🌐 Mailman 3 December 2014 - Python-list - python.org

Re: converting PDF files is now an easy task
by David H. Lipman Dec. 26, 2014

Dec. 26, 2014

From: "joshuaemsteves" <joshuaemsteves(a)yahoo.com> > There is a familiar chance Pdf to Doc wasn't going to take off as long as > don't be an idiot and use this pdf to docx converter. I expect this was a > bad hypothesis, but you really have to open your mind. While Pdf to Doc is > used in those situations, it was shown that Pdf to Doc can provide more > reliable Pdf to Doc results as under any circumstances, this is simply a > rare Pdf to Doc. This way usually needs Pdf to Doc. Many punks may want to > know how to get a Pdf to Doc. > How is Python used in that conversion process ? -- Dave Multi-AV Scanning Tool - http://multi-av.thespykiller.co.uk http://www.pctipp.ch/downloads/dl/35905.asp

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Future of python on android
by Fetchinson . Dec. 26, 2014

Dec. 26, 2014

Hi all, I was using sl4a for quite some time on android and it basically worked very well although some features are missing. It looks like sl4a is dead, although I could be wrong. Does anyone knowledgeable have any further info on the future of sl4a? For instance it doesn't work with android 5 and there doesn't seem to be any activity around it for the past couple of years. If sl4a is out, what's the preferred way of running python on android? Guido is still working at google, right? This makes me suspect that python is ought to be part of the android ecosystem but can't find anything else beyond sl4a. So what's the future proof way of writing/deploying/installing python programs on android? Cheers, Daniel -- Psss, psss, put it down! - http://www.cafepress.com/putitdown

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Re: question on string object handling in Python 2.7.8
by Denis McMahon Dec. 25, 2014

Dec. 25, 2014

On Tue, 23 Dec 2014 20:28:30 -0500, Dave Tian wrote: > Hi, > > There are 2 statements: > A: a = ‘h’ > B: b = ‘hh’ > > According to me understanding, A should be faster as characters would > shortcut this 1-byte string ‘h’ without malloc; B should be slower than > A as characters does not work for 2-byte string ‘hh’, which triggers the > malloc. However, when I put A/B into a big loop and try to measure the > performance using cProfile, B seems always faster than A. > Testing code: > for i in range(0, 100000000): > a = ‘h’ #or b = ‘hh’ > Testing cmd: python -m cProfile test.py > > So what is wrong here? B has one more malloc than A but is faster than > B? Your understanding. The first time through the loop, python creates a string object "h" or "hh" and creates a pointer (a or b) and assigns it to the string object. The next range(1, 100000000) times through the loop, python re-assigns the existing pointer to the existing string object. Maybe a 2 character string is faster to locate in the object table than a 1 character string, so that in the 2 character case, the lookup is faster. -- Denis McMahon, denismfmcmahon(a)gmail.com

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Re: If One Line
by Rick Johnson Dec. 25, 2014

Dec. 25, 2014

On Thursday, December 25, 2014 10:16:54 AM UTC-6, Jacob Kruger wrote: > One line assignment is ok, but, seems like you can't perform actions. > > #the following will work: > I = 1 if True else 2 > > #but the following will generate an error: > if I == 1: print("one") Only if "I" is undefined.

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Re: learning to use iterators
by Rustom Mody Dec. 25, 2014

Dec. 25, 2014

On Wednesday, December 24, 2014 8:42:32 PM UTC+5:30, Vito De Tullio wrote: > Seb wrote: > > >>>> def n_grams(a, n): > > ... z = (islice(a, i, None) for i in range(n)) > > ... return zip(*z) > > ... > > > > I'm impressed at how succinctly this islice helps to build a list of > > tuples with indices for all the required windows. > > If you want it succinctly, there is this variation on the theme: > > n_grams = lambda a, n: zip(*(a[i:] for i in range(n))) +3 but -1 +1 for using an inlining the z +1 for showing that the * can take any expression [I did not know that] +1 for the slice in succinct form -1 for the lambda -- quite unnecessary and a red-herring If squeezing out the last char is the idea n_grams = lambda a, n: zip(*(a[i:] for i in range(n))) is same as def n_grams(a,n):return zip(*(a[i:]for i in range(n)))

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converting PDF files is now an easy task
by joshuaemsteves Dec. 25, 2014

Dec. 25, 2014

There is a familiar chance Pdf to Doc wasn't going to take off as long as don't be an idiot and use this pdf to docx converter. I expect this was a bad hypothesis, but you really have to open your mind. While Pdf to Doc is used in those situations, it was shown that Pdf to Doc can provide more reliable Pdf to Doc results as under any circumstances, this is simply a rare Pdf to Doc. This way usually needs Pdf to Doc. Many punks may want to know how to get a Pdf to Doc. http://sourceforge.net/projects/pdftodoc -- View this message in context: http://python.6.x6.nabble.com/converting-PDF-files-is-now-an-easy-task-tp50… Sent from the Python - python-list mailing list archive at Nabble.com.

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Re: question on string object handling in Python 2.7.8
by Gregory Ewing Dec. 25, 2014

Dec. 25, 2014

Dave Tian wrote: > A: a = ‘h’ > B: b = ‘hh’ > > According to me understanding, A should be faster as characters would > shortcut this 1-byte string ‘h’ without malloc; It sounds like you're expecting characters to be stored "unboxed" like in Java. That's not the way Python works. Objects are used for everything, including numbers and characters (there is no separate character type in Python, they're just length-1 strings). > for i in range(0, 100000000): > a = ‘h’ #or b = ‘hh’ > Testing cmd: python -m cProfile test.py Since you're assigning a string literal, there's just one string object being allocated (at the time the code is read in and compiled). All the loop is doing is repeatedly assigning a reference to that object to a or b, which doesn't require any further mallocs; all it does is adjust reference counts. This will be swamped by the overhead of the for-loop itself, which is allocating and deallocating 100 million integer objects. I would expect both of these to be exactly the same speed, within measurement error. Any difference you're seeing is probably just noise, or the result of some kind of warming-up effect. -- Greg

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is pathlib Path.resolve working as intended?
by Chris Cioffi Dec. 25, 2014

Dec. 25, 2014

Ok, I'm guessing I'm doing something wrong, but I can't see what. I'm playing around with pathlib (Python 3.4.2) on Mac OSX, Yosemite. In the past I've used os.path.expanduser() to expand paths with ~. Based on the description, I would have expected .resolve to do that automatically, but it doesn't seem to work. Is this a bug, oversight or design choice? Here's my example of what I was doing: >>> p = Path('~/.profile') >>> p PosixPath('~/.profile') >>> p.resolve <bound method PosixPath.resolve of PosixPath('~/.profile')> >>> >>> p.resolve() Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/usr/local/Cellar/python3/3.4.2_1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/pathlib.py", line 1031, in resolve s = self._flavour.resolve(self) File "/usr/local/Cellar/python3/3.4.2_1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/pathlib.py", line 297, in resolve return _resolve(base, str(path)) or sep File "/usr/local/Cellar/python3/3.4.2_1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/pathlib.py", line 282, in _resolve target = accessor.readlink(newpath) File "/usr/local/Cellar/python3/3.4.2_1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/pathlib.py", line 374, in readlink return os.readlink(path) FileNotFoundError: [Errno 2] No such file or directory: '/Users/chris/~' >>> Chris

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missing os.lchmod, os.lchflags
by Ethan Furman Dec. 25, 2014

Dec. 25, 2014

According to the docs [1] these functions should be available as of 2.6, yet they are missing on a 2.7, 3.2, and 3.4 install (ubuntu 12.10 and 14.04) Any ideas why? -- ~Ethan~ [1] https://docs.python.org/2/library/os.html#os.lchmod

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Re: question on string object handling in Python 2.7.8
by Steven D'Aprano Dec. 24, 2014

Dec. 24, 2014

Dave Tian wrote: > Hi, > > There are 2 statements: > A: a = ‘h’ > B: b = ‘hh’ > > According to me understanding, A should be faster as characters would > shortcut this 1-byte string ‘h’ without malloc; B should be slower than A > as characters does not work for 2-byte string ‘hh’, which triggers the > malloc. However, when I put A/B into a big loop and try to measure the > performance using cProfile, B seems always faster than A. > > Testing code: > for i in range(0, 100000000): a = ‘h’ #or b = ‘hh’ > Testing cmd: python -m cProfile test.py Any performance difference is entirely an artifact of your testing method. You have completely misinterpreted what this piece of code will do. What happens here is that you time a piece of code to: - Build a large list containing 100 million individual int objects. Each int object has to be allocated at run time, as does the list. Each int object is about 12 bytes in size. - Then, the name i is bound to one of those int objects. This is a fast pointer assignment. - Then, a string object containing either 'h' or 'hh' is allocated. In either case, that requires 21 bytes, plus one byte per character. So either 22 or 23 bytes. - The name a is bound to that string object. This is also a fast pointer assignment. - The loop returns to the top, and the name i is bound to the next int object. - Then, the name a is bound *to the same string object*, since it will have been cached. No further malloc will be needed. So as you can see, the time you measure is dominated by allocating a massive list containing 100 million int objects. Only a single string object is allocated, and the time difference between creating 'h' versus 'hh' is insignificant. The byte code can be inspected like this: py> code = compile("for i in range(100000000): a = 'h'", '', 'exec') py> from dis import dis py> dis(code) 1 0 SETUP_LOOP 26 (to 29) 3 LOAD_NAME 0 (range) 6 LOAD_CONST 0 (100000000) 9 CALL_FUNCTION 1 12 GET_ITER >> 13 FOR_ITER 12 (to 28) 16 STORE_NAME 1 (i) 19 LOAD_CONST 1 ('h') 22 STORE_NAME 2 (a) 25 JUMP_ABSOLUTE 13 >> 28 POP_BLOCK >> 29 LOAD_CONST 2 (None) 32 RETURN_VALUE Notice instruction 19 and 22: 19 LOAD_CONST 1 ('h') 22 STORE_NAME 2 (a) The string object is built at compile time, not run time, and Python simply binds the name a to the pre-existing string object. If you looked at the output of your timing code, you would see something like this (only the times would be much larger, I cut the loop down from 100 million to only a few tens of thousands): [steve@ando ~]$ python -m cProfile /tmp/x.py 3 function calls in 0.062 seconds Ordered by: standard name ncalls tottime percall cumtime percall filename:lineno(function) 1 0.051 0.051 0.062 0.062 x.py:1(<module>) 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects} 1 0.012 0.012 0.012 0.012 {range} The profiler doesn't even show the time required to bind the name to the string object. Here is a better way of demonstrating the same thing: py> from timeit import Timer py> t = Timer("a = 'h'") py> min(t.repeat()) 0.0508120059967041 py> t = Timer("a = 'hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh'") py> min(t.repeat()) 0.050585031509399414 No meaningful difference in time. What difference you do see is a fluke of timing. -- Steven

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