Mailman 3 July 2009 - SciPy-User

Re: [SciPy-user] Finding Gaps in an Array
by Lorenzo Isella July 6, 2009

July 6, 2009

> Date: Mon, 06 Jul 2009 10:18:07 +0200 > From: Lorenzo Isella <lorenzo.isella(a)gmail.com> > Subject: [SciPy-user] Finding Gaps in an Array > To: scipy-user(a)scipy.org > Message-ID: <4A51B33F.7060306(a)gmail.com> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > Dear All, > Suppose you have an integer array whose entries are like the one below > (no entry is ever repeated and the array is sorted in increasing order) > > 1240070020 > 1240070040 > 1240070060 > 1240070080 > 1240070100 > 1240070180 > 1240070200 > 1240070620 > 1240070640 > 1240070880 > 1240070900 > 1240070940 > 1240070980 > 1240071000 > 1240071020 > 1240071040 > 1240071060 > 1240071080 > 1240071100 > 1240071120 > 1240071140 > 1240071160 > 1240071180 > 1240071200 > 1240071220 > 1240071240 > 1240071260 > 1240071280 > 1240071300 > > Let us assume that the entries represent contact times (in seconds) > between 2 objects (one does not need to know anything else for this > post). The value of the initial time does not matter (i.e. you could > subtract it to all the elements in the array to have a list starting > from zero). Now, you define the contact duration as the time spanned by > consecutive entries evenly spaced by 20 seconds. > E.g. in the array above, the first 5 elements define a contact of > duration 1240070100-1240070020=80. Then there is an 80-sec gap > (1240070180-1240070100) followed by a contact of duration 20 > (1240070200-1240070180). > How can you automatically calculate these intervals? > Any suggestion is welcome. > Cheers > > Lorenzo > > > > > Date: Mon, 6 Jul 2009 05:01:14 -0700 (PDT) > From: David Baddeley <david_baddeley(a)yahoo.com.au> > Subject: Re: [SciPy-user] Finding Gaps in an Array > To: SciPy Users List <scipy-user(a)scipy.org> > Message-ID: <672254.30337.qm(a)web33004.mail.mud.yahoo.com> > Content-Type: text/plain; charset=utf-8 > > > Hi Lorenzo, > > numpy.diff(your_array) > > will give you the times between entries, > > numpy.where(numpy.diff(your_array) > 20) > > will give the indices where the gap was longer than 20 > > cheers, > David > > > > Thanks David, But I also need something more. Let us make it real easy. Consider the array import numpy as n f=n.array([0,1,2,4,9,22,23,24,32,33,59,60,76]) I want to calculate the length of the subarrays consisting of numbers evenly spaced by 1 (like in the example, but now the gap is 1). In the case of f, these subarrays are: [0,1,2], [22,23,24], [32,33] and [59,60]. Hence, I would like to get back an array giving me the counts [3,3,2,2]. Does anyone know how to do this efficiently? Cheers Lorenzo

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Finding Gaps in an Array
by Lorenzo Isella July 6, 2009

July 6, 2009

Dear All, Suppose you have an integer array whose entries are like the one below (no entry is ever repeated and the array is sorted in increasing order) 1240070020 1240070040 1240070060 1240070080 1240070100 1240070180 1240070200 1240070620 1240070640 1240070880 1240070900 1240070940 1240070980 1240071000 1240071020 1240071040 1240071060 1240071080 1240071100 1240071120 1240071140 1240071160 1240071180 1240071200 1240071220 1240071240 1240071260 1240071280 1240071300 Let us assume that the entries represent contact times (in seconds) between 2 objects (one does not need to know anything else for this post). The value of the initial time does not matter (i.e. you could subtract it to all the elements in the array to have a list starting from zero). Now, you define the contact duration as the time spanned by consecutive entries evenly spaced by 20 seconds. E.g. in the array above, the first 5 elements define a contact of duration 1240070100-1240070020=80. Then there is an 80-sec gap (1240070180-1240070100) followed by a contact of duration 20 (1240070200-1240070180). How can you automatically calculate these intervals? Any suggestion is welcome. Cheers Lorenzo

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The baffling behavior that just won't die
by David Goldsmith July 5, 2009

July 5, 2009

Hi, folks. I'm having a problem using numpy.lookfor() that is very reminiscent of this thread: http://mail.scipy.org/pipermail/scipy-user/2008-June/017294.html The simplest reproduction of the problem is illustrated thusly: c:\Python25\Lib\site-packages\numpy>dir random.py Volume in drive C is SQ004816V03 Volume Serial Number is 8E67-90B7 Directory of c:\Python25\Lib\site-packages\numpy File Not Found c:\Python25\Lib\site-packages\numpy>python Python 2.5.4 (r254:67916, Dec 23 2008, 15:10:54) [MSC v.1310 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> import numpy as np >>> 'lookfor' in dir(np) True >>> help(np.lookfor) cannot import name Random And that's all she wrote. I read the thread cited above and it led me to examine the goings-on using python -v; as warned, there was a waterfall of imports, but at the end of it all was: . : # C:\Python25\lib\repr.pyc matches C:\Python25\lib\repr.py import repr # precompiled from C:\Python25\lib\repr.pyc import collections # builtin # C:\Python25\lib\tempfile.pyc matches C:\Python25\lib\tempfile.py import tempfile # precompiled from C:\Python25\lib\tempfile.pyc import random # directory random # random\__init__.pyc matches random\__init__.py import random # precompiled from random\__init__.pyc # random\info.pyc matches random\info.py import random.info # precompiled from random\info.pyc import random.mtrand # dynamically loaded from random\mtrand.pyd cannot import name Random Please help. Thanks! DG

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Scipy weave errors: undefined symbols when importing compiled extension
by Craig Finch July 5, 2009

July 5, 2009

I have two computers that have almost identical Python installations, but one of them has problems with inline Weave code. Here is a simple test script that I am using: #!/usr/bin/env python from scipy import weave input_val = 10 code=r"""return_val = 10*input_val;""" print weave.inline(code, ['input_val'], headers=[], compiler='gcc') It runs successfully on one computer with the following configuration: Python 2.5.2 (r252:60911, Apr 17 2009, 18:42:17) [GCC 4.1.1 (Gentoo 4.1.1-r3)] on linux2 Scipy 0.7.0 Numpy 1.3.0 On the other computer, it fails with the following message: Traceback (most recent call last): File "test.py", line 12, in <module> print weave.inline(code, ['input_val'], headers=[], compiler='gcc') File "/home/cfinch/lib/python2.5/site-packages/scipy/weave/inline_tools.py", line 335, in inline **kw) File "/home/cfinch/lib/python2.5/site-packages/scipy/weave/inline_tools.py", line 468, in compile_function exec 'import ' + module_name File "<string>", line 1, in <module> ImportError: /home/cfinch/.python25_compiled/sc_71b2502f9a0b0ca9f89b0cdc7ad3819e0.so: undefined symbol: _ZNSt8ios_base4InitD1Ev I used nm to check, and that symbol is indeed present in the compiled .so file. The configuration of this computer is: Python 2.5.4 (r254:67916, Apr 22 2009, 15:52:10) [GCC 4.1.1 20070105 (Red Hat 4.1.1-52)] on linux2 Scipy 0.7.0 Numpy 1.3.0 One complication is that the default compiler on this system is icc, and numpy distutils doesn't offer a way to force it to use gcc for everything. I have tried everything I can think of--deleting the files in .python25_compiled, rebuilding Scipy to ensure it was built with gcc, even hacking Numpy's distutils to force it to compile the extension with gcc. I also ran the unit tests that come with weave. Running weave.test() produces no errors, but running weave.test("full") produces lots of errors: Ran 390 tests in 145.246s FAILED (KNOWNFAIL=6, errors=196) <nose.result.TextTestResult run=390 errors=196 failures=0> There are no errors on the other system. Can anyone suggest a way to fix this? Craig

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Creating a Matrix from a Sum
by Lorenzo Isella July 4, 2009

July 4, 2009

Dear All, I need some help to efficiently write an array manipulation. I suspect this could be a one-liner. Assume that you have a histogram of observations that you store in a vector x. Let us say that its i-th entry, x_i, corresponds to the number of observations in the i-th channel, for i=1,2...N. (or 0,1...N-1, please let me know if there is any potential 0/1 pitfall in the following). Now, for any two channels i and j, I want to calculate the probability of having an observation in any other channel k, where k>=max(i,j). That is to say P(i,j)=sum_{k=max(i,j)}^N x_k /C, where C=sum_{i=1}^N x_i is just a normalization factor. Does anyone know a good way of writing this in a code to get the P(i,j) matrix? Many thanks Lorenzo

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Re: [SciPy-user] Access to IPython's timeit result
by Joe Kington July 4, 2009

July 4, 2009

Hi Tom, Have you looked at the timeit module? http://docs.python.org/library/timeit.html At least in 2.6, timeit.timeit() is builtin. In older versions, you just need to build a timeit.Time class instance and call it's timeit() function, which returns an average over a number of executions (default=1000000). If you really want the "best of 3" like ipython gives you, you can use the repeat function of the timeit.Time module to return a list of the execution times for each run. Hope that helps, -Joe On Sat, Jul 4, 2009 at 12:02 PM, Joe Kington <jkington(a)geology.wisc.edu>wrote: > Hi Tom, > > Have you looked at the timeit module? > > http://docs.python.org/library/timeit.html > > At least in 2.6, timeit.timeit() is builtin. In older versions, you just > need to build a timeit.Time class instance and call it's timeit() function, > which returns an average over a number of executions (default=1000000). > > If you really want the "best of 3" like ipython gives you, you can use the > repeat function of the timeit.Time module to return a list of the execution > times for each run. > > Hope that helps, > -Joe > > On Sat, Jul 4, 2009 at 8:31 AM, Tom K. <tpk(a)kraussfamily.org> wrote: > >> >> Hi, >> Looking at timeit in IPython, you get a nice output e.g. >> >> In [14]: timeit x=range(100000) >> 100 loops, best of 3: 5.75 ms per loop >> >> but what if I want to get access to that 5.75 ms number so I can write a >> program to capture these over a list of "N" values and make a plot of time >> versus "N"? >> >> - Tom K. >> -- >> View this message in context: >> http://www.nabble.com/Access-to-IPython%27s-timeit-result-tp24334619p243346… >> Sent from the Scipy-User mailing list archive at Nabble.com. >> >> _______________________________________________ >> SciPy-user mailing list >> SciPy-user(a)scipy.org >> http://mail.scipy.org/mailman/listinfo/scipy-user >> > >

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Access to IPython's timeit result
by Tom K. July 4, 2009

July 4, 2009

Hi, Looking at timeit in IPython, you get a nice output e.g. In [14]: timeit x=range(100000) 100 loops, best of 3: 5.75 ms per loop but what if I want to get access to that 5.75 ms number so I can write a program to capture these over a list of "N" values and make a plot of time versus "N"? - Tom K. -- View this message in context: http://www.nabble.com/Access-to-IPython%27s-timeit-result-tp24334619p243346… Sent from the Scipy-User mailing list archive at Nabble.com.

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Add a column to masked array.
by Dharhas Pothina July 2, 2009

July 2, 2009

Hi, I have a masked array 'data' where data.shape = (1805,4) and another masked array 'a' where a.shape = (1805,) I would like to generate a new masked array with the 'a' as the last column , i.e. the final shape should be (1805,5). I've tried np.hstack which gave an error 'arrays must have same number of dimensions' and ma.hstack just combines both masked arrays into a list. how would I go about doing what I need? thanks - dharhas

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