Re: Easy-to-use Python GUI
On 25 дек, 06:47, "Joel Koltner" wrote:
> Is there an easy-to-use, "function"-based cross-platform GUI toolkit for
> Python out there that's a little more sophisticated than EasyGui? EasyGui
> looks good, but it's a little more restrictive than what I'd like to have, yet
> I'm (stubbornly :-) ) resistant to stepping up to a "full service" GUI toolkit
> such as pyGTK or wxPython where it's all about event loops and callbacks and
> you need to start planning how the GUI affects the overall program flow rather
> than just using a "forms" (or "Wizard")-type approach where you put up a few
> dialogs, users fill in some variables, and your program just sits around
> waiting until "OK" or "Cancel" is clicked.
>
> One approach that I like comes from SAX BASIC/WinWrap, which is more or less a
> clone of Microsoft's Visual BASIC for Applications, but they (apparently)
> wanted everything to still be human-readable, so they have a simple GUI
> ("form") builder that generates code that looks like this:
>
> ---
>
> Begin Dialog UserDialog 850,497,"Export Control" ' %GRID:10,7,1,1
>
> GroupBox 20,7,360,217,"Drill File Generation",.GroupBox1
> CheckBox 40,35,130,14,"Output drill file(s)",.genDrill
> Text 40,63,270,28,"Identify via layers as any that contain this text in
> their names:",.Text
> TextBox 40,98,220,21,.viaLayerName
> Text 40,140,100,14,"Output method:",.Text8
> DropListBox 160,140,180,21,DrillStyle(),.drillStyle
> Text 40,175,130,28,"Select drill table units:",.Text2
> ListBox 200,175,120,28,unitNames(),.unitName
>
> OKButton 310,469,90,21
> CancelButton 410,469,90,21
>
> End Dialog
>
> ' GUI builder generates or modifies everything above, but can also be edited
> by hand
> ' You write the following code...
>
> Dim dlg As UserDialog
>
> dlg.genDrill = 1
> ReDim DrillStyle(1)
> DrillStyle(0) = "All Via Layers In One File"
> DrillStyle(1) = "One File Per Via Layer"
> dlg.drillStyle = 1
>
> func=Dialog(dlg)
>
> ---
>
> This is pretty darned easy for me understand and modify either by hand or with
> the GUI builder. Still, it's quite powerful, since it supports all the common
> GUI elements (text, group boxes, checkboxes, drop-down lists, text boxes,
> buttons, etc.). This is about the level of sophistication I'm looking for.
>
> Anything like this for Python?
>
> Thanks,
> ---Joel
Try PyScripter :)
http://mmm-experts.com/Products.aspx?ProductId=4
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cxFreeze executable linked to /usr/lib/libpython2.3.so
Hi, I compiled a python script using cxFreeze because I need a standalone application, while the Windows version runs without any python installation the linux version of the executable is linked to libpython2.3.so.1.0 => /usr/lib/libpython2.3.so.1.0 thus the end user have to install python 2.3 in order to run the binary. I would like to know if it's possible, using cxFreeze to create a binary that is not linked to any python system library, for example by putting the library in the directory where the binary file is located. Thank Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: wxTreeCtrl Q?
I hope this can help: http://wxwidgets.org/manuals/2.6.1/wx_wxtreectrl.html#wxtreectrlsetitemfont http://wxwidgets.org/manuals/2.6.1/wx_wxtreectrl.html#wxtreectrlsetitembold This pages are from the wxwidgets api reference but the functions are the same in wxPython Bye Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Cameron Laird wrote: > This thread confuses me. > > I've lost track of the real goal. If it's an exact calculation of > binomial coefficients--or even one of several other potential > targets mentioned--I echo Steven D'Aprano, and ask, are you *sure* > the suggestions already offered aren't adequate? Hi Cameron, my real goal was to calculate the hypergeometric distribution. The problem was that the function for hypergeometric calculation from scipy uses the scipy.comb function which by default uses floats so for large numbers comb(n,r) returns inf. and hence the hypergeometric returns nan. The first suggestion, the one by Robert Kern, resolved my problem: Raven wrote: >Thanks to all of you guys, I could resolve my problem using the >logarithms as proposed by Robert. Then the other guys gave alternative solutions so I tried them out. So form me the suggestions offered are more than adequate :-) Cameron Laird wrote: >Also, I think you > might not realize how accurate Stirling's approximation (perhaps to > second order) is in the range of interest. The problem with Stirling's approximation is that I need to calculate the hypergeometric hence the factorial for numbers within a large range e.g. choose(14000,170) or choose(5,2) Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Travis E. Oliphant wrote: > Notice the keyword for the comb function (in scipy) lets you use it to > compute exact values. SciPy does not just automatically use the long > integer because this will always slow you down. > > comb(N, k, exact=0) > > Combinations of N things taken k at a time. > > If exact==0, then floating point precision is used, otherwise > exact long integer is computed. > > Notes: >- Array arguments accepted only for exact=0 case. >- If k > N, N < 0, or k < 0, then a 0 is returned. > > -Travis Oliphant Great, thanks Travis. Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Bengt Richter wrote: > ISTM you wouldn't get zero if you scaled by 10**significant_digits (however > many > you require) before dividing. E.g., expected hits per trillion (or septillion > or whatever) > expresses probability too. Perhaps that could work in your calculation? > > Regards, > Bengt Richter Sorry but I can't figure out how to do it, can you give me an example please? Thnx Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Bengt Richter wrote: > ISTM you wouldn't get zero if you scaled by 10**significant_digits (however > many > you require) before dividing. E.g., expected hits per trillion (or septillion > or whatever) > expresses probability too. Perhaps that could work in your calculation? > > Regards, > Bengt Richter Sorry Bengt but I can't figure out how to do it, can you give me an example please? Thanks in advance. Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Scott David Daniels ha scritto: > You should really look into the timeit module -- you'll get nice > solid timings slightly easier to tweak. This seems a very interesting module, I will give it a try as soon as possible. Thanks Scott. Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Well, what to say? I am very happy for all the solutions you guys have posted :-) For Paul: I would prefer not to use Stirling's approximation The problem with long integers is that to calculate the hypergeometric I need to do float division and multiplication because integer division returns 0. A solution could be to calculate log(Long_Factorial_Integer) and finally calculate the hypergeometric with the logarithmic values. I've done a test: iterated 1000 times two different functions for the hypergeometric, the first one based on scipy.special.gammaln: from scipy.special import gammaln def lnchoose(n, m): nf = gammaln(n + 1) mf = gammaln(m + 1) nmmnf = gammaln(n - m + 1) return nf - (mf + nmmnf) def hypergeometric_gamma(k, n1, n2, t): if t > n1 + n2: t = n1 + n2 if k > n1 or k > t: return 0 elif t > n2 and ((k + n2) < t): return 0 else: c1 = lnchoose(n1,k) c2 = lnchoose(n2, t - k) c3 = lnchoose(n1 + n2 ,t) return exp(c1 + c2 - c3) and the second one based on the code by Steven and Scott: import time from math import log, exp def bincoeff1(n, r): if r < n - r: r = n - r x = 1 for i in range(n, r, -1): x *= i for i in range(n - r, 1, -1): x /= i return x def hypergeometric(k, n1, n2, t): if t > n1 + n2: t = n1 + n2 if k > n1 or k > t: return 0 elif t > n2 and ((k + n2) < t): return 0 else: c1 = log(raw_bincoeff1(n1,k)) c2 = log(raw_bincoeff1(n2, t - k)) c3 = log(raw_bincoeff1(n1 + n2 ,t)) return exp(c1 + c2 - c3) def main(): t = time.time() for i in range(1000): r = hypergeometric(6,6,30,6) print time.time() - t t = time.time() for i in range(1000): r = hypergeometric_gamma(6,6,30,6) print time.time() - t if __name__ == "__main__": main() and the result is: 0.0386447906494 0.192448139191 The first approach is faster so I think I will adopt it. -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Thanks Steven for your very interesting post. This was a critical instance from my problem: >>>from scipy import comb >>> comb(14354,174) inf The scipy.stats.distributions.hypergeom function uses the scipy.comb function, so it returned nan since it tries to divide an infinite. I did not tried to write a self-made function using standard python as I supposed that the scipy functions reached python's limits but I was wrong, what a fool :-) >If you are calculating hundreds of hypergeometric probabilities, 30 >seconds each could be quite painful, but it certainly shows that Python is >capable of doing it without resorting to logarithms which may lose some >significant digits. Although, in fairness, the log function doesn't seem >to lose much accuracy for arguments in the range you are dealing with. Yes I am calculating hundreds of hypergeometric probabilities so I need fast calculations Ale -- http://mail.python.org/mailman/listinfo/python-list
Re: Hypergeometric distribution
Thanks to all of you guys, I could resolve my problem using the logarithms as proposed by Robert. I needed to calculate the factorial for genomic data, more specifically for the number of genes in the human genome i.e. about 30.000 and that is a big number :-) I didn't know gmpy Thanks a lot, really Ale -- http://mail.python.org/mailman/listinfo/python-list
Hypergeometric distribution
Hi to all, I need to calculate the hpergeometric distribution: choose(r, x) * choose(b, n-x) p(x; r,b,n) = - choose(r+b, n) choose(r,x) is the binomial coefficient I use the factorial to calculate the above formula but since I am using large numbers, the result of choose(a,b) (ie: the binomial coefficient) is too big even for large int. I've tried the scipy library, but this library calculates the hypergeometric using the factorials too, so the problem subsist. Is there any other libray or an algorithm to calculate the hypergeometric distribution? The statistical package R can handle such calculations but I don't want to use python R binding since I want a standalone app. Thanks a lot Ale -- http://mail.python.org/mailman/listinfo/python-list
