locate items in matrix (index of lists of lists)
Hello there, let's suppose I have the following matrix: mat = [[1,2,3], [3,2,4], [7,8,9], [6,2,9]] where [.. , .. , ..] are the rows. I am interested into getting the "row index" of all the matrix rows where a certain number occurs. For example for 9 I should get 2 and 3 (starting from 0). For 10 I should get an error msg (item not found) and handle it. How to get the"row indexes" of found items? In practice I am looking for an equivalent to "list.index(x)" for the case "lists of lists" Many Thanks! Alex PS: this is just a simplified example, but I have actually to deal with large matrices [~50 * 4] -- http://mail.python.org/mailman/listinfo/python-list
get rid of duplicate elements in list without set
Hello there, I'd like to get the same result of set() but getting an indexable object. How to get this in an efficient way? Example using set A = [1, 2, 2 ,2 , 3 ,4] B= set(A) B = ([1, 2, 3, 4]) B[2] TypeError: unindexable object Many thanks, alex -- http://mail.python.org/mailman/listinfo/python-list
replacing numbers in LARGE MATRIX with criterium in 2 columns (a--> b)
Hello, I have this matrix [20*4 - but it could be n*4 , with n~100,000] in file "EL_list" like this: 1, 1, 2, 3 2, 4, 1, 5 3, 5, 1, 6 4, 7, 5, 6 5, 8, 7, 9 6, 8, 5, 7 7, 10, 9, 7 8, 10, 11, 12 9, 7, 13, 10 10, 14, 15, 16 11, 14, 17, 15 12, 17, 14, 18 13, 13, 16, 10 14, 16, 15, 11 15, 16, 11, 10 16, 19, 20, 21 17, 22, 15, 20 18, 17, 20, 15 19, 23, 20, 24 20, 25, 24, 20 I would like to replace some numbers in "EL_list" but only in columns 2,3,4 using the criterium by line in file "criterium" (column 1 are the new numbers, column 2 the old ones). 1 1 2 3 3 5 4 12 5 13 6 14 7 15 8 16 9 17 10 18 11 19 12 10 13 21 14 22 15 23 16 24 17 25 For example all the 7 have to replaced by 15 and so on.. -- How to implement it in a fast end efficient code? many thanks, Alex -- http://mail.python.org/mailman/listinfo/python-list
writing on file not until the end
Hello,
I am a newby with python. I wrote the following code to extract a text
from a file and write it to another file:
linestring = open(path, 'r').read() #read all the inp file in
linestring
i=linestring.index("*NODE")
i=linestring.index("E",i)
e=linestring.index("*",i+10)
textN = linestring[i+2:e-1] # crop the ELement+nodes list
Nfile = open("N.txt", "w")
Nfile.write(textN)
unfortunately when I check N.txt some lines are missing (it only crop
until "57, 0.231749688431, 0.0405121944142" but I espect the final
line to be "242, 0.2979675, 0.224605896461". I check textN and it has
all the lines until "242..")
when I try Nfile.write(textN) again it writes some more lines but
still not all!
what is wrong whit this?
Many thanks for your help!
Alex
the original file to crop (in path) looks like this:
*HEADING
ABAQUS-style file created by OOF2 on 2009-03-05 03:23:49.204607 from a
mesh of the microstructure A.
** Materials defined by OOF2:
** C:
** W:
** Master elements used in OOF2:
** 2: D2_2, Isoparametric 2-noded edge element
** 3: T3_3, Isoparametric 3 noded triangle with linear interpolation
for both fields and positions.
** 4: Q4_4, Isoparametric 4 noded quadrilateral with bilinear
interpolation for both positions and fields
** Boundary Conditions:
** Notes:
** The set of nodes and elements may be different from the set
created from a skeleton
**depending on the element type and if the mesh was refined.
** The materials and boundary conditions provided by OOF2 may be
translated into ABAQUS by the user.
** The element type provided below should be verified and modified
accordingly.
** Only elements (and nodes of such elements) that have an
associated material are included in this file.
*NODE
6, 0.0, 0.0
1, 0.031365, 0.0
9, 0.06273, 0.0
10, 0.094095, 0.0
17, 0.12546, 0.0
18, 0.156825, 0.0
29, 0.18819, 0.0
33, 0.219555, 0.0
37, 0.25092, 0.0
38, 0.282285, 0.0
245, 0.31365, 0.0
4, 0.0, 0.027505735912
7, 0.0283153247289, 0.0247564240049
8, 0.06273, 0.0236123
13, 0.126933916405, 0.0219324392255
22, 0.156825, 0.0236123
25, 0.185442602873, 0.0174188364874
35, 0.216158167035, 0.0260087067213
41, 0.25092, 0.0236123
65, 0.31365, 0.0265806479154
45, 0.0, 0.0472246
44, 0.0270022417013, 0.0457250093507
46, 0.06273, 0.0472246
47, 0.098090467, 0.0476663143118
48, 0.123432920369, 0.0493041533953
50, 0.157517619392, 0.0491485665468
54, 0.18819, 0.0472246
56, 0.21829479153, 0.0489914989989
59, 0.255396164365, 0.053642386456
62, 0.283534894472, 0.0482125495477
66, 0.31365, 0.0473757108379
68, 0.0, 0.0708369
71, 0.031365, 0.0708369
73, 0.0667269869261, 0.0724057787737
74, 0.094095, 0.0708369
75, 0.12203525, 0.0683265441516
77, 0.154069643632, 0.067833220225
79, 0.192676294482, 0.0697610582702
80, 0.218433034079, 0.0733140035725
81, 0.250592391777, 0.0705659765909
115, 0.31365, 0.0749934030765
86, 0.0, 0.0937712452856
87, 0.0327416679167, 0.0988320724937
92, 0.0642181016634, 0.0996865210439
94, 0.0939583966251, 0.0960413983543
98, 0.12277254344, 0.0956157005759
101, 0.156825, 0.0944492
107, 0.190080813107, 0.0989004943145
109, 0.219555, 0.0944492
111, 0.25092, 0.0944492
116, 0.31365, 0.0913902112346
117, 0.0, 0.118185969919
118, 0.031365, 0.1180615
119, 0.0657061477957, 0.117247512024
120, 0.095407726, 0.117579706786
122, 0.120457073874, 0.118852698163
125, 0.150412085671, 0.119805228459
128, 0.188401966746, 0.118578758447
132, 0.219555, 0.1180615
131, 0.25092, 0.1180615
134, 0.282285, 0.1180615
138, 0.0, 0.1416738
137, 0.031365, 0.1416738
139, 0.06273, 0.1416738
140, 0.098131408, 0.141731507791
160, 0.156825, 0.1416738
145, 0.18819, 0.1416738
144, 0.21849787291, 0.144222847202
146, 0.25214522485, 0.139721206044
147, 0.287968204391, 0.138882348861
151, 0.0, 0.1652861
152, 0.031365, 0.1652861
153, 0.06273, 0.1652861
154, 0.094095, 0.1652861
156, 0.124683249222, 0.165861357563
189, 0.156825, 0.1652861
163, 0.189978244061, 0.168552116135
193, 0.219555, 0.1652861
170, 0.251953134991, 0.166067888686
175, 0.31365, 0.1652861
179, 0.0, 0.1888984
178, 0.031365, 0.1888984
180, 0.06273, 0.1888984
181, 0.0948683944207, 0.186105991855
184, 0.12546, 0.1888984
185, 0.156825, 0.1888984
191, 0.185226863639, 0.189033619679
192, 0.21459450643, 0.190219519385
196, 0.248868626255, 0.189540639649
197, 0.277613280215, 0.188996385139
200, 0.31365, 0.188840621897
201, 0.0, 0.2125107
202, 0.031365, 0.2125107
203, 0.061560970472, 0.211435761567
206, 0.094095, 0.2125107
208, 0.123520746115, 0.21724383
212, 0.156824305885, 0.210136954906
215, 0.19080765368, 0.21182785412
216, 0.219555, 0.2125107
217, 0.24652142594, 0.211374186361
219, 0.282347639617, 0.2121001649
220, 0.31365, 0.213088792923
222, 0.0, 0.236123
221, 0.031365, 0.236123
224, 0.06273, 0.236123
226, 0.094095, 0.236123
230, 0.12546, 0.236123
236, 0.156825, 0.236123
237, 0.18819, 0.236123
238, 0.219555, 0.236123
246, 0.25092, 0.236123
239, 0.282285, 0.236123
244, 0.31365, 0.236123
2, 0.0150409063102, 0.0175241129844
5, 0.0, 0.0143072034694
3, 0.0156
improving a huge double-for cycle
Hello there :) , I am a python newbie and need to run following code for a task in an external simulation programm called "Abaqus" which makes use of python to access the mesh (ensamble of nodes with xy coordinates) of a certain geometrical model. [IN is the starting input containing the nodes to be check, there are some double nodes with the same x and y coordinates which need to be removed. SN is the output containing such double nodes] Code: Select all for i in range(len(IN)): #scan all elements of the list IN for j in range(len(IN)): if i <> j: if IN[i].coordinates[0] == IN[j].coordinates[0]: if IN[i].coordinates[1] == IN[j].coordinates[1]: SN.append(IN[i].label) Unfortunately my len(IN) is about 100.000 and the running time about 15h :( Any idea to improve it? I have already tried to group the "if statements" in a single one: Code: Select all if i <> j and if IN[i].coordinates[0] == IN[j].coordinates[0] and if IN[i].coordinates[1] == IN[j].coordinates[1]: but no improvements. Many thanks, Alex -- http://mail.python.org/mailman/listinfo/python-list
how to improve this cycle (extracting data from structured array)?
Hello guys, I am just wondering if there is a quick way to improve this algorithm [N is a structured array which hold info about the nodes n of a finite element mesh, and n is about 300.000). I need to extract info from N and put it in to a 3*n matrix NN which I reshape then with numpy. I think to "append" is quite unefficient: *** N = odb.rootAssembly.instances['PART-1-1'].nodes NN=[] B=[0,0,0] #auxsiliar vector for i in range(len(N)): B[0] = N[i].label B[1] = N[i].coordinates[0] B[2] = N[i].coordinates[1] NN = append(NN,B) NN=NN.reshape(-1,3) Many Thanks in advance! Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: how to improve this cycle (extracting data from structured array)?
On Feb 11, 1:08 pm, Tim Chase wrote: > Alexzive wrote: > > I am just wondering if there is a quick way to improve this algorithm > > [N is a structured array which hold info about the nodes n of a finite > > element mesh, and n is about 300.000). I need to extract info from N > > and put it in to a 3*n matrix NN which I reshape then with numpy. I > > think to "append" is quite unefficient: > > > *** > > N = odb.rootAssembly.instances['PART-1-1'].nodes > > > NN=[] > > > B=[0,0,0] > > #auxsiliar vector > > for i in range(len(N)): > > B[0] = N[i].label > > B[1] = N[i].coordinates[0] > > B[2] = N[i].coordinates[1] > > NN = append(NN,B) > > Usually this would be written with a list-comprehension, > something like > > nn = [(x.label, x.coordinates[0], x.coordinates[1]) > for x in N] > > or if you really need lists-of-lists instead of lists-of-tuples: > > nn = [[x.label, x.coordinates[0], x.coordinates[1]] > for x in N] > > -tkc yeah, much better thanks! Alex -- http://mail.python.org/mailman/listinfo/python-list
speed up a numpy code with huge array
Hello Pythonguys! is there a way to improve the performance of the attached code ? it takes about 5 h on a dual-core (using only one core) when len(V) ~1MIL. V is an array which is supposed to store all the volumes of tetrahedral elements of a grid whose coord. are stored in NN (accessed trough the list of tetraelements --> EL) Thanks in advance! Alex print 'start ' + nameodb #path = '/windows/D/SIM-MM/3D/E_ortho/' + nameodb + '.odb' path = pt + nameodb + '.odb' odb = openOdb(path) N = odb.rootAssembly.instances['PART-1-1'].nodes if loadV==1: pathV=pt+vtet V=numpy.loadtxt(pathV) VTOT = V[0] L3 = V[1] print 'using ' + vtet else: NN=[] B=[0,0,0,0] for i in range(len(N)): B[0] = N[i].label B[1] = N[i].coordinates[0] B[2] = N[i].coordinates[1] B[3] = N[i].coordinates[2] NN = append(NN,B) NN=NN.reshape(-1,4) EL = odb.rootAssembly.instances['PART-1-1'].elements L1 = max(NN[:,1])-min(NN[:,1]) L2 = max(NN[:,2])-min(NN[:,2]) L3 = max(NN[:,3])-min(NN[:,3]) VTOT=L1*L2*L3 print 'VTOT: [mm³]' + str(VTOT) V = array([]) print 'calculating new Vtet ' V = range(len(EL)+2) V[0] = VTOT V[1] = L3 for j in range(0,len(EL)): Va = EL[j].connectivity[0] Vb = EL[j].connectivity[1] Vc = EL[j].connectivity[2] Vd = EL[j].connectivity[3] ix = where(NN[:,0] == Va) Xa = NN[ix,1][0][0] Ya = NN[ix,2][0][0] Za = NN[ix,3][0][0] ix = where(NN[:,0] == Vb) Xb = NN[ix,1][0][0] Yb = NN[ix,2][0][0] Zb = NN[ix,3][0][0] ix = where(NN[:,0] == Vc) Xc = NN[ix,1][0][0] Yc = NN[ix,2][0][0] Zc = NN[ix,3][0][0] ix = where(NN[:,0] == Vd) Xd = NN[ix,1][0][0] Yd = NN[ix,2][0][0] Zd = NN[ix,3][0][0] a = [Xa,Ya,Za] b = [Xb,Yb,Zb] c = [Xc,Yc,Zc] d = [Xd,Yd,Zd] aa = numpy.diff([b,a],axis=0)[0] bb = numpy.diff([c,b],axis=0)[0] cc = numpy.diff([d,c],axis=0)[0] D=array([aa,bb,cc]) det=numpy.linalg.det(D) V[j+2] = abs(det)/6 pathV = pt + vtet savetxt(pathV, V, fmt='%.3e') ### -- http://mail.python.org/mailman/listinfo/python-list
Re: speed up a numpy code with huge array
thank you all for the tips. I 'll try them soon. I also notice another bottleneck, when python tries to access some array data stored in the odb files (---> in text below), even before starting the algoritm: ### EPS_nodes = range(len(frames)) for f in frames: ... sum = 0 --->UN = F[f].fieldOutputs['U'].getSubset(region=TOP).values <--- ... EPS_nodes[f] = UN[10].data[Scomp-1]/L3 ### unfortunately I don't have time to learn cython. Using dictionaries sounds promising. Thanks! Alex On May 26, 8:14 am, Stefan Behnel wrote: > Alexzive, 25.05.2010 21:05: > > > is there a way to improve the performance of the attached code ? it > > takes about 5 h on a dual-core (using only one core) when len(V) > > ~1MIL. V is an array which is supposed to store all the volumes of > > tetrahedral elements of a grid whose coord. are stored in NN (accessed > > trough the list of tetraelements --> EL) > > Consider using Cython for your algorithm. It has direct support for NumPy > arrays and translates to fast C code. > > Stefan -- http://mail.python.org/mailman/listinfo/python-list
Re: speed up a numpy code with huge array
sorry it was just bullshit what I wrote about the second bottleneck, it seemed to hang up but it was just me forgetting to double-enter during debugging after "for cycle". On May 26, 1:43 pm, Alexzive wrote: > thank you all for the tips. > I 'll try them soon. > > I also notice another bottleneck, when python tries to access some > array data stored in the odb files (---> in text below), even before > starting the algoritm: > > ### > EPS_nodes = range(len(frames)) > for f in frames: > ... sum = 0 > ---> UN = F[f].fieldOutputs['U'].getSubset(region=TOP).values <--- > ... EPS_nodes[f] = UN[10].data[Scomp-1]/L3 > > ### > > unfortunately I don't have time to learn cython. Using dictionaries > sounds promising. > Thanks! > Alex > > On May 26, 8:14 am, Stefan Behnel wrote: > > > > > Alexzive, 25.05.2010 21:05: > > > > is there a way to improve the performance of the attached code ? it > > > takes about 5 h on a dual-core (using only one core) when len(V) > > > ~1MIL. V is an array which is supposed to store all the volumes of > > > tetrahedral elements of a grid whose coord. are stored in NN (accessed > > > trough the list of tetraelements --> EL) > > > Consider using Cython for your algorithm. It has direct support for NumPy > > arrays and translates to fast C code. > > > Stefan -- http://mail.python.org/mailman/listinfo/python-list
how to build with 2.4 having 2.6 as main python
Hello there, my Mandriva has the 2.6.4 python pre-installed (in /usr/lib64/ python2.6/) I need to install numpy 1.4 for python 2.4.3 (I installed it separately from source on/usr/local/lib/python2.4/ ) but still typing "python" I get: Python 2.6.4 (r264:75706, Jan 8 2010, 18:59:59) [GCC 4.4.1] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> what to change in order to get "python" calling python 2.4.3 instead of 2.6.4 (at least during python setup.py build)? I suppose I need something like changing the link to /usr/local/bin/ python.. but I fear to do something bad by myself.. please help! -- http://mail.python.org/mailman/listinfo/python-list
Re: how to build with 2.4 having 2.6 as main python
thanks guys, the solution for me was python2.4 setup.py install --prefix=/usr/local cheers, AZ On Jun 14, 11:00 am, Steven D'Aprano wrote: > On Mon, 14 Jun 2010 01:30:09 -0700, Alexzive wrote: > > what to change in order to get "python" calling python 2.4.3 instead of > > 2.6.4 (at least during python setup.py build)? > > That will do bad things to your system, which will be expecting the > system Python to be 2.6 and instead will be 2.4. You will probably find > system tools will start to fail. > > > I suppose I need something like changing the link to /usr/local/bin/ > > python.. > > but I fear to do something bad by myself.. please help! > > Yes, that will do it, but if you do, you will probably break things. Best > to just call the python2.4 binary directly. > > If you call > > python2.4 > > from the command line, what happens? > > -- > Steven -- http://mail.python.org/mailman/listinfo/python-list
