced to using a loop or a list comprehension: In [38]: cdic = dict(zip(ctab, range(ctab.size))) In [39]: cdic Out[39]: {48: 0, 49: 1, 50: 2, 51: 3, 52: 4, 53: 5, 54: 6, 55: 7, 56: 8, 57: 9, 58: 10, 59: 11, 60: 12, 61: 13, 62: 14} A_remapped = np.asarray([cdic[x] for x in A]) Am I overlooking a better way of doing this? Thanks, Jorge

2 4

[ANN] EuroPython 2010 registration and talk submissions now open!
by Ciarán Mooney March 16, 2010

March 16, 2010

EuroPython 2010 - 17th to 24th July 2010 ---------------------------------------- EuroPython is a conference for the Python programming language community, including the Django, Zope and Plone communities. It is aimed at everyone in the Python community, of all skill levels, both users and programmers. Last year's conference was the largest open source conference in the UK and one of the largest community organised software conferences in Europe. This year EuroPython will be held from the 17th to 24th July in Birmingham, UK. It will include over 100 talks, tutorials, sprints and social events. Registration ------------ Registration is open now at: http://www.europython.eu/registration/ For the best registration rates, book as soon as you can! Extra Early Bird closes soon, after which normal Early Bird rate will apply until 10th May Talks, Activities and Events ---------------------------- Do you have something you wish to present at EuroPython? You want to give a talk, run a tutorial or sprint? Go to http://www.europython.eu/talks/cfp/ for information and advice! Go to http://wiki.europython.eu/Sprints to plan a sprint! Help Us Out ----------- EuroPython is run by volunteers, like you! We could use a hand, and any contribution is welcome. Go to http://wiki.europython.eu/Helping to join us! Go to http://www.europython.eu/contact/ to contact us directly! Sponsors -------- Sponsoring EuroPython is a unique opportunity to affiliate with this prestigious conference and to reach a large number of Python users from computing professionals to academics, from entrepreneurs to motivated and well-educated job seekers. http://www.europython.eu/sponsors/ Spread the Word --------------- We are a community-run not-for-profit conference. Please help to spread the word by distributing this announcement to colleagues, project mailing lists, friends, your blog, Web site, and through your social networking connections. Take a look at our publicity resources: http://wiki.europython.eu/Publicity General Information ------------------- For more information about the conference, please visit the official site: http://www.europython.eu/ Looking forward to see you! The EuroPython Team

1 0

problem with applying patch
by z99719 z99719 March 15, 2010

March 15, 2010

I have installed latest numpy from svn on ubuntu karmic(9.10) and compiled the code in my sand box. I downloaded two patches (test_umath_complex.py.patch, build_clib.py.patch) from the review patch directory on http://projects.scipy.org/numpy/report/12 page for testing. The following patch command gives me patch -p0 -i test_umath_complex.py.patch patch unexpectedly ends in middle of line patch: **** Only garbage was found in the patch input. The patch files are placed in the top level numpy sandbox directory and are in xml format rather than a diff file. I used "save link as" on the link to download the patch on the "Attachments" section of the link page. Checked the test_umath_complex.py in the numpy/core/test directory and found significant differences with the listing on the patch link (both old, and the new versions). Don't think this is limited to just these patches any patch that I tried gives the same error message. Thank you regards alex

3 2