Mailman 3 December 2006 - NumPy-Discussion

equivalent to isempty command in matlab (newbie question)
by Giorgio Luciano Dec. 6, 2006

Dec. 6, 2006

Today I've also posted a question to scipy groups because I've thought I've found a solution but this work good bar(N, b1[:,0], width, color='r', yerr=binterv) ############ s3=find(sig1[:,arange(ini,c)]<=0.001) b1=b.flatten() #if s3!=[]: for i3 in arange(len(s3)): text(s3[i3], b1[s3[i3]+ini],'***') s2=find(logical_and(sig1[:,arange(ini,c)]>0.001,sig1[:,arange(ini,c)]<=0.01)) for i2 in arange(len(s2)): text(s2[i2], b1[s2[i2]+ini],'**') s1=find(logical_and(sig1[:,arange(ini,c)]>0.01,sig1[:,arange(ini,c)]<=0.05)) for i1 in arange(len(s1)): text(s1[i1], b1[s1[i1]+ini],'*') title('Plot of the coefficients of the model') and when i uncomment the ifs3!=[] part it does not... so in this case I've solve the problem.. but is there an equivalent for isempty matlab command in numpy ? Thanks in advance for the reply Giorgio

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Re: [Numpy-discussion] How to speed up this function?
by Nadav Horesh Dec. 6, 2006

Dec. 6, 2006

You can speed it up easily by avoiding the loop. The main idea is to replace the indexing of the type [i+1,j], [i-1,j], [i,j+1], [i,j-1] by the appropriate slicing. For example for i in xrange(1,n): for j in xrange(1,m) a[i,j] = b[i-1,j] + c[i,j+1] can be replaced by a[1:,:-1] = b[:-1] + c[:,1:] Nadav. -----Original Message----- From: numpy-discussion-bounces(a)scipy.org [mailto:numpy-discussion-bounces@scipy.org] On Behalf Of fsenkel(a)verizon.net Sent: Monday, December 04, 2006 05:28 To: numpy-discussion(a)scipy.org Subject: [Numpy-discussion] How to speed up this function? Hello, I'm taking a CFD class, one of the codes I wrote runs very slow. When I look at hotshot is says the function below is the problem. Since this is an explicit step, the for loops are only traversed once, so I think it's caused by memory usage, but I'm not sure if it's the local variables or the loop? I can vectorize the inner loop, would declaring the data structures in the calling routine and passing them in be a better idea than using local storage? I'm new at python and numpy, I need to look at how to get profiling information for the lines within a function. Thank you, Frank PS I tried to post this via google groups, but it didn't seem to go through, sorry if it ends up as multiple postings def findw(wnext,wprior,phiprior,uprior,vprior): #format here is x[i,j] where i's are rows, j's columns, use flipud() to get the #print out consistent with the spacial up-down directions #assign local names that are more #inline with the class notation w = wprior phi = phiprior u = uprior v = vprior #three of the BC are known so just set them #symetry plane wnext[0,0:gcols] = 0.0 #upper wall wnext[gN,0:gcols] = 2.0/gdy**2 * (phi[gN,0:gcols] - phi[gN-1,0:gcols]) #inlet, off the walls wnext[1:grows-1,0] = 0.0 upos = where(u>0) vpos = where(v>0) Sx = ones_like(u) Sx[upos] = 0.0 Sy = ones_like(v) Sy[vpos] = 0.0 uw = u*w vw = v*w #interior nodes for j in range(1,gasizej-1): for i in range(1,gasizei-1): wnext[i,j] =( w[i,j] + gnu*gdt/gdx**2 * (w[i,j-1] - 2.0*w[i,j] + w[i,j+1]) + gnu*gdt/gdy**2 * (w[i-1,j] - 2.0*w[i,j] + w[i+1,j]) - (1.0 - Sx[i,j]) * gdt/gdx * (uw[i,j] - uw[i,j-1]) - Sx[i,j] * gdt/gdx * (uw[i,j+1] - uw[i,j]) - (1.0 - Sy[i,j]) * gdt/gdy * (vw[i,j] - vw[i-1,j]) - Sy[i,j] * gdt/gdy * (vw[i+1,j] - vw[i,j]) ) ## print "***wnext****" ## print "i: ", i, "j: ", j, "wnext[i,j]: ", wnext[i,j] #final BC at outlet, off walls wnext[1:grows-1,gM] = wnext[1:grows-1,gM-1] _______________________________________________ Numpy-discussion mailing list Numpy-discussion(a)scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion

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Re: [Numpy-discussion] Precision in Python
by Nadav Horesh Dec. 6, 2006

Dec. 6, 2006

-----Original Message----- From: numpy-discussion-bounces(a)scipy.org [mailto:numpy-discussion-bounces@scipy.org] On Behalf Of Elton Mendes Sent: Monday, November 27, 2006 13:57 To: numpy-discussion(a)scipy.org Subject: [Numpy-discussion] Precision in Python Hi. I'm having a precision problem in python Example: >>> a = 5.14343434 >>> b = round(a,1) >>> b 5.0999999999999996 >>> It´s possible to round the number exactly to 5.1 NO. 5.1 can not be represented exactly as a machine native float. The only way I know to represent this value exactly is to use the decimal module. Usually you do not want to do this. Nadav.

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Re: [Numpy-discussion] dot operations on multidimensional arrays
by Nadav Horesh Dec. 6, 2006

Dec. 6, 2006

Try numpy.tensordot Nadav -----Original Message----- From: numpy-discussion-bounces(a)scipy.org [mailto:numpy-discussion-bounces@scipy.org] On Behalf Of Carsten Rostgaard Sent: Thursday, November 23, 2006 10:06 To: numpy-discussion(a)scipy.org Subject: [Numpy-discussion] dot operations on multidimensional arrays Hi! I am trying to use the "dot" method on multi-(more than 2)-dimensional arrays. Specifically I do >> y = dot(a, b) where a is a 2D array and b is a 3D array. using numpy I get the the help: " dot(...) dot(a,v) returns matrix-multiplication between a and b. The product-sum is over the last dimension of a and the second-to-last dimension of b. " I then expect that >> y[i, j, k] = sum(a[i, :] * b[j, :, k]) which is actually what I get. The question is then: 1) Is there any way to change the axis for which the product-sum is performed. This can of course be done by a swapaxis before and after the operation, but this makes the array non-contiguous, in which case the dot operation often makes bugs (at least in Numeric). 2) For complicated reasons we still use Numeric in our software package, and in this, "dot" behaves very strangely. According to the Numeric help: " dot(a, b) dot(a,b) returns matrix-multiplication between a and b. The product-sum is over the last dimension of a and the second-to-last dimension of b. " so I would have expected again that y[i, j, k] = sum(a[i, :] * b[j, :, k]), and the dimensions actually fit, i.e. y.shape = (a.shape[0], b.shape[0], b.shape[2]), but only some rows of the result has these values!! Does anyone know what Numeric.dot(a, b) actually does when b has more than two dimensions? I use the following test script: ---------------------BEGIN SCRIPT----------------------- import Numeric as num # import numpy as num # make 'random' input arrays a = num.zeros((2, 5)) b = num.zeros((3, 5, 4)) a.flat[:] = num.arange(len(a.flat)) - 3 b.flat[:] = num.arange(len(b.flat)) + 5 # built-in dot product y1 = num.dot(a, b) # manual dot product y2 = num.zeros((a.shape[0], b.shape[0], b.shape[2])) for i in range(a.shape[0]): for j in range(b.shape[0]): for k in range(b.shape[2]): y2[i, j, k] = num.sum(a[i,:] * b[j, :, k]) # test for consistency print y1 == y2 ---------------------END SCRIPT------------------------ with the result: [[[1 1 1 1] [0 0 0 0] [0 0 0 0]] [[1 1 1 1] [0 0 0 0] [0 0 0 0]]] thanks a lot, Carsten Rostgaard Carsten.Rostgaard(a)fysik.dtu.dk _______________________________________________ Numpy-discussion mailing list Numpy-discussion(a)scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion

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take semantics (bug?)
by Michael McNeil Forbes Dec. 6, 2006

Dec. 6, 2006

What are the semantics of the "take" function? I would have expected that the following have the same shape and size: >>> a = array([1,2,3]) >>> inds = a.nonzero() >>> a[inds] array([1, 2, 3]) >>> a.take(inds) array([[1, 2, 3]]) Is there a bug somewhere here or is this intentional? Michael.

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numpy in debian
by Keith Goodman Dec. 5, 2006

Dec. 5, 2006

I'm impressed with how easy it is to compile numpy. I'm even more impressed with how easy it is let someone else compile it and just apt-get it. Does anyone know the numpy plans for debian? It is currently at 1.0rc1. I'm afraid to ask the debian numpy maintainers since they have already done me such a big favor by packaging numpy and I haven't even thanked them yet. Oh, yeah. Then there's that other project I should thank. Thank you. Numpy is a joy to use.

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dot operations on multidimensional arrays
by Carsten Rostgaard Dec. 5, 2006

Dec. 5, 2006

Hi! I am trying to use the "dot" method on multi-(more than 2)-dimensional arrays. Specifically I do >> y = dot(a, b) where a is a 2D array and b is a 3D array. using numpy I get the the help: " dot(...) dot(a,v) returns matrix-multiplication between a and b. The product-sum is over the last dimension of a and the second-to-last dimension of b. " I then expect that >> y[i, j, k] = sum(a[i, :] * b[j, :, k]) which is actually what I get. The question is then: 1) Is there any way to change the axis for which the product-sum is performed. This can of course be done by a swapaxis before and after the operation, but this makes the array non-contiguous, in which case the dot operation often makes bugs (at least in Numeric). 2) For complicated reasons we still use Numeric in our software package, and in this, "dot" behaves very strangely. According to the Numeric help: " dot(a, b) dot(a,b) returns matrix-multiplication between a and b. The product-sum is over the last dimension of a and the second-to-last dimension of b. " so I would have expected again that y[i, j, k] = sum(a[i, :] * b[j, :, k]), and the dimensions actually fit, i.e. y.shape = (a.shape[0], b.shape[0], b.shape[2]), but only some rows of the result has these values!! Does anyone know what Numeric.dot(a, b) actually does when b has more than two dimensions? I use the following test script: ---------------------BEGIN SCRIPT----------------------- import Numeric as num # import numpy as num # make 'random' input arrays a = num.zeros((2, 5)) b = num.zeros((3, 5, 4)) a.flat[:] = num.arange(len(a.flat)) - 3 b.flat[:] = num.arange(len(b.flat)) + 5 # built-in dot product y1 = num.dot(a, b) # manual dot product y2 = num.zeros((a.shape[0], b.shape[0], b.shape[2])) for i in range(a.shape[0]): for j in range(b.shape[0]): for k in range(b.shape[2]): y2[i, j, k] = num.sum(a[i,:] * b[j, :, k]) # test for consistency print y1 == y2 ---------------------END SCRIPT------------------------ with the result: [[[1 1 1 1] [0 0 0 0] [0 0 0 0]] [[1 1 1 1] [0 0 0 0] [0 0 0 0]]] thanks a lot, Carsten Rostgaard Carsten.Rostgaard(a)fysik.dtu.dk

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Precision in Python
by Elton Mendes Dec. 5, 2006

Dec. 5, 2006

Hi. I'm having a precision problem in python Example: >>> a = 5.14343434 >>> b = round(a,1) >>> b 5.0999999999999996 >>> It´s possible to round the number exactly to 5.1

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How to speed up this function?
by fsenkel＠verizon.net Dec. 5, 2006

Dec. 5, 2006

Hello, I'm taking a CFD class, one of the codes I wrote runs very slow. When I look at hotshot is says the function below is the problem. Since this is an explicit step, the for loops are only traversed once, so I think it's caused by memory usage, but I'm not sure if it's the local variables or the loop? I can vectorize the inner loop, would declaring the data structures in the calling routine and passing them in be a better idea than using local storage? I'm new at python and numpy, I need to look at how to get profiling information for the lines within a function. Thank you, Frank PS I tried to post this via google groups, but it didn't seem to go through, sorry if it ends up as multiple postings def findw(wnext,wprior,phiprior,uprior,vprior): #format here is x[i,j] where i's are rows, j's columns, use flipud() to get the #print out consistent with the spacial up-down directions #assign local names that are more #inline with the class notation w = wprior phi = phiprior u = uprior v = vprior #three of the BC are known so just set them #symetry plane wnext[0,0:gcols] = 0.0 #upper wall wnext[gN,0:gcols] = 2.0/gdy**2 * (phi[gN,0:gcols] - phi[gN-1,0:gcols]) #inlet, off the walls wnext[1:grows-1,0] = 0.0 upos = where(u>0) vpos = where(v>0) Sx = ones_like(u) Sx[upos] = 0.0 Sy = ones_like(v) Sy[vpos] = 0.0 uw = u*w vw = v*w #interior nodes for j in range(1,gasizej-1): for i in range(1,gasizei-1): wnext[i,j] =( w[i,j] + gnu*gdt/gdx**2 * (w[i,j-1] - 2.0*w[i,j] + w[i,j+1]) + gnu*gdt/gdy**2 * (w[i-1,j] - 2.0*w[i,j] + w[i+1,j]) - (1.0 - Sx[i,j]) * gdt/gdx * (uw[i,j] - uw[i,j-1]) - Sx[i,j] * gdt/gdx * (uw[i,j+1] - uw[i,j]) - (1.0 - Sy[i,j]) * gdt/gdy * (vw[i,j] - vw[i-1,j]) - Sy[i,j] * gdt/gdy * (vw[i+1,j] - vw[i,j]) ) ## print "***wnext****" ## print "i: ", i, "j: ", j, "wnext[i,j]: ", wnext[i,j] #final BC at outlet, off walls wnext[1:grows-1,gM] = wnext[1:grows-1,gM-1]

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ValueError: dimensions too large.
by koara＠atlas.cz Dec. 4, 2006

Dec. 4, 2006

Hello, i tried to create a 2d array, but encountered: ValueError: dimensions too large. Does this refer to insufficient memory, or is there really a limit on dimension sizes? Cheers.

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