Re: Performance of int/long in Python 3
On Mon, Apr 1, 2013 at 4:33 PM, rusi wrote: > So I really wonder: Is python losing more by supporting SMP with > performance hit on BMP? If your strings fit entirely within the BMP, then you should see no penalty compared to previous versions of Python. If they happen to fit inside ASCII, then there may well be significant improvements. But regardless, what you gain is the ability to work with *any* string, regardless of its content, without worrying about it. You can count characters regardless of their content. Imagine if a tuple of integers behaved differently if some of those integers flipped to being long ints: x = (1, 2, 4, 8, 1<<30, 1<<300, 1<<10) Wouldn't you be surprised if len(x) returned 8? I certainly would be. And that's what a narrow build of Python does with Unicode. Unicode strings are approximately comparable to tuples of integers. In fact, they can be interchanged fairly readily: string = "Treble clef: \U0001D11E" array = tuple(map(ord,string)) assert(len(array) == 14) out_string = ''.join(map(chr,array)) assert(out_string == string) This doesn't work in Python 2.6 on Windows, partly because of surrogates, but also because chr() isn't designed for Unicode strings. There's probably a solution to the second, but not really to the first. The tuple of ords should match the way the characters are laid out to a human. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Performance of int/long in Python 3
On Sun, Mar 31, 2013 at 11:33 PM, rusi wrote: > > So I really wonder: Is python losing more by supporting SMP with > performance hit on BMP? I don't believe so. Although performance is undeniably worse for some benchmarks, it is also better for some others. Nobody has yet demonstrated an actual, real-world program that is affected negatively by the Unicode change. All of jmf's complaints amount to cherry-picking data and leaping to conclusions. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help please
On Monday, April 1, 2013 1:24:52 AM UTC-4, Chris Angelico wrote:
> On Mon, Apr 1, 2013 at 4:15 PM, wrote:
>
> > integer = input("Enter a positive integer: ")
>
> > again = raw_input("Again? (Y/N): ")
>
>
>
> Okay, the first thing I'm going to say is: Don't use input() in Python
>
> 2. It's dangerous in ways you won't realize. Use int(raw_input(...))
>
> for something like this, which will guarantee you an integer.
>
>
>
> I'm guessing this is homework. Please be honest about that; we'll help
>
> you learn but won't just give you the answers.
>
>
>
> All you need to do is initialize something to zero, and then keep
>
> adding 'number' onto it in the loop. You should be able to sort that
>
> out.
>
>
>
> ChrisA
Elaborate, please.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Performance of int/long in Python 3
On Mar 31, 5:55 pm, Mark Lawrence wrote: > I'm feeling very sorry for this horse, it's been flogged so often it's > down to bare bones. While I am now joining the camp of those fed up with jmf's whining, I do wonder if we are shooting the messenger… >From a recent Roy mysqldb-unicode thread: > My unicode-fu is a bit weak. Are we looking at a Python problem, a > MySQLdb problem, or a problem with the underlying MySQL server? We've > certainly inserted utf-8 data before without any problems. It's > possible this is the first time we've tried to handle a character > outside the BMP. : : > OK, that leads to the next question. Is there anyway I can (in Python > 2.7) detect when a string is not entirely in the BMP? If I could find > all the non-BMP characters, I could replace them with U+FFFD > (REPLACEMENT CHARACTER) and life would be good (enough). Steven's: > But it means that if you're one of the 99.9% of users who mostly use > characters in the BMP, … And from http://www.tlg.uci.edu/~opoudjis/unicode/unicode_astral.html > The informal name for the supplementary planes of Unicode is "astral planes", > since > (especially in the late '90s) their use seemed to be as remote as > the theosophical "great beyond". … > As of this writing for instance, Dreamweaver MX for MacOSX (which I am > currently using > to prepare this) will let you paste BMP text into its WYSIWYG window; but > pasting > Supplementary Plane text there will make it crash. So I really wonder: Is python losing more by supporting SMP with performance hit on BMP? The problem as I see it is that a choice that is sufficiently skew is no more a straightforward choice. An example will illustrate: I can choose to drive or not -- a choice. Statistics tell me that on average there are 3 fatalities every day; I am very concerned that I could get killed so I choose not to drive. Which neglects that there are a couple of million safe-drives at the same time as the '3 fatalities' [What if anything this has to do with jmf's rants I dont know because I dont know if anyone (including jmf) knows what he is ranting about. ] -- http://mail.python.org/mailman/listinfo/python-list
Re: Help please
On Mon, Apr 1, 2013 at 4:15 PM, wrote:
> integer = input("Enter a positive integer: ")
> again = raw_input("Again? (Y/N): ")
Okay, the first thing I'm going to say is: Don't use input() in Python
2. It's dangerous in ways you won't realize. Use int(raw_input(...))
for something like this, which will guarantee you an integer.
I'm guessing this is homework. Please be honest about that; we'll help
you learn but won't just give you the answers.
All you need to do is initialize something to zero, and then keep
adding 'number' onto it in the loop. You should be able to sort that
out.
ChrisA
--
http://mail.python.org/mailman/listinfo/python-list
Help please
I want to add up the integers of this code in one line. For example, if I had
the code
integer = 0
denom = 10
again = "y" #sentinel:
while again == "y" or again == "Y":
integer = input("Enter a positive integer: ")
while denom <= integer:
denom = denom*10
while denom > 1:
denom = denom/10
number = integer/denom
integer = integer%denom
print str(number)
again = raw_input("Again? (Y/N): ")
and inputted "54321," it would look like:
Enter a positive integer: 54321
5
4
3
2
1
Again? (Y/N): n
What I want to do is add up the "54321" so it comes out with
"Sum of digits is 15." on one line.
Thanks!
--
http://mail.python.org/mailman/listinfo/python-list
executor.map() TypeError: zip argument #2 must support iteration
executor.map()TypeError: zip argument #2 must support iteration when I run it ,just generated TypeError: zip argument #2 must support iteration. can anyone help me fix this problem ? import time, concurrent.futures lst100=[i for i in range(100)] t1=time.clock() print(list(map(str,lst100))) t2=time.clock() print(t2-t1) t3=time.clock() with concurrent.futures.ThreadPoolExecutor(max_workers=100) as executor: future_to_url = executor.map(str,lst100, 60) print(list(future_to_url)) t4=time.clock() print(t4-t3) -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On Mon, 01 Apr 2013 00:39:56 +, Alex wrote: > Chris Angelico wrote: > > >> Opening paragraph, "... exponentiation, which groups from right to >> left". It follows the obvious expectation from mathematics. (The OP is >> using Python 2, but the same applies.) > > Thanks. I did miss that parenthetical comment in para 6.15, and that > would have been the correct place to look, since it appears that > operators are not parts of expressions, but rather separate them. Is > that the "obvious expectation from mathematics," though? Given that > > 3 > 5 > 4 > > (i.e.: 4**5**3) is transitive, I would have expected Python to exhibit > more consistency with the other operators. I guess that is one of the > foolish consistencies that comprise the hobgoblins of my little mind, > though. I don't think you mean "transitive" here. Transitivity refers to relations, not arbitrary operators. If ≎ is some relation, then it is transitive if and only if: x ≎ y and y ≎ z implies that x ≎ y. http://en.wikipedia.org/wiki/Transitive_relation Concrete examples of transitive relations: greater than, equal to, less than and equal to. On the other hand, "unequal to" is not a transitive relation. Nor is "approximately equal to". Suppose we say that two values are approximately equal if their difference is less than 0.5: 2.1 ≈ 2.4 and 2.4 ≈ 2.7 but 2.1 ≉ 2.7 Exponentiation is not commutative: 2**3 != 3**2 nor is it associative: 2**(3**2) != (2**3)**2 so I'm not really sure what you are trying to say here. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Sudoku
Sorry. Won't happen again. signing off this topic. Eric. -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On Mon, Apr 1, 2013 at 11:39 AM, Alex wrote: > Given that > > 3 > 5 > 4 > > (i.e.: 4**5**3) is transitive, I would have expected Python to exhibit > more consistency with the other operators. I guess that is one of the > foolish consistencies that comprise the hobgoblins of my little mind, > though. Not sure what you mean by transitive here. Certainly (4**5)**3 and 4**(5**3) are not the same value. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
Chris Angelico wrote: > > Opening paragraph, "... exponentiation, which groups from right to > left". It follows the obvious expectation from mathematics. (The OP is > using Python 2, but the same applies.) Thanks. I did miss that parenthetical comment in para 6.15, and that would have been the correct place to look, since it appears that operators are not parts of expressions, but rather separate them. Is that the "obvious expectation from mathematics," though? Given that 3 5 4 (i.e.: 4**5**3) is transitive, I would have expected Python to exhibit more consistency with the other operators. I guess that is one of the foolish consistencies that comprise the hobgoblins of my little mind, though. -- http://mail.python.org/mailman/listinfo/python-list
Re: collections.Iterator __subclasshook__ does not check if next() is callable
On 03/31/2013 10:57 AM, Byron Ruth wrote: I submitted this as bug last night: http://bugs.python.org/issue17584 and was *honored* to be rejected by Raymond Hettinger. However, I would like feedback on whether my concern (this bug) is justified and clarity if not. Consider: ```python class A(object): def __init__(self): self.r = iter(range(5)) def __iter__(self): return self @property def next(self): return next(self.r) ``` The `next` method is a property, however: ```python from collections import Iterator a = A() isinstance(a, Iterator) # True next(a) # TypeError: 'int' object is not callable ``` I am using `collections.Iterator` as the means to check if the object is an iterator, however I am not sure if that is _root_ problem here. My understanding of the iterator protocol is that is assumes the __iter__ and next *methods* are implemented. In the example, `A.next` is defined as a property, but is still identified as an iterator. To me, this is incorrect behavior since it's not conforming to the iterator protocol requirements (i.e. a `next` method, not a property). The root problem is that whoever implemented A did so incorrectly. Raymond stated: "The design of ABCs are to check for the existence to required named; none of them verify the signature." I think I understand _why_ this is the case.. but I downstream libraries use `collections.Iterator` to determine if an object _is one_: see https://github.com/django/django/blob/master/django/utils/itercompat.py#L22-L31 Who's job is it to check if `next` (and technically `__iter__`) are methods? The programmer who wrote it, with good tests. Technically, the issue is not that 'next' is a property, but that it is not returning a callable. If it did (and the callable was appropriate for the iterator), that would also work. -- ~Ethan~ -- http://mail.python.org/mailman/listinfo/python-list
Re: Sudoku
On 31 March 2013 23:34, Dave Angel wrote: >[...] With my Python > 2.7.2, exit(something) with something being a string prints the string and > then exits. Nowhere have I seen that documented, and I thought it either > took an int or nothing. It is documented, just not exactly where you'd expect it (exit is a constant, not a function): http://docs.python.org/2/library/constants.html#constants-added-by-the-site-module As for the behaviour when you pass a string, it's documented here: http://docs.python.org/2/library/exceptions.html#exceptions.SystemExit -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: Sudoku
On Mon, Apr 1, 2013 at 9:27 AM, Eric Parry wrote: > [ chomp 128 lines of quoted text ] > > I tried all those things. The program keeps running after the solution in > every case. Never mind. It won't do that in VBA when I finish it. > Eric. You have just spammed us with, and I counted them, one hundred and twenty-eight lines of quoted text. And you posted to both comp.lang.python and python-list, so some of us will have seen those lines twice. Please don't do this. If you MUST use Google Groups, check out this page, and try to adjust your posts to be more courteous. http://wiki.python.org/moin/GoogleGroupsPython Thanks! ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Sudoku
On 03/31/2013 06:03 PM, Eric Parry wrote: I think in the original it was exit(a). That did not work either. There you go again. "Did not work" tells us very little. With my Python 2.7.2, exit(something) with something being a string prints the string and then exits. Nowhere have I seen that documented, and I thought it either took an int or nothing. So please be much more specific. Did it give some exception, or did it not print the right result (I don't see any print statement in the code), or what? -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On 03/31/2013 06:06 PM, Alex wrote: Dave Angel wrote: On 03/31/2013 02:56 AM, morphex wrote: 1**2 1 1**2**3 1 1**2**3**4 1L 1**2**3**4**5 Traceback (most recent call last): File "", line 1, in MemoryError Does anyone know why this raises a MemoryError? Doesn't make sense to me. Perhaps you didn't realize that the expression will be done from right to left. Really? The Python 3 documentation (http://docs.python.org/3/reference/expressions.html) says in section 6.14 (Evaluation order) that "Python evaluates expressions from left to right" (an exception being when evaluating assignments, in which case the RHS of the assignment is calculated first, in left-to-right order). Section 6.4 discusses the power operator specifically and does not contradict 6.14 except that the power operator uses right-to-left evaluation in the presence of unparenthesized unary operators. Neither of these two exception cases appear to apply here, so I think the OP is reasonable in expecting Python to do the operation left-to-right. Am I missing something written somewhere else in the docs? Are the docs I quoted wrong? Please help me understand the discrepancy I am perceiving here. Alex On the page you reference, in section 6.14, see the 4th entry: expr1 + expr2 * (expr3 - expr4) expr1 is evaluated before expr2, but the multiply happens before the add. Now see the following paragraph (6.15): """ The following table summarizes the operator precedences in Python, from lowest precedence (least binding) to highest precedence (most binding). Operators in the same box have the same precedence. Unless the syntax is explicitly given, operators are binary. Operators in the same box group left to right (except for comparisons, including tests, which all have the same precedence and chain from left to right — see section Comparisons — and exponentiation, which groups from right to left). """ What this paragraph refers to as "grouping" is commonly called associativity in other languages. There are three different concepts here that apply whenever an expression has more than one term: 1) evaluation order is important for terms lie function calls which have side effects 2) precedence, lie where multiply will happen before add 3) associativity, where the order of operations for operators with the same precedence are done either left to right (usually), or right to left (like exponentiation) -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On Mon, Apr 1, 2013 at 9:28 AM, Chris Angelico wrote: > On Mon, Apr 1, 2013 at 9:06 AM, Alex wrote: >> Really? >> >> The Python 3 documentation >> (http://docs.python.org/3/reference/expressions.html) says in section >> 6.14 (Evaluation order) that "Python evaluates expressions from left to >> right" (an exception being when evaluating assignments, in which case >> the RHS of the assignment is calculated first, in left-to-right order). >> >> Section 6.4 discusses the power operator specifically and does not >> contradict 6.14 except that the power operator uses right-to-left >> evaluation in the presence of unparenthesized unary operators. > > http://docs.python.org/3/reference/expressions.html#operator-precedence > > Opening paragraph, "... exponentiation, which groups from right to > left". It follows the obvious expectation from mathematics. (The OP is > using Python 2, but the same applies.) Though your point about 6.14 is still true. It states the order that the integers will be evaluated in. Note one of the examples given: expr1 + expr2 * (expr3 - expr4) The evaluation of the expression starts with 3 and 4, then picks up 2, then 1, but the operands themselves are evaluated left to right. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On Mon, Apr 1, 2013 at 9:06 AM, Alex wrote: > Dave Angel wrote: > >> On 03/31/2013 02:56 AM, morphex wrote: >> > > > > 1**2 >> > 1 >> > > > > 1**2**3 >> > 1 >> > > > > 1**2**3**4 >> > 1L >> > > > > 1**2**3**4**5 >> > Traceback (most recent call last): >> > File "", line 1, in >> > MemoryError >> > > > > >> > >> > Does anyone know why this raises a MemoryError? Doesn't make sense >> > to me. >> >> Perhaps you didn't realize that the expression will be done from >> right to left. > > Really? > > The Python 3 documentation > (http://docs.python.org/3/reference/expressions.html) says in section > 6.14 (Evaluation order) that "Python evaluates expressions from left to > right" (an exception being when evaluating assignments, in which case > the RHS of the assignment is calculated first, in left-to-right order). > > Section 6.4 discusses the power operator specifically and does not > contradict 6.14 except that the power operator uses right-to-left > evaluation in the presence of unparenthesized unary operators. http://docs.python.org/3/reference/expressions.html#operator-precedence Opening paragraph, "... exponentiation, which groups from right to left". It follows the obvious expectation from mathematics. (The OP is using Python 2, but the same applies.) ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Sudoku
On Monday, April 1, 2013 8:33:47 AM UTC+10:30, Eric Parry wrote: > On Sunday, March 31, 2013 9:45:36 AM UTC+10:30, Dave Angel wrote: > > > On 03/30/2013 06:06 PM, Eric Parry wrote: > > > > > > > On Saturday, March 30, 2013 8:41:08 AM UTC+10:30, Dave Angel wrote: > > > > > > >> On 03/29/2013 05:47 PM, Eric Parry wrote: > > > > > > >> > > > > > > >>> > > > > > > >> > > > > > > >> Sometimes a bug in such a function will cause it to run indefinitely, > > > > > > >> > > > > > > >> and/or to overflow the stack. I don't see such a bug in this function. > > > > > > >> > > > > > > >> > > > > > > >> > > > > > > >> > > > > > > >> > > > > > > >> -- > > > > > > >> > > > > > > >> DaveA > > > > > > > > > > > > > > The exit() did not work. > > > > > > > > > > > > Would you like to elaborate? exit() is supposed to take an int > > > > > > parameter, but the author apparently didn't notice that. So perhaps you > > > > > > got an exception of some sort. Change it to exit() or exit(0) and it > > > > > > might solve the problem, depending on what the problem was. > > > > > > > > > > > > > I replaced it with return = 0, and that does work. > > > > > > > > > > > > No it doesn't. return = 0 is a syntax error in both Python 2.x and 3.x > > > > > > > > > > > > But if you changed it to a valid return statement, then that's why it > > > > > > doesn't stop on the first solution. > > > > > > > > > > > > > > > > > > -- > > > > > > DaveA > > > > I think in the original it was exit(a). That did not work either. > > I'll try the others. > > Eric. I tried all those things. The program keeps running after the solution in every case. Never mind. It won't do that in VBA when I finish it. Eric. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Mon, Apr 1, 2013 at 9:02 AM, Mark Lawrence wrote: > On 31/03/2013 22:21, Chris Angelico wrote: >>> >>>sue = time.mktime( >>> (int(m.group(7)), int(months[m.group(2)]), int(m.group(3)), >>>int(m.group(4)), int(m.group(5)), int(m.group(6)), >>>int(days[m.group(1)]), 0, 0) >>> ) >>> expire_time = (sue current_time)/60/60/24 >> >> >> Here's a likely problem. There's supposed to be an operator - probably >> a plus sign - between sue and current_time. >> > > There is actually a minus sign there which showed up when I pasted the code > into the Eclipse/Pydev editor. That sadly doesn't fix the problem with the > call to mktime, 12 open round brackets to 14 close if I've counted > correctly. Oh, of course, since sue comes from mktime. But yeah, definitely needs an operator there. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
Dave Angel wrote: > On 03/31/2013 02:56 AM, morphex wrote: > > > > > 1**2 > > 1 > > > > > 1**2**3 > > 1 > > > > > 1**2**3**4 > > 1L > > > > > 1**2**3**4**5 > > Traceback (most recent call last): > > File "", line 1, in > > MemoryError > > > > > > > > > Does anyone know why this raises a MemoryError? Doesn't make sense > > to me. > > Perhaps you didn't realize that the expression will be done from > right to left. Really? The Python 3 documentation (http://docs.python.org/3/reference/expressions.html) says in section 6.14 (Evaluation order) that "Python evaluates expressions from left to right" (an exception being when evaluating assignments, in which case the RHS of the assignment is calculated first, in left-to-right order). Section 6.4 discusses the power operator specifically and does not contradict 6.14 except that the power operator uses right-to-left evaluation in the presence of unparenthesized unary operators. Neither of these two exception cases appear to apply here, so I think the OP is reasonable in expecting Python to do the operation left-to-right. Am I missing something written somewhere else in the docs? Are the docs I quoted wrong? Please help me understand the discrepancy I am perceiving here. Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Sudoku
On Sunday, March 31, 2013 9:45:36 AM UTC+10:30, Dave Angel wrote: > On 03/30/2013 06:06 PM, Eric Parry wrote: > > > On Saturday, March 30, 2013 8:41:08 AM UTC+10:30, Dave Angel wrote: > > >> On 03/29/2013 05:47 PM, Eric Parry wrote: > > >> > > >>> > > >> > > >> Sometimes a bug in such a function will cause it to run indefinitely, > > >> > > >> and/or to overflow the stack. I don't see such a bug in this function. > > >> > > >> > > >> > > >> > > >> > > >> -- > > >> > > >> DaveA > > > > > > The exit() did not work. > > > > Would you like to elaborate? exit() is supposed to take an int > > parameter, but the author apparently didn't notice that. So perhaps you > > got an exception of some sort. Change it to exit() or exit(0) and it > > might solve the problem, depending on what the problem was. > > > > > I replaced it with return = 0, and that does work. > > > > No it doesn't. return = 0 is a syntax error in both Python 2.x and 3.x > > > > But if you changed it to a valid return statement, then that's why it > > doesn't stop on the first solution. > > > > > > -- > > DaveA I think in the original it was exit(a). That did not work either. I'll try the others. Eric. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On 31/03/2013 22:21, Chris Angelico wrote: sue = time.mktime( (int(m.group(7)), int(months[m.group(2)]), int(m.group(3)), int(m.group(4)), int(m.group(5)), int(m.group(6)), int(days[m.group(1)]), 0, 0) ) expire_time = (sue current_time)/60/60/24 Here's a likely problem. There's supposed to be an operator - probably a plus sign - between sue and current_time. There is actually a minus sign there which showed up when I pasted the code into the Eclipse/Pydev editor. That sadly doesn't fix the problem with the call to mktime, 12 open round brackets to 14 close if I've counted correctly. -- If you're using GoogleCrap™ please read this http://wiki.python.org/moin/GoogleGroupsPython. Mark Lawrence -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Sunday, March 31, 2013 3:27:06 PM UTC-6, Roy Smith wrote: > If this is for an interview, you really should be doing this on your > own. I assume the point of the interview is to see how well you know > Python. Please don't expect people here to take your interview for you. Maybe the interviewer gives higher ratings to someone who knows how to take advantage of all available resources than someone who goes off in a corner and tries to do it alone? -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Sunday, March 31, 2013 5:35:38 PM UTC-4, Chris Angelico wrote: > On Mon, Apr 1, 2013 at 8:21 AM, jojo wrote: > > > Thanks for your replies. Just to be clear this is for a interview and they > > would like me to figure out what the code does and come back with some test > > cases > > > > That explains the utter lack of comments, then. In well-maintained > > code, you would simply read through the comments to get an idea of > > what it does. > > > > A couple of key things to look up: glob.glob and os.popen. When you > > know what they do, you should be able to get a broad understanding of > > the whole program. > > > > ChrisA OK perfect Chris. Thanks for your reply. -- http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On 03/31/2013 02:41 PM, Roy Smith wrote:
In article ,
Dave Angel wrote:
On 03/31/2013 12:52 PM, C.T. wrote:
On Sunday, March 31, 2013 12:20:25 PM UTC-4, zipher wrote:
Thank you, Mark! My problem is the data isn't consistently ordered. I can
use slicing and indexing to put the year into a tuple, but because a car
manufacturer could have two names (ie, Aston Martin) or a car model could
have two names(ie, Iron Duke), its harder to use slicing and indexing for
those two. I've added the following, but the output is still not what I
need it to be.
So the correct answer is "it cannot be done," and an explanation.
Many times I've been given impossible conditions for a problem. And
invariably the correct solution is to press [back] on the supplier of the
constraints.
In real life, you often have to deal with crappy input data (and bogus
project requirements). Sometimes you just need to be creative.
There's only a small set of car manufacturers. A good start would be
mining wikipedia's [[List of automobile manufacturers]]. Once you've
got that list, you could try matching portions of the input against the
list.
Depending on how much effort you wanted to put into this, you could
explore all sorts of fuzzy matching (ie "delorean" vs "delorean motor
company"), but even a simple search is better than giving up.
And, this is a good excuse to explore some of the interesting
third-party modules. For example, mwclient ("pip install mwclient")
gives you a neat Python interface to wikipedia. And there's a whole
landscape of string matching packages to explore.
We deal with this every day at Songza. Are Kesha and Ke$ha the same
artist? Pushing back on the record labels to clean up their catalogs
isn't going to get us very far.
I agree with everything you've said, although in your case, presumably
the record labels are not your client/boss, so that's not who you push
back against. The client should know when the data is being fudged, and
have a say in how it's to be done.
But this is a homework assignment. I think the OP is learning Python,
not how to second-guess a client.
--
DaveA
--
http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Sunday, March 31, 2013 5:27:06 PM UTC-4, Roy Smith wrote: > In article <4455829d-5b4a-44ee-b65f-5f72d429b...@googlegroups.com>, > > jojo wrote: > > > > > Thanks for your replies. Just to be clear this is for a interview and they > > > would like me to figure out what the code does and come back with some test > > > cases. I don't need to code the tests, just give some high level tests. As > > > far as I can make out it is some system where you input your name and it > > will > > > bring back your details plus how much time you have left on your card. Have > > > to say I find the code extremely confusing, hopefully all python isn't like > > > this!! > > > > If this is for an interview, you really should be doing this on your > > own. I assume the point of the interview is to see how well you know > > Python. Please don't expect people here to take your interview for you. Where did I ask people to take the interview for me? I asked for some tips on interpreting the code something which you have been unable to give me. If a senior dev (I am assuming you are a python dev) is unable to figure out what the code does, then I think me, been a newbie to new language (with horrible syntax) is entitled to ask for help. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Mon, Apr 1, 2013 at 8:21 AM, jojo wrote: > Thanks for your replies. Just to be clear this is for a interview and they > would like me to figure out what the code does and come back with some test > cases That explains the utter lack of comments, then. In well-maintained code, you would simply read through the comments to get an idea of what it does. A couple of key things to look up: glob.glob and os.popen. When you know what they do, you should be able to get a broad understanding of the whole program. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
In article <4455829d-5b4a-44ee-b65f-5f72d429b...@googlegroups.com>, jojo wrote: > Thanks for your replies. Just to be clear this is for a interview and they > would like me to figure out what the code does and come back with some test > cases. I don't need to code the tests, just give some high level tests. As > far as I can make out it is some system where you input your name and it will > bring back your details plus how much time you have left on your card. Have > to say I find the code extremely confusing, hopefully all python isn't like > this!! If this is for an interview, you really should be doing this on your own. I assume the point of the interview is to see how well you know Python. Please don't expect people here to take your interview for you. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Sunday, March 31, 2013 5:13:49 PM UTC-4, Roy Smith wrote: > In article <2912c674-e30b-4339-9344-1f460cb96...@googlegroups.com>, > > jojo wrote: > > > > > for fname in dirList: > > > cmd = "keytool �printcert �file " + fname > > > for line in os.popen(cmd).readlines(): > > >line = line.rstrip() > > >m = p.search(line) > > >if m: > > > sue = time.mktime( > > > (int(m.group(7)), int(months[m.group(2)]), int(m.group(3)), > > > int(m.group(4)), int(m.group(5)), int(m.group(6)), > > > int(days[m.group(1)]), 0, 0) > > > ) > > > expire_time = (sue � current_time)/60/60/24 > > > if expire_time < 0: > > > print cert_name + " has already expired!" > > > elif expire_time < 31: > > > print cert_name + " expires in " +str(int(expire_time)) + " days" > > >else: > > > m = q.search(line) > > > if m: > > > cert_name = m.group(1) > > > > Was this code really indented like this when you got it? You've got (at > > least) three different indent sizes. I see 1, 2, and 3 space indents in > > different places in the code. > > > > I'm not even sure if this is legal, but even if it is, it's really bad > > form. Pick an indent, and stick with it uniformly. 4 spaces seems to > > be pretty standard. > > > > That being said, I'm going to return to my previous statement that until > > you know what the code is *supposed* to do, trying to test it is > > meaningless. Hi Rob. Thanks for your replies. Just to be clear this is for a interview and they would like me to figure out what the code does and come back with some test cases. I don't need to code the tests, just give some high level tests. As far as I can make out it is some system where you input your name and it will bring back your details plus how much time you have left on your card. Have to say I find the code extremely confusing, hopefully all python isn't like this!! -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Sunday, March 31, 2013 5:21:00 PM UTC-4, Chris Angelico wrote: > On Mon, Apr 1, 2013 at 8:06 AM, jojo wrote: > > > On Sunday, March 31, 2013 4:39:11 PM UTC-4, Chris Angelico wrote: > > >> On Mon, Apr 1, 2013 at 7:10 AM, jojo wrote: > > >> > > >> > Im used to C# so the syntax looks bizarre to me! Any help would be great. > > >> > > >> > > >> > > >> The first thing you'll need to understand about Python syntax is that > > >> > > >> indentation is important. By posting this code flush-left, you've > > >> > > >> actually destroyed its block structure. Could you post it again, with > > >> > > >> indentation, please? We'd then be in a much better position to help. > > >> > > >> > > >> > > >> Chris Angelico > > > > > > > > > Hi Chris, thanks for your reply. See code below... > > > > Ah, you appear to be posting from Google Groups. You may want to check > > this page out, as a lot of people rather dislike GG posts. > > > > http://wiki.python.org/moin/GoogleGroupsPython > > > > The best method is simply to avoid Google Groups altogether. > > > > Anyway, some code comments. (Though the biggest comment to make about > > the code is its utter lack of comments. Not a good idea in any > > language, for anything more than the most trivial script.) > > > > > current_time = time.time() + 60*60+24*30 > > > > This line doesn't, quite frankly, make a lot of sense; time.time() > > returns the current time already, but then an offset of one hour and > > twelve minutes is added. > > > > >if m: > > > sue = time.mktime( > > > (int(m.group(7)), int(months[m.group(2)]), int(m.group(3)), > > > int(m.group(4)), int(m.group(5)), int(m.group(6)), > > > int(days[m.group(1)]), 0, 0) > > > ) > > > expire_time = (sue current_time)/60/60/24 > > > > Here's a likely problem. There's supposed to be an operator - probably > > a plus sign - between sue and current_time. > > > > >else: > > > m = q.search(line) > > > if m: > > > cert_name = m.group(1) > > > > And this last line needs indentation. > > > > The very easiest way to debug Python code is to run it. If it runs, > > great! See what output it made and whether it's correct or not. If it > > doesn't, Python will give you an exception traceback that points you > > to the failing line. Get familiar with them, as you'll be seeing them > > a lot :) > > > > Chris Angelico Ok, thanks Chris!! -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Mon, Apr 1, 2013 at 8:06 AM, jojo wrote: > On Sunday, March 31, 2013 4:39:11 PM UTC-4, Chris Angelico wrote: >> On Mon, Apr 1, 2013 at 7:10 AM, jojo wrote: >> >> > Im used to C# so the syntax looks bizarre to me! Any help would be great. >> >> >> >> The first thing you'll need to understand about Python syntax is that >> >> indentation is important. By posting this code flush-left, you've >> >> actually destroyed its block structure. Could you post it again, with >> >> indentation, please? We'd then be in a much better position to help. >> >> >> >> Chris Angelico > > > Hi Chris, thanks for your reply. See code below... Ah, you appear to be posting from Google Groups. You may want to check this page out, as a lot of people rather dislike GG posts. http://wiki.python.org/moin/GoogleGroupsPython The best method is simply to avoid Google Groups altogether. Anyway, some code comments. (Though the biggest comment to make about the code is its utter lack of comments. Not a good idea in any language, for anything more than the most trivial script.) > current_time = time.time() + 60*60+24*30 This line doesn't, quite frankly, make a lot of sense; time.time() returns the current time already, but then an offset of one hour and twelve minutes is added. >if m: > sue = time.mktime( > (int(m.group(7)), int(months[m.group(2)]), int(m.group(3)), > int(m.group(4)), int(m.group(5)), int(m.group(6)), > int(days[m.group(1)]), 0, 0) > ) > expire_time = (sue current_time)/60/60/24 Here's a likely problem. There's supposed to be an operator - probably a plus sign - between sue and current_time. >else: > m = q.search(line) > if m: > cert_name = m.group(1) And this last line needs indentation. The very easiest way to debug Python code is to run it. If it runs, great! See what output it made and whether it's correct or not. If it doesn't, Python will give you an exception traceback that points you to the failing line. Get familiar with them, as you'll be seeing them a lot :) Chris Angelico -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
In article <2912c674-e30b-4339-9344-1f460cb96...@googlegroups.com>, jojo wrote: > for fname in dirList: > cmd = "keytool printcert file " + fname > for line in os.popen(cmd).readlines(): >line = line.rstrip() >m = p.search(line) >if m: > sue = time.mktime( > (int(m.group(7)), int(months[m.group(2)]), int(m.group(3)), > int(m.group(4)), int(m.group(5)), int(m.group(6)), > int(days[m.group(1)]), 0, 0) > ) > expire_time = (sue current_time)/60/60/24 > if expire_time < 0: > print cert_name + " has already expired!" > elif expire_time < 31: > print cert_name + " expires in " +str(int(expire_time)) + " days" >else: > m = q.search(line) > if m: > cert_name = m.group(1) Was this code really indented like this when you got it? You've got (at least) three different indent sizes. I see 1, 2, and 3 space indents in different places in the code. I'm not even sure if this is legal, but even if it is, it's really bad form. Pick an indent, and stick with it uniformly. 4 spaces seems to be pretty standard. That being said, I'm going to return to my previous statement that until you know what the code is *supposed* to do, trying to test it is meaningless. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
In article <37f23623-8bf5-421a-ab6a-34ff622c6...@googlegroups.com>, jojo wrote: > Hi - I am a newbie to python and was wondering can someone tell me what the > following code does. I need to figure out how to test it I know this is going to sound unhelpful, but if your task is to test the code, what good does it do to know what it does? Clearly, it does what it does. What you really want to know is "What is it *supposed* to do?" Until you know what it's supposed to do, it is meaningless to even think about testing it. Seriously. I'm not trying to make life difficult for you. I've seen too much time and effort wasted on useless testing because there wasn't any specification for what the code was supposed to do. I will give you one hint, however. You've got stuff like: for line in os.popen(cmd).readlines(): line = line.rstrip() This is a syntax error because the indenting is wrong. Unlike C#, indenting is significant in Python. I don't know if the code you've got really is indented wrong, or it's just an artifact of how you posted it. But before anything else, you need to sort that out. Unless, of course, the specification for this code is, "It is supposed to raise IndentationError", in which case it passes the test :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Sunday, March 31, 2013 4:39:11 PM UTC-4, Chris Angelico wrote:
> On Mon, Apr 1, 2013 at 7:10 AM, jojo wrote:
>
> > Im used to C# so the syntax looks bizarre to me! Any help would be great.
>
>
>
> The first thing you'll need to understand about Python syntax is that
>
> indentation is important. By posting this code flush-left, you've
>
> actually destroyed its block structure. Could you post it again, with
>
> indentation, please? We'd then be in a much better position to help.
>
>
>
> Chris Angelico
Hi Chris, thanks for your reply. See code below...
import time
import glob
import re
import os
current_time = time.time() + 60*60+24*30
dirList = glob.glob('\content\paytek\ejbProperties\cybersource\*.crt')
q = re.compile('^Owner:.*CN=([^\s\,]+)')
p = re.compile('until: (\w+) (\w+) (\d+) (\d+):(\d+):(\d+) \w+ (\d+)')
cert_name = ""
days = {"Mon":0, "Tue":1, "Wed":2, "Thu":3, "Fri":4, "Sat":5, "Sun":6}
months = {"Jan":1, "Feb":2, "Mar":3, "Apr":4,
"May":5, "Jun":6, "Jul":7, "Aug":8,
"Sep":9, "Oct":10, "Nov":11, "Dec":12}
for fname in dirList:
cmd = "keytool printcert file " + fname
for line in os.popen(cmd).readlines():
line = line.rstrip()
m = p.search(line)
if m:
sue = time.mktime(
(int(m.group(7)), int(months[m.group(2)]), int(m.group(3)),
int(m.group(4)), int(m.group(5)), int(m.group(6)),
int(days[m.group(1)]), 0, 0)
)
expire_time = (sue current_time)/60/60/24
if expire_time < 0:
print cert_name + " has already expired!"
elif expire_time < 31:
print cert_name + " expires in " +str(int(expire_time)) + " days"
else:
m = q.search(line)
if m:
cert_name = m.group(1)
--
http://mail.python.org/mailman/listinfo/python-list
Re: Help with python code!
On Mon, Apr 1, 2013 at 7:10 AM, jojo wrote: > Im used to C# so the syntax looks bizarre to me! Any help would be great. The first thing you'll need to understand about Python syntax is that indentation is important. By posting this code flush-left, you've actually destroyed its block structure. Could you post it again, with indentation, please? We'd then be in a much better position to help. Chris Angelico -- http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
Τη Κυριακή, 31 Μαρτίου 2013 11:21:21 μ.μ. UTC+3, ο χρήστης ru...@yahoo.com έγραψε: > On 03/31/2013 02:08 PM, Νίκος Γκρ33κ wrote: > > > > > But i look the code and run python via interactive prompt and it says > > > it has no error. > > > > Does it produce any output? Is that output the right html? That is, if > > you save the html to a file and open that file in a browser, does it look > > right? > > > > > So i don't have a clue on what i still need to try since i get no > > > error via cmd or via browser(blank page) > > > > I have no clue either because I don't know what output you are getting > > when you run the script interactively, nor do I know what the code is > > that you are running. I think i have mase some progress and it seems that the reason *perhaps* is encoding issues. look at http://superhost.gr now and biew its source too. -- http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
On 03/31/2013 02:08 PM, Νίκος Γκρ33κ wrote: > But i look the code and run python via interactive prompt and it says > it has no error. Does it produce any output? Is that output the right html? That is, if you save the html to a file and open that file in a browser, does it look right? > So i don't have a clue on what i still need to try since i get no > error via cmd or via browser(blank page) I have no clue either because I don't know what output you are getting when you run the script interactively, nor do I know what the code is that you are running. -- http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
On 03/31/2013 01:19 PM, Νίκος Γκρ33κ wrote: > I just tried the testmysql.py script: >[...snip code...] I hope no one who reads this list also has access to your database and that you don't use that username/password anyplace else. > it works, as you can see at: > http://superhost.gr/cgi-bin/testmysql.py > so MySQLdb mpodule does work for Python3 ! > so mysql connector is not the problem. > but hwta is it then i get no errors!! Get no errors from what? The original script you were having trouble with? Are you running it by using your browser or from interactively in a terminal window? I probably should have asked this before, but can do you have shell access to the script? Can you run it interactively in a terminal window? -- http://mail.python.org/mailman/listinfo/python-list
Help with python code!
Hi - I am a newbie to python and was wondering can someone tell me what the
following code does. I need to figure out how to test it
import time
import glob
import re
import os
current_time = time.time() + 60*60+24*30
dirList = glob.glob('\content\paytek\ejbProperties\cybersource\*.crt')
q = re.compile('^Owner:.*CN=([^\s\,]+)')
p = re.compile('until: (\w+) (\w+) (\d+) (\d+):(\d+):(\d+) \w+ (\d+)')
cert_name = ""
days = {"Mon":0, "Tue":1, "Wed":2, "Thu":3, "Fri":4, "Sat":5, "Sun":6}
months = {"Jan":1, "Feb":2, "Mar":3, "Apr":4,
"May":5, "Jun":6, "Jul":7, "Aug":8,
"Sep":9, "Oct":10, "Nov":11, "Dec":12}
for fname in dirList:
cmd = "keytool printcert file " + fname
for line in os.popen(cmd).readlines():
line = line.rstrip()
m = p.search(line)
if m:
sue = time.mktime(
(int(m.group(7)), int(months[m.group(2)]), int(m.group(3)),
int(m.group(4)), int(m.group(5)), int(m.group(6)),
int(days[m.group(1)]), 0, 0)
)
expire_time = (sue current_time)/60/60/24
if expire_time < 0:
print cert_name + " has already expired!"
elif expire_time < 31:
print cert_name + " expires in " +str(int(expire_time)) + " days"
else:
m = q.search(line)
if m:
cert_name = m.group(1)
Im used to C# so the syntax looks bizarre to me! Any help would be great.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
Τη Κυριακή, 31 Μαρτίου 2013 10:46:57 μ.μ. UTC+3, ο χρήστης ru...@yahoo.com έγραψε: > On 03/31/2013 01:12 PM, Νίκος Γκρ33κ wrote: > > > Firsly, thank you for your willing to help me. i wrote, uploaded an > > > chmoded test.py and you can see the cgi enviromental table here: > > > http://superhost.gr/cgi-bin/test.py All values seem okey, so it > > > really isnt somehting wrong with the cgi enviroment. Also i chnagen > > > the host line to what you suggestes so interactive prompts will not > > > give errors. Please take a look the values to see if something not > > > look ok and i 'am about to create another trest script to check if i > > > can pefrom a simple mysql query with python 3.2.3 just to make sure > > > the MySQLdb connector work ok with 3.x > > > > I didn't mean that there was anything *wrong* with your cgi > > environment, only that it is *different* than your interactive > > environment. > > > > It is easier to debug code in an interactive environment than a > > cgi one so you want to do as much debugging interactively as possible,. > > But because the environments are different your code will behave > > differently. Knowing what is different between the two environments > > will help you know if an error is due to a real problem in your > > code, or is just due to the different environment. And it will > > help you setup your interactive environment to be as close as > > possible to the cgi one. But i look the code and run python via interactive prompt and it says it has no error. So i don't have a clue on what i still need to try since i get no error via cmd or via browser(blank page) -- http://mail.python.org/mailman/listinfo/python-list
Re: collections.Iterator __subclasshook__ does not check if next() is callable
Thanks for responding Terry. I can assure you I did not initially realize both the `next` and the `__iter__` methods were implemented when I ran into my original problem. I saw a behavior and had to work backwards to realize why it was behaving the way it was (the comparison against Iterator). Once I realized this, the behavior made complete sense. It just dawned on me the fact that `next` was not being checked to be callable (I was surprised by this at the time) which is why I investigated the `Iterator.__subclasshook__` and assumed it was behaving incorrectly based on my assumptions. On Sunday, March 31, 2013 3:47:07 PM UTC-4, Terry Jan Reedy wrote: > On 3/31/2013 1:57 PM, Byron Ruth wrote: > > > I submitted this as bug last night: http://bugs.python.org/issue17584 and > > was *honored* to be rejected by Raymond Hettinger. However, I would like > > feedback on whether my concern (this bug) is justified and clarity if not. > > > > > > Consider: > > > > > > ```python > > > class A(object): > > > def __init__(self): > > > self.r = iter(range(5)) > > > def __iter__(self): > > > return self > > > @property > > > def next(self): > > > return next(self.r) > > > ``` > > > > > > The `next` method is a property, however: > > > > A competent Python programmer should not do that. In Py3, the method is > > properly renamed '__next__', which should make doing that accidentally > > even less likely. > > > > > > > > ```python > > > from collections import Iterator > > > a = A() > > > isinstance(a, Iterator) # True > > > next(a) # TypeError: 'int' object is not callable > > > ``` > > > > > > I am using `collections.Iterator` as the means to check if the object is an > > iterator, > > > > Being an Iterator only means that it *might* be an iterator. > > > > > however I am not sure if that is _root_ problem here. My > > understanding of the iterator protocol is that is assumes the __iter__ > > and next *methods* are implemented. In the example, `A.next` is defined > > as a property, but is still identified as an iterator. To me, this is > > incorrect behavior since it's not conforming to the iterator protocol > > requirements (i.e. a `next` method, not a property). > > > > There is more to any protocol than can be statically checked. > > > > > Raymond stated: "The design of ABCs are to check for the existence to > > required named; none of them verify the signature." > > > > Having the required attributes is currently the definition of being an > > instance of an ABC. Adding 'not a property' would be possible. but > > hardly worthwhile. Checking signatures would be worthwhile, but > > signatures are not yet available to Python for C-coded methods, let > > alone other implementations. > > > > I think I understand _why_ this is the case.. but I downstream > > libraries use `collections.Iterator` to determine if an object _is one_: > > see > > https://github.com/django/django/blob/master/django/utils/itercompat.py#L22-L31 > > > > > > Who's job is it to check if `next` (and technically `__iter__`) are methods? > > > > The programmer, and a user who does not trust the competence of the > > programmer. But this is the least of the possible errors. > > > > -- > > Terry Jan Reedy -- http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
On 03/31/2013 01:12 PM, Νίκος Γκρ33κ wrote: > Firsly, thank you for your willing to help me. i wrote, uploaded an > chmoded test.py and you can see the cgi enviromental table here: > http://superhost.gr/cgi-bin/test.py All values seem okey, so it > really isnt somehting wrong with the cgi enviroment. Also i chnagen > the host line to what you suggestes so interactive prompts will not > give errors. Please take a look the values to see if something not > look ok and i 'am about to create another trest script to check if i > can pefrom a simple mysql query with python 3.2.3 just to make sure > the MySQLdb connector work ok with 3.x I didn't mean that there was anything *wrong* with your cgi environment, only that it is *different* than your interactive environment. It is easier to debug code in an interactive environment than a cgi one so you want to do as much debugging interactively as possible,. But because the environments are different your code will behave differently. Knowing what is different between the two environments will help you know if an error is due to a real problem in your code, or is just due to the different environment. And it will help you setup your interactive environment to be as close as possible to the cgi one. -- http://mail.python.org/mailman/listinfo/python-list
Re: collections.Iterator __subclasshook__ does not check if next() is callable
On 3/31/2013 1:57 PM, Byron Ruth wrote: I submitted this as bug last night: http://bugs.python.org/issue17584 and was *honored* to be rejected by Raymond Hettinger. However, I would like feedback on whether my concern (this bug) is justified and clarity if not. Consider: ```python class A(object): def __init__(self): self.r = iter(range(5)) def __iter__(self): return self @property def next(self): return next(self.r) ``` The `next` method is a property, however: A competent Python programmer should not do that. In Py3, the method is properly renamed '__next__', which should make doing that accidentally even less likely. ```python from collections import Iterator a = A() isinstance(a, Iterator) # True next(a) # TypeError: 'int' object is not callable ``` I am using `collections.Iterator` as the means to check if the object is an iterator, Being an Iterator only means that it *might* be an iterator. > however I am not sure if that is _root_ problem here. My understanding of the iterator protocol is that is assumes the __iter__ and next *methods* are implemented. In the example, `A.next` is defined as a property, but is still identified as an iterator. To me, this is incorrect behavior since it's not conforming to the iterator protocol requirements (i.e. a `next` method, not a property). There is more to any protocol than can be statically checked. Raymond stated: "The design of ABCs are to check for the existence to required named; none of them verify the signature." Having the required attributes is currently the definition of being an instance of an ABC. Adding 'not a property' would be possible. but hardly worthwhile. Checking signatures would be worthwhile, but signatures are not yet available to Python for C-coded methods, let alone other implementations. I think I understand _why_ this is the case.. but I downstream libraries use `collections.Iterator` to determine if an object _is one_: see https://github.com/django/django/blob/master/django/utils/itercompat.py#L22-L31 Who's job is it to check if `next` (and technically `__iter__`) are methods? The programmer, and a user who does not trust the competence of the programmer. But this is the least of the possible errors. -- Terry Jan Reedy -- http://mail.python.org/mailman/listinfo/python-list
Re: collections.Iterator __subclasshook__ does not check if next() is callable
Raymond's replied to my follow-up and made me realize that the `next` property could return a callable and it would be transparent to the caller. On Sunday, March 31, 2013 1:57:08 PM UTC-4, Byron Ruth wrote: > I submitted this as bug last night: http://bugs.python.org/issue17584 and was > *honored* to be rejected by Raymond Hettinger. However, I would like feedback > on whether my concern (this bug) is justified and clarity if not. > > > > Consider: > > > > ```python > > class A(object): > > def __init__(self): > > self.r = iter(range(5)) > > def __iter__(self): > > return self > > @property > > def next(self): > > return next(self.r) > > ``` > > > > The `next` method is a property, however: > > > > ```python > > from collections import Iterator > > a = A() > > isinstance(a, Iterator) # True > > next(a) # TypeError: 'int' object is not callable > > ``` > > > > I am using `collections.Iterator` as the means to check if the object is an > iterator, however I am not sure if that is _root_ problem here. My > understanding of the iterator protocol is that is assumes the __iter__ and > next *methods* are implemented. In the example, `A.next` is defined as a > property, but is still identified as an iterator. To me, this is incorrect > behavior since it's not conforming to the iterator protocol requirements > (i.e. a `next` method, not a property). > > > > Raymond stated: "The design of ABCs are to check for the existence to > required named; none of them verify the signature." I think I understand > _why_ this is the case.. but I downstream libraries use > `collections.Iterator` to determine if an object _is one_: see > https://github.com/django/django/blob/master/django/utils/itercompat.py#L22-L31 > > > > Who's job is it to check if `next` (and technically `__iter__`) are methods? -- http://mail.python.org/mailman/listinfo/python-list
interacting with an opened libreoffice calc file
Hi everyone, i'm new to the newsgroup and to python allthough (thanks to internet and the helpfull people i find) i've done a few scripts in python working like a charm. First of all i have to say i'm working on linux with python 2.3.7 (hope it's right) and libreoffice calc. My calc file gathers info from the web and adjusts it like i want and everything works fine there. My final result is a sheet called RESULTS where i have 6 columns : 1st column has BUY MARKET NAMES 2nd column has EURO PRICES (relative to the market's name in EURO currency) 3rd column has ORIG PRICE (value in the original currency) 4th column has SELL MARKET NAMES 5th column has EURO PRICES (relative to the market's name in EURO currency) 6th column has ORIG PRICE (value in the original currency) I would like to have another sheet or place (let's call it ALERTS) where i can make a list of alerts i would like to receive ... for example : MARKET1 BUY <= 20 MARKET2 SELL = 90 MARKET3 BUY > 59 What i would like is a script in python that reads all lines in the sheet ALERTS every n. (minutes or seconds) and compares the values from the RESULTS sheet and if the condition is TRUE then shoots an ALARM (soundfile would be great, maybe a customizable one) Is this possible? How can i acheive this? Thanks to anyone that can give me a solution, hint o show me the right direction. Cheers! Marco. -- http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
I just tried the testmysql.py script:
#!/usr/bin/python3
# coding=utf-8
import cgitb; cgitb.enable()
import cgi, re, os, sys, socket, datetime, MySQLdb, locale, random, subprocess
# connect to database
con = MySQLdb.connect( db = 'nikos_metrites', host = 'localhost', user =
'nikos_nikos', passwd = 'tiabhp2r#', init_command='SET NAMES UTF8' )
cur = con.cursor()
print ( "Content-type: text/html; charset=utf-8\n" )
#
=
# if extra string is attached to the URL is 'log' then show explicit page log
and exit
#
=
cur.execute('''SELECT hits FROM counters''')
data = cur.fetchall()#URL is unique, so should only be one
for item in data:
print( item )
===
it works, as you can see at: http://superhost.gr/cgi-bin/testmysql.py
so MySQLdb mpodule does work for Python3 !
so mysql connector is not the problem.
but hwta is it then i get no errors!!
--
http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
Τη Κυριακή, 31 Μαρτίου 2013 9:14:43 μ.μ. UTC+3, ο χρήστης ru...@yahoo.com
έγραψε:
> On 03/31/2013 08:03 AM, Νίκος Γκρ33κ wrote:
>
> > Hello all,
>
> >
>
> > i need some help
>
> > i recently changes pythoon 2.6 code => python 3.2.3 but my script although
> > not producing any errors now doesnt display anything else but a blank page
> > at htp://superhost.gr
>
> > can you help?
>
> >
>
> > I tried MySQLdb, pymysql, oursql, but nothing happens.
>
> > i still get a blank page. I dont know what else to try since i see no error.
>
>
>
> When I look at your page and do a "show source" in my browser
>
> it shows me the following:
>
>
>
>
>
>--> -->
>
>
>
>
>
>
>
> One obvious error is that you are writing html text before the
>
> "Content-Type: text/html" line. And whatever code is supposed
>
> to create the content is not getting run.
>
>
>
> > can anyone suggest anything?
>
>
>
> It is hard to help you because there is no code to look at.
>
> Also, I looked at your web site a couple days ago after you
>
> said you were getting a blank page, but I got a traceback
>
> page. I realize your are working on the problem and changing
>
> things but nobody wants to spend time looking for a problem,
>
> only to find out you've changed the code and their time was
>
> wasted.
>
>
>
> Also, your requests for help are spread out over multiple
>
> threads making it hard for anyone to figure out what you've
>
> already tried and what the current state of your code and
>
> problems are.
>
>
>
> I haven't used MySql so I can't offer any specific help regarding
>
> that, but I can say how I would try to attack your problem if it
>
> were my problem.
>
>
>
> First, Python 2 and Python 3 are two different languages. Code
>
> written for Python 2 will not run under Python 3 (in general)
>
> unless it is changed. So first, are you sure that the database
>
> connector (dbi module) runs under Python 3? Are you sure it is
>
> installed properly?
>
>
>
> To answer this question definitively you can write a simple
>
> Python program to import the dbi module, open a connection to
>
> your database and execute a simple query.
>
>
>
> Once you are sure you can access the database with Python 3
>
> code, next is your cgi script code.
>
>
>
> Try running your cgi script interactively.
>
>
>
> But when you do this, remember that the environment that your
>
> script sees is different when you run it interactively than
>
> when Apache runs your script. When Apache runs your script
>
> it might be using a different userid (which can affect what
>
> files your script can open), there different environment variables
>
> (like REMOTE_HOST which will be missing when you run interactively).
>
>
>
> You can see the environment that exists when Apache is running
>
> your script by creating a cgi script like:
>
>
>
>
>
> #!/bin/env python3
>
> import cgi, os, sys
>
>
>
> print("Content-type: text/html\n")
>
> print("""\
>
>
> PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
>
> "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
>
> http://www.w3.org/1999/xhtml"; lang="en-US" xml:lang="en-US">
>
>
>
> Python CGI Environment
>
>
>
>
>
>
>
> Python CGI Environment
>
> """)
>
> print("Python %s" % sys.version)
>
> print("Current directory: %s" % os.getcwd())
>
> print("Environment Variables:")
>
> for k in sorted(os.environ.keys()):
>
> print(" %s: %s" % (k, cgi.escape(os.environ[k])))
>
>
>
> print("")
>
>
>
>
>
> Open that cgi script in your browser. You can also run it
>
> interactively and you'll see many differences with what
>
> you see in your browser.
>
>
>
> When you run your failing cgi script interactively, you can
>
> give it an argument that is the url of the cgi script. For
>
> example, if you do:
>
>
>
> cd ~me/public_html/cgi_bin/
>
> python3 myscript.py 'http://superhost.gr/myscript.cgi'
>
>
>
> the cgi module will parse the url argument and setup the right
>
> values for the QUERY_STRING environment value that Apache would
>
> normally set up.
>
>
>
> There may be other problems resulting from the different interactive
>
> environment. If I recall, at one point you were having an error
>
> because REMOTE_ADDR was not present in the interactive environment.
>
>
>
> Perhaps the best way to deal with that is to modify your code.
>
> Foe example replace
>
>
>
> host = socket.gethostbyaddr( os.environ['REMOTE_ADDR'] )[0]
>
>
>
> with
>
> try:
>
> host = socket.gethostbyaddr( os.environ['REMOTE_ADDR'] )[0]
>
> except KeyError:
>
> host = 'not available'
>
>
>
> Then your script will be able to run interactively and you can
>
> continue debugging the real problems in it. Of course there
>
> may be other issue you have to fix as well but find and fix
Re: Creating a dictionary from a .txt file
On 3/31/2013 11:52 AM, C.T. wrote:
Hello,
I'm currently working on a homework problem that requires me to create a
dictionary from a .txt file that contains some of the worst cars ever made. The
file looks something like this:
1958 MGA Twin Cam
1958 Zunndapp Janus
1961 Amphicar
1961 Corvair
1966 Peel Trident
1970 AMC Gremlin
1970 Triumph Stag
1971 Chrysler Imperial LeBaron Two-Door Hardtop
The car manufacturer should be the key and a tuple containing the year and the
model should be the key's value. I tried the following to just get the contents
of the file into a list, but only the very last line in the txt file is shown
as a list with three elements (ie, ['2004', 'Chevy', 'SSR']) when I print temp.
d={}
car_file = open('worstcars.txt', 'r')
for line in car_file:
temp = line.split()
If all makers are one word (Austen-Martin would be ok, and if the file
is otherwise consistently year maker model words, then adding
'maxsplit=3' to the split call would be all the parsing you need.
--
http://mail.python.org/mailman/listinfo/python-list
"Laws of Form" are a notation for the SK calculus, demo in Python.
I was investigating G. Spencer-Brown's Laws of Form[1] by implementing it
in Python. You can represent the "marks" of LoF as datastructures in
Python composed entirely of tuples.
For example:
A mark: ()
A mark next to a mark: (), ()
A mark within a mark: ((),)
and so on...
It is known that the propositional calculus can be represented by the
"arithmetic of the mark". The two rules suffice:
((),) == nothing
() == (), ()
There are details, but essentially math and logic can be derived from the
behaviour of "the mark".
After reading "Every Bit Counts"[2] I spent some time trying to come up
with an EBC coding for the circle language (Laws of Form expressed as
tuples, See Burnett-Stuart's "The Markable Mark"[3])
With a little skull-sweat I found the following encoding:
For the empty state (no mark) write '0'.
Start at the left, if you encounter a boundary (one "side" of a mark, or
the left, opening, parenthesis) write a '1'.
Repeat for the contents of the mark, then the neighbors.
_ -> 0
() -> 100
(), () -> 10100
((),) -> 11000
and so on...
I recognized these numbers as the patterns of the language called
'Iota'.[4]
Briefly, the SK combinators:
S = λx.λy.λz.xz(yz)
K = λx.λy.x
Or, in Python:
S = lambda x: lambda y: lambda z: x(z)(y(z))
K = lambda x: lambda y: x
can be used to define the combinator used to implement Iota:
i = λc.cSK
or,
i = lambda c: c(S)(K)
And the bitstrings are decoded like so: if you encounter '0' return i,
otherwise decode two terms and apply the first to the second.
In other words, the empty space, or '0', corresponds to i:
_ -> 0 -> i
and the mark () corresponds to i applied to itself:
() -> 100 -> i(i)
which is an Identity function I.
The S and K combinators can be "recovered" by application of i to itself
like so (this is Python code, note that I am able to use 'is' instead of
the weaker '==' operator. The i combinator is actually recovering the
very same lambda functions used to create it. Neat, eh?):
K is i(i(i(i))) is decode('1010100')
S is i(i(i(i(i is decode('101010100')
Where decode is defined (in Python) as:
decode = lambda path: _decode(path)[0]
def _decode(path):
bit, path = path[0], path[1:]
if bit == '0':
return i, path
A, path = _decode(path)
B, path = _decode(path)
return A(B), path
(I should note that there is an interesting possibility of encoding the
tuples two ways: contents before neighbors (depth-first) or neighbors
before content (breadth-first). Here we look at the former.)
So, in "Laws of Form" K is ()()() and S is ()()()() and, amusingly, the
identity function I is ().
The term '(())foo' applies the identity function to foo, which matches the
behaviour of the (()) form in the Circle Arithmetic (()) == _ ("nothing".)
(())A -> i(i)(A) -> I(A) -> A
I just discovered this (that the Laws of Form have a direct mapping to
the combinator calculus[5] by means of λc.cSK) and I haven't found anyone
else mentioning yet (although [6] might, I haven't worked my way all the
way through it yet.)
There are many interesting avenues to explore from here, and I personally
am just beginning, but this seems like something worth reporting.
Warm regards,
~Simon Peter Forman
[1] http://en.wikipedia.org/wiki/Laws_of_Form
[2] research.microsoft.com/en-us/people/dimitris/every-bit-counts.pdf
[3] http://www.markability.net/
[4] http://semarch.linguistics.fas.nyu.edu/barker/Iota/
[5] http://en.wikipedia.org/wiki/SKI_combinator_calculus have
[6]
http://memristors.memristics.com/Combinatory%20Logic%20and%20LoF/Combinatory%20Logic%20and%20the%20Laws%20of%20Form.html
--
http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
In article ,
Dave Angel wrote:
> On 03/31/2013 12:52 PM, C.T. wrote:
> > On Sunday, March 31, 2013 12:20:25 PM UTC-4, zipher wrote:
> >>
> >>
> >
> > Thank you, Mark! My problem is the data isn't consistently ordered. I can
> > use slicing and indexing to put the year into a tuple, but because a car
> > manufacturer could have two names (ie, Aston Martin) or a car model could
> > have two names(ie, Iron Duke), its harder to use slicing and indexing for
> > those two. I've added the following, but the output is still not what I
> > need it to be.
>
> So the correct answer is "it cannot be done," and an explanation.
>
> Many times I've been given impossible conditions for a problem. And
> invariably the correct solution is to press [back] on the supplier of the
> constraints.
In real life, you often have to deal with crappy input data (and bogus
project requirements). Sometimes you just need to be creative.
There's only a small set of car manufacturers. A good start would be
mining wikipedia's [[List of automobile manufacturers]]. Once you've
got that list, you could try matching portions of the input against the
list.
Depending on how much effort you wanted to put into this, you could
explore all sorts of fuzzy matching (ie "delorean" vs "delorean motor
company"), but even a simple search is better than giving up.
And, this is a good excuse to explore some of the interesting
third-party modules. For example, mwclient ("pip install mwclient")
gives you a neat Python interface to wikipedia. And there's a whole
landscape of string matching packages to explore.
We deal with this every day at Songza. Are Kesha and Ke$ha the same
artist? Pushing back on the record labels to clean up their catalogs
isn't going to get us very far.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
On 03/31/2013 08:03 AM, Νίκος Γκρ33κ wrote:
> Hello all,
>
> i need some help
> i recently changes pythoon 2.6 code => python 3.2.3 but my script although
> not producing any errors now doesnt display anything else but a blank page at
> htp://superhost.gr
> can you help?
>
> I tried MySQLdb, pymysql, oursql, but nothing happens.
> i still get a blank page. I dont know what else to try since i see no error.
When I look at your page and do a "show source" in my browser
it shows me the following:
--> -->
One obvious error is that you are writing html text before the
"Content-Type: text/html" line. And whatever code is supposed
to create the content is not getting run.
> can anyone suggest anything?
It is hard to help you because there is no code to look at.
Also, I looked at your web site a couple days ago after you
said you were getting a blank page, but I got a traceback
page. I realize your are working on the problem and changing
things but nobody wants to spend time looking for a problem,
only to find out you've changed the code and their time was
wasted.
Also, your requests for help are spread out over multiple
threads making it hard for anyone to figure out what you've
already tried and what the current state of your code and
problems are.
I haven't used MySql so I can't offer any specific help regarding
that, but I can say how I would try to attack your problem if it
were my problem.
First, Python 2 and Python 3 are two different languages. Code
written for Python 2 will not run under Python 3 (in general)
unless it is changed. So first, are you sure that the database
connector (dbi module) runs under Python 3? Are you sure it is
installed properly?
To answer this question definitively you can write a simple
Python program to import the dbi module, open a connection to
your database and execute a simple query.
Once you are sure you can access the database with Python 3
code, next is your cgi script code.
Try running your cgi script interactively.
But when you do this, remember that the environment that your
script sees is different when you run it interactively than
when Apache runs your script. When Apache runs your script
it might be using a different userid (which can affect what
files your script can open), there different environment variables
(like REMOTE_HOST which will be missing when you run interactively).
You can see the environment that exists when Apache is running
your script by creating a cgi script like:
#!/bin/env python3
import cgi, os, sys
print("Content-type: text/html\n")
print("""\
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
http://www.w3.org/1999/xhtml"; lang="en-US" xml:lang="en-US">
Python CGI Environment
Python CGI Environment
""")
print("Python %s" % sys.version)
print("Current directory: %s" % os.getcwd())
print("Environment Variables:")
for k in sorted(os.environ.keys()):
print(" %s: %s" % (k, cgi.escape(os.environ[k])))
print("")
Open that cgi script in your browser. You can also run it
interactively and you'll see many differences with what
you see in your browser.
When you run your failing cgi script interactively, you can
give it an argument that is the url of the cgi script. For
example, if you do:
cd ~me/public_html/cgi_bin/
python3 myscript.py 'http://superhost.gr/myscript.cgi'
the cgi module will parse the url argument and setup the right
values for the QUERY_STRING environment value that Apache would
normally set up.
There may be other problems resulting from the different interactive
environment. If I recall, at one point you were having an error
because REMOTE_ADDR was not present in the interactive environment.
Perhaps the best way to deal with that is to modify your code.
Foe example replace
host = socket.gethostbyaddr( os.environ['REMOTE_ADDR'] )[0]
with
try:
host = socket.gethostbyaddr( os.environ['REMOTE_ADDR'] )[0]
except KeyError:
host = 'not available'
Then your script will be able to run interactively and you can
continue debugging the real problems in it. Of course there
may be other issue you have to fix as well but find and fix
them one at a time.
Add print statements to your code to find out where things are
going wrong. Don't trust your own thinking about the code!
If you think a variable should be 3 at some point in the code,
print it out and verify it!
After you get the program to generate html that looks ok, comment
out your debugging print statements (so that only html is printed),
redirect the html output into a file, them open the file in your
browser and see if it looks ok. If not, modify the code until
it does.
After you've got the script producing good output interactively
see if works from Apache as cgi script. If not, it should be
much easier now to find the problem since most of the other
problems have been fixed.
> I can even provide host:port user & pass f
collections.Iterator __subclasshook__ does not check if next() is callable
I submitted this as bug last night: http://bugs.python.org/issue17584 and was *honored* to be rejected by Raymond Hettinger. However, I would like feedback on whether my concern (this bug) is justified and clarity if not. Consider: ```python class A(object): def __init__(self): self.r = iter(range(5)) def __iter__(self): return self @property def next(self): return next(self.r) ``` The `next` method is a property, however: ```python from collections import Iterator a = A() isinstance(a, Iterator) # True next(a) # TypeError: 'int' object is not callable ``` I am using `collections.Iterator` as the means to check if the object is an iterator, however I am not sure if that is _root_ problem here. My understanding of the iterator protocol is that is assumes the __iter__ and next *methods* are implemented. In the example, `A.next` is defined as a property, but is still identified as an iterator. To me, this is incorrect behavior since it's not conforming to the iterator protocol requirements (i.e. a `next` method, not a property). Raymond stated: "The design of ABCs are to check for the existence to required named; none of them verify the signature." I think I understand _why_ this is the case.. but I downstream libraries use `collections.Iterator` to determine if an object _is one_: see https://github.com/django/django/blob/master/django/utils/itercompat.py#L22-L31 Who's job is it to check if `next` (and technically `__iter__`) are methods? -- http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On 03/31/2013 12:52 PM, C.T. wrote: On Sunday, March 31, 2013 12:20:25 PM UTC-4, zipher wrote: Thank you, Mark! My problem is the data isn't consistently ordered. I can use slicing and indexing to put the year into a tuple, but because a car manufacturer could have two names (ie, Aston Martin) or a car model could have two names(ie, Iron Duke), its harder to use slicing and indexing for those two. I've added the following, but the output is still not what I need it to be. So the correct answer is "it cannot be done," and an explanation. Many times I've been given impossible conditions for a problem. And invariably the correct solution is to press bac on the supplier of the constraints. Unless there are some invisible characters in that file, lie tabs in between the fields, it loocs liec you're out of luc. Or you could manually edit the file before running the program. [The character after 'j' is broccen on this cceyboard.] -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On Sunday, March 31, 2013 12:38:56 PM UTC-4, Roy Smith wrote:
> In article ,
>
> "C.T." wrote:
>
>
>
> > Hello,
>
> >
>
> > I'm currently working on a homework problem that requires me to create a
>
> > dictionary from a .txt file that contains some of the worst cars ever made.
>
> > The file looks something like this:
>
> >
>
> > 1958 MGA Twin Cam
>
> > 1958 Zunndapp Janus
>
> > 1961 Amphicar
>
> > 1961 Corvair
>
> > 1966 Peel Trident
>
> > 1970 AMC Gremlin
>
> > 1970 Triumph Stag
>
> > 1971 Chrysler Imperial LeBaron Two-Door Hardtop
>
> >
>
> > The car manufacturer should be the key and a tuple containing the year and
>
> > the model should be the key's value. I tried the following to just get the
>
> > contents of the file into a list, but only the very last line in the txt
> > file
>
> > is shown as a list with three elements (ie, ['2004', 'Chevy', 'SSR']) when
> > I
>
> > print temp.
>
> >
>
> > d={}
>
> > car_file = open('worstcars.txt', 'r')
>
> > for line in car_file:
>
> > temp = line.split()
>
> > print (temp)
>
> > car_file.close()
>
>
>
> Yup. Because you run through the whole file, putting each line into
>
> temp, overwriting the previous temp value.
>
>
>
> > d=[]
>
> > car_file = open('worstcars.txt', 'r')
>
> > for line in car_file:
>
> > d.append(line.strip('\n'))
>
> > print (d)
>
> > car_file.close()
>
>
>
> You could do most of that with just:
>
>
>
> car_file = open('worstcars.txt', 'r')
>
> d = car_file.readlines()
>
>
>
> but there's no real reason to read the whole file into a list. What you
>
> probably want to do is something like:
>
>
>
> d = {}
>
> car_file = open('worstcars.txt', 'r')
>
> for line in car_file:
>
>year, manufacturer, model = parse_line(line)
>
>d[manufacturer] = (year, model)
>
>
>
> One comment about the above; it assumes that there's only a single entry
>
> for a given manufacturer in the file. If that's not true, the above
>
> code will only keep the last one. But let's assume it's true for the
>
> moment.
>
>
>
> Now, we're just down to writing parse_line(). This takes a string and
>
> breaks it up into 3 strings. I'm going to leave this as an exercise for
>
> you to work out. The complicated part is going to be figuring out some
>
> logic to deal with anything from multi-word model names ("Imperial
>
> LeBaron Two-Door Hardtop"), to lines like the Corvair where there is no
>
> manufacturer (or maybe there's no model?).
Roy, thank you so much! I'll do some more research to see how I can achieve
this. Thank you!
--
http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On Mon, Apr 1, 2013 at 4:19 AM, C.T. wrote: > Thank you, Chris! I could use slicing and indexing to build the dictionary > but the problem is with the car manufacturer an the car model. Either or both > could be multiple names. Then you're going to need some other form of magic to recognize where the manufacturer ends and the model starts. Do you have, say, tabs between the fields and spaces within? ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On Sunday, March 31, 2013 12:06:18 PM UTC-4, Chris Angelico wrote:
> On Mon, Apr 1, 2013 at 2:52 AM, C.T.
>
> > After playing around with the code, I came up with the following code to
> > get everything into a list:
>
> >
>
> > d=[]
>
> > car_file = open('worstcars.txt', 'r')
>
> > for line in car_file:
>
> > d.append(line.strip('\n'))
>
> > print (d)
>
> > car_file.close()
>
> >
>
> > Every line is now an element in list d. The question I have now is how can
> > I make a dictionary out of the list d with the car manufacturer as the key
> > and a tuple containing the year and the model should be the key's value.
>
>
>
> Ah, a nice straight-forward text parsing problem!
>
>
>
> The question is how to recognize the manufacturer. Is it guaranteed to
>
> be the second blank-delimited word, with the year being the first? If
>
> so, you were almost there with .split().
>
>
>
> car_file = open('worstcars.txt', 'r')
>
> # You may want to consider the 'with' statement here - no need to close()
>
> for line in car_file:
>
> temp = line.split(None, 2)
>
> if len(temp)==3:
>
> year, mfg, model = temp
>
> # Now do something with these three values
>
> print("Manufacturer: %s Year: %s Model: %s"%(mfg,year,model))
>
>
>
> That's sorted out the parsing side of things. Do you know how to build
>
> up the dictionary from there?
>
>
>
> What happens if there are multiple entries in the file for the same
>
> manufacturer? Do you need to handle that?
>
>
>
> ChrisA
Thank you, Chris! I could use slicing and indexing to build the dictionary but
the problem is with the car manufacturer an the car model. Either or both could
be multiple names.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On Sunday, March 31, 2013 12:20:25 PM UTC-4, zipher wrote:
> Every line is now an element in list d. The question I have now is how can I
> make a dictionary out of the list d with the car manufacturer as the key and
> a tuple containing the year and the model should be the key's value. Here is
> a sample of what list d looks like:
>
>
>
>
> ['1899 Horsey Horseless', '1909 Ford Model T', '1911 Overland OctoAuto',
> '2003 Hummer H2', '2004 Chevy SSR']
>
>
>
> Any help would be appreciated!
>
>
>
>
> As long as your data is consistently ordered, just use list indexing. d[2]
> is your key, and (d[1],d[3]) the key's value.
>
>
>
> Mark
> Tacoma, Washington
Thank you, Mark! My problem is the data isn't consistently ordered. I can use
slicing and indexing to put the year into a tuple, but because a car
manufacturer could have two names (ie, Aston Martin) or a car model could have
two names(ie, Iron Duke), its harder to use slicing and indexing for those two.
I've added the following, but the output is still not what I need it to be.
t={}
for i in d :
t[d[d.index(i)][5:]]= tuple(d[d.index(i)][:4])
print (t)
The output looks something like this:
{'Ford Model T': ('1', '9', '0', '9'), 'Mosler Consulier GTP': ('1', '9', '8',
'5'), 'Scripps-Booth Bi-Autogo': ('1', '9', '1', '3'), 'Morgan Plus 8 Propane':
('1', '9', '7', '5'), 'Fiat Multipla': ('1', '9', '9', '8'), 'Ford Pinto':
('1', '9', '7', '1'), 'Triumph Stag': ('1', '9', '7', '0'), 'BMW 7-series':
('2', '0', '0', '2')}
Here the key is the car manufacturer and car model and the value is a tuple
containing the year separated by a comma.( Not sure why that is ?)
--
http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
In article ,
"C.T." wrote:
> Hello,
>
> I'm currently working on a homework problem that requires me to create a
> dictionary from a .txt file that contains some of the worst cars ever made.
> The file looks something like this:
>
> 1958 MGA Twin Cam
> 1958 Zunndapp Janus
> 1961 Amphicar
> 1961 Corvair
> 1966 Peel Trident
> 1970 AMC Gremlin
> 1970 Triumph Stag
> 1971 Chrysler Imperial LeBaron Two-Door Hardtop
>
> The car manufacturer should be the key and a tuple containing the year and
> the model should be the key's value. I tried the following to just get the
> contents of the file into a list, but only the very last line in the txt file
> is shown as a list with three elements (ie, ['2004', 'Chevy', 'SSR']) when I
> print temp.
>
> d={}
> car_file = open('worstcars.txt', 'r')
> for line in car_file:
> temp = line.split()
> print (temp)
> car_file.close()
Yup. Because you run through the whole file, putting each line into
temp, overwriting the previous temp value.
> d=[]
> car_file = open('worstcars.txt', 'r')
> for line in car_file:
> d.append(line.strip('\n'))
> print (d)
> car_file.close()
You could do most of that with just:
car_file = open('worstcars.txt', 'r')
d = car_file.readlines()
but there's no real reason to read the whole file into a list. What you
probably want to do is something like:
d = {}
car_file = open('worstcars.txt', 'r')
for line in car_file:
year, manufacturer, model = parse_line(line)
d[manufacturer] = (year, model)
One comment about the above; it assumes that there's only a single entry
for a given manufacturer in the file. If that's not true, the above
code will only keep the last one. But let's assume it's true for the
moment.
Now, we're just down to writing parse_line(). This takes a string and
breaks it up into 3 strings. I'm going to leave this as an exercise for
you to work out. The complicated part is going to be figuring out some
logic to deal with anything from multi-word model names ("Imperial
LeBaron Two-Door Hardtop"), to lines like the Corvair where there is no
manufacturer (or maybe there's no model?).
--
http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
> > Every line is now an element in list d. The question I have now is how can > I make a dictionary out of the list d with the car manufacturer as the key > and a tuple containing the year and the model should be the key's value. > Here is a sample of what list d looks like: > > ['1899 Horsey Horseless', '1909 Ford Model T', '1911 Overland OctoAuto', > '2003 Hummer H2', '2004 Chevy SSR'] > > Any help would be appreciated! > > As long as your data is consistently ordered, just use list indexing. d[2] is your key, and (d[1],d[3]) the key's value. Mark Tacoma, Washington -- http://mail.python.org/mailman/listinfo/python-list
Re: Creating a dictionary from a .txt file
On Mon, Apr 1, 2013 at 2:52 AM, C.T. wrote:
> After playing around with the code, I came up with the following code to get
> everything into a list:
>
> d=[]
> car_file = open('worstcars.txt', 'r')
> for line in car_file:
> d.append(line.strip('\n'))
> print (d)
> car_file.close()
>
> Every line is now an element in list d. The question I have now is how can I
> make a dictionary out of the list d with the car manufacturer as the key and
> a tuple containing the year and the model should be the key's value.
Ah, a nice straight-forward text parsing problem!
The question is how to recognize the manufacturer. Is it guaranteed to
be the second blank-delimited word, with the year being the first? If
so, you were almost there with .split().
car_file = open('worstcars.txt', 'r')
# You may want to consider the 'with' statement here - no need to close()
for line in car_file:
temp = line.split(None, 2)
if len(temp)==3:
year, mfg, model = temp
# Now do something with these three values
print("Manufacturer: %s Year: %s Model: %s"%(mfg,year,model))
That's sorted out the parsing side of things. Do you know how to build
up the dictionary from there?
What happens if there are multiple entries in the file for the same
manufacturer? Do you need to handle that?
ChrisA
--
http://mail.python.org/mailman/listinfo/python-list
Creating a dictionary from a .txt file
Hello,
I'm currently working on a homework problem that requires me to create a
dictionary from a .txt file that contains some of the worst cars ever made. The
file looks something like this:
1958 MGA Twin Cam
1958 Zunndapp Janus
1961 Amphicar
1961 Corvair
1966 Peel Trident
1970 AMC Gremlin
1970 Triumph Stag
1971 Chrysler Imperial LeBaron Two-Door Hardtop
The car manufacturer should be the key and a tuple containing the year and the
model should be the key's value. I tried the following to just get the contents
of the file into a list, but only the very last line in the txt file is shown
as a list with three elements (ie, ['2004', 'Chevy', 'SSR']) when I print temp.
d={}
car_file = open('worstcars.txt', 'r')
for line in car_file:
temp = line.split()
print (temp)
car_file.close()
After playing around with the code, I came up with the following code to get
everything into a list:
d=[]
car_file = open('worstcars.txt', 'r')
for line in car_file:
d.append(line.strip('\n'))
print (d)
car_file.close()
Every line is now an element in list d. The question I have now is how can I
make a dictionary out of the list d with the car manufacturer as the key and a
tuple containing the year and the model should be the key's value. Here is a
sample of what list d looks like:
['1899 Horsey Horseless', '1909 Ford Model T', '1911 Overland OctoAuto', '2003
Hummer H2', '2004 Chevy SSR']
Any help would be appreciated!
--
http://mail.python.org/mailman/listinfo/python-list
Re: Python 3.2.3 and my blank page
I like to feed trolls :-) On 31 Mrz., 16:03, Νίκος Γκρ33κ wrote: > Hello all, Hello Ferrous Cranus [3]! > ... > I tried MySQLdb, pymysql, oursql, but nothing happens. > i still get a blank page. I dont know what else to try since i see no error. Well, the output of your cgi is: --> --> This is not valid html. Therefore the browser does not render anything. > > can anyone suggest anything? Learn html [1] and python [2], read your code. Find out, why most of your html-output does not get into the response. Solve the reasons. Done. > > I can even provide host:port user & pass for someone willing to take a look > from the inside. I'm sure it depends what you are willing to pay. [Sorry, I forgot. You are looking for someone, who does your work (which you are payed for because aou are a hoster) for free.] [1] https://www.google.at/search?q=learn+html+online [2] https://www.google.at/search?q=learn+python+online [3] http://redwing.hutman.net/~mreed/warriorshtm/ferouscranus.htm -- http://mail.python.org/mailman/listinfo/python-list
A Healthy Alternative to Takeaway Regret
A Healthy Alternative to Takeaway Regret http://natigtas7ab.blogspot.com/2013/03/a-healthy-alternative-to-takeaway-regret.html -- http://mail.python.org/mailman/listinfo/python-list
Re: Convert Latitude, Longitude To TimeZone
In article ,
Steve B wrote:
> I found a piece of code
> [http://blog.pamelafox.org/2012/04/converting-addresses-to-timezones-in.html
> ] which uses the function [https://gist.github.com/pamelafox/2288222]
>
> When I try to run the code, I get the error geonames is not defined (This is
> the function previously linked to)
The best thing to do when asking questions like this is to copy-paste
the exact error message you got. It was probably something like:
Traceback (most recent call last):
File "", line 1, in
NameError: name 'geonames' is not defined
In this example, it's easy enough to figure out what went wrong from
your description, but in general, giving the exact error helps in the
diagnosis.
> geonames_client = geonames.GeonamesClient('Username_alpha')
>
> geonames_result = geonames_client.find_timezone({'lat': 48.871236, 'lng':
> 2.77928})
>
> user.timezone = geonames_result['timezoneId']
You simply never imported the geonames module. Somewhere before your
first line of code, you want to add:
import geonames
This assumes that you've saved the module in a file named "geonames.py".
--
http://mail.python.org/mailman/listinfo/python-list
Convert Latitude, Longitude To TimeZone
Hi All
I'm new to python (4 days J) and was wondering if anyone out there can help
me
I am trying to get the time zones for latitude and longitude coordinates
but am having a few problems
The mistakes are probably very basic
I have a table in a database with around 600 rows. Each row contains a lat
long coordinate for somewhere in the world
I want to feed these co-ordinates into a function and then retrieve the time
zone. The aim being to convert events which have a local time stamp within
each of these 600 places into UTC time
I found a piece of code
[http://blog.pamelafox.org/2012/04/converting-addresses-to-timezones-in.html
] which uses the function [https://gist.github.com/pamelafox/2288222]
When I try to run the code, I get the error geonames is not defined (This is
the function previously linked to)
I have applied for an account with geonames, I think I have just saved the
function file in the wrong directory or something simple. Can anyone help
#---
# Converts latitude longitude into a time zone
# REF: https://gist.github.com/pamelafox/2288222
# REF:
http://blog.pamelafox.org/2012/04/converting-addresses-to-timezones-in.html
#---
geonames_client = geonames.GeonamesClient('Username_alpha')
geonames_result = geonames_client.find_timezone({'lat': 48.871236, 'lng':
2.77928})
user.timezone = geonames_result['timezoneId']
Thanks
--
http://mail.python.org/mailman/listinfo/python-list
Python 3.2.3 and my blank page
Hello all, i need some help i recently changes pythoon 2.6 code => python 3.2.3 but my script although not producing any errors now doesnt display anything else but a blank page at htp://superhost.gr can you help? I tried MySQLdb, pymysql, oursql, but nothing happens. i still get a blank page. I dont know what else to try since i see no error. can anyone suggest anything? I can even provide host:port user & pass for someone willing to take a look from the inside. -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On Sun, Mar 31, 2013 at 9:15 AM, Roy Smith wrote: > > > $ prtstat 29937 > > Process: mongodState: S (sleeping) > > [...] > > Memory > > Vsize: 1998285 MB > > RSS: 5428 MB > > RSS Limit: 18446744073709 MB > > If I counted the digits right, that 1.9 TB. I love the RSS Limit. I'll > be really impressed when somebody builds a machine with enough RAM to > reach that :-) > http://www.ncsa.illinois.edu/UserInfo/Resources/Hardware/SGIAltix/TechSummary/ Look at co-compute2. The indicated memory is available as shared memory across all 512 cores. And that's nothing---it can be configured with up to 64 TB of global shared memory: http://www.sgi.com/products/servers/uv/configs.html Impressed? :) All the best, Jason -- http://mail.python.org/mailman/listinfo/python-list
Re: Python GUI questions
On 19.03.2013 21:01, maiden129 wrote: Hello, I'm using python 3.2.3 and I'm making a program that show the of occurrences of the character in the string in Tkinter. My questions are: How can I make an empty Entry object that will hold a word that a user will enter? How to make an empty Entry object that will hold a single character that the user will enter? [..] Hello, here is a very good documentation about Tkinter and its most likely that the same principals of coding apply to Python 3.x when it comes down to the Tkinter part: http://www.pythonware.com/library/tkinter/introduction/index.htm Also as an tip, you can make use of a StringVar object for example - those are (if Im not completely wrong) meant to hold values which you can then access from the interface. Alike a binding to any element holding text. Which should take care of the updating part if you call a proper "StringVar/yourVariableName".set method of that particular class. As for the UI creation, have a look at that documentation, its very easy to navigate inside if you know what you are looking for. Regards Jan -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
In article , Dave Angel wrote: > I'm typing this while a terminal is open doing the particular operation, > and the system doesn't seem in the least sluggish. > > Currently the memory used is at 10gig, and while there are some pauses > in my typing, the system has not died. This is on Linux Ubuntu 12.04. > > At 15gig, there are some blockages, of maybe 5 secs each. 15 gig? Feh. > $ prtstat 29937 > Process: mongodState: S (sleeping) > [...] > Memory > Vsize: 1998285 MB > RSS: 5428 MB > RSS Limit: 18446744073709 MB If I counted the digits right, that 1.9 TB. I love the RSS Limit. I'll be really impressed when somebody builds a machine with enough RAM to reach that :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
In article <8276eff6-9e5c-4060-b9e8-94fab6062...@googlegroups.com>, morphex wrote: > Aha, OK. Thought I found a bug but yeah that makes sense ;) > > While we're on the subject, wouldn't it be nice to have some cap there so > that it isn't possible to more or less block the system with large > exponentiation? Every time we think we know a good upper bound to something (i.e. "nobody will ever need more than 64k of memory"), it turns out that limit is wrong. There's great value in writing code which continues to work until it runs out of some external resource (i.e. memory, disk space, whatever). Eventually, somebody will want to do some calculation which you previously thought was absurd but turns out to be useful. I don't use Python bugnums very often, but when I do, I'm really happy they're there. -- http://mail.python.org/mailman/listinfo/python-list
Re: flaming vs accuracy [was Re: Performance of int/long in Python 3]
On 31/03/2013 08:35, jmfauth wrote:
--
Neil Hodgson:
"The counter-problem is that a French document that needs to include
one mathematical symbol (or emoji) outside Latin-1 will double in size
as a Python string."
Serious developers/typographers/users know that you can not compose
a text in French with "latin-1". This is now also the case with
German (Germany).
---
Neil's comment is correct,
sys.getsizeof('a' * 1000 + 'z')
1026
sys.getsizeof('a' * 1000 + '€')
2040
This is not really the problem. "Serious users" may
notice sooner or later, Python and Unicode are walking in
opposite directions (technically and in spirit).
timeit.repeat("'a' * 1000 + 'ẞ'")
[1.1088995672090292, 1.0842266613261913, 1.1010779011941594]
timeit.repeat("'a' * 1000 + 'z'")
[0.6362570846925735, 0.6159128762502917, 0.6200501673623791]
(Just an opinion)
jmf
I'm feeling very sorry for this horse, it's been flogged so often it's
down to bare bones.
--
If you're using GoogleCrap™ please read this
http://wiki.python.org/moin/GoogleGroupsPython.
Mark Lawrence
--
http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
In article <5157e6cc$0$29974$c3e8da3$54964...@news.astraweb.com>, Steven D'Aprano wrote: > For what it's worth, that last intermediate result (two to the power of > the 489-digit number) has approximately a billion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion digits. > > (American billion and trillion, 10**9 and 10**12 respectively.) What's that in crore? -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On 03/31/2013 08:07 AM, morphex wrote: Aha, OK. Thought I found a bug but yeah that makes sense ;) While we're on the subject, wouldn't it be nice to have some cap there so that it isn't possible to more or less block the system with large exponentiation? There's an assumption there. The Operating System should defend itself against starvation by any single process. Besides, there are many ways for a process to run out of memory, and exponentiation is probably the least likely of them. In general, an application cannot tell whether a particular memory allocation will succeed or not without actually trying the allocation. If it fails, you get the exception. I'm typing this while a terminal is open doing the particular operation, and the system doesn't seem in the least sluggish. Currently the memory used is at 10gig, and while there are some pauses in my typing, the system has not died. This is on Linux Ubuntu 12.04. At 15gig, there are some blockages, of maybe 5 secs each. -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
Aha, OK. Thought I found a bug but yeah that makes sense ;) While we're on the subject, wouldn't it be nice to have some cap there so that it isn't possible to more or less block the system with large exponentiation? On Sunday, March 31, 2013 9:33:32 AM UTC+2, Steven D'Aprano wrote: > On Sat, 30 Mar 2013 23:56:46 -0700, morphex wrote: > > > > > Hi. > > > > > > I was just doodling around with the python interpreter today, and here > > > is the dump from the terminal: > > > > > > morphex@laptop:~$ python > > > Python 2.7.3 (default, Sep 26 2012, 21:53:58) [GCC 4.7.2] on linux2 > > > Type "help", "copyright", "credits" or "license" for more information. > > 1**2 > > > 1 > > 1**2**3 > > > 1 > > 1**2**3**4 > > > 1L > > 1**2**3**4**5 > > > Traceback (most recent call last): > > > File "", line 1, in > > > MemoryError > > > > > > > Does anyone know why this raises a MemoryError? Doesn't make sense to > > > me. > > > > Because exponentiation is right-associative, not left. > > > > 1**2**3**4**5 is calculated like this: > > > > 1**2**3**4**5 > > => 1**2**3**1024 > > => 1**2**373...481 # 489-digit number > > => 1**(something absolutely humongous) > > => 1 > > > > except of course you get a MemoryError in calculating the intermediate > > values. > > > > In other words, unlike you or me, Python is not smart enough to realise > > that 1**(...) is automatically 1, it tries to calculate the humongous > > intermediate result, and that's what fails. > > > > For what it's worth, that last intermediate result (two to the power of > > the 489-digit number) has approximately a billion trillion trillion > > trillion trillion trillion trillion trillion trillion trillion trillion > > trillion trillion trillion trillion trillion trillion trillion trillion > > trillion trillion trillion trillion trillion trillion trillion trillion > > trillion trillion trillion trillion trillion trillion trillion trillion > > trillion trillion trillion trillion trillion trillion digits. > > > > (American billion and trillion, 10**9 and 10**12 respectively.) > > > > > > > > -- > > Steven -- http://mail.python.org/mailman/listinfo/python-list
ASCII versus non-ASCII [was Re: flaming vs accuracy [was Re: Performance of int/long in Python 3]]
On Sun, 31 Mar 2013 00:35:23 -0700, jmfauth wrote:
> This is not really the problem. "Serious users" may notice sooner or
> later, Python and Unicode are walking in opposite directions
> (technically and in spirit).
>
timeit.repeat("'a' * 1000 + 'ẞ'")
> [1.1088995672090292, 1.0842266613261913, 1.1010779011941594]
timeit.repeat("'a' * 1000 + 'z'")
> [0.6362570846925735, 0.6159128762502917, 0.6200501673623791]
Perhaps you should stick to Python 3.2, where ASCII strings are no faster
than non-ASCII strings.
Python 3.2 versus Python 3.3, no significant difference:
# 3.2
py> timeit.repeat("'a' * 1000 + 'ẞ'")
[1.7418999671936035, 1.7198870182037354, 1.763346004486084]
# 3.3
py> timeit.repeat("'a' * 1000 + 'ẞ'")
[1.8083378580026329, 1.818592812011484, 1.7922867869958282]
Python 3.2, ASCII vs Non-ASCII:
py> timeit.repeat("'a' * 1000 + 'z'")
[1.756322135925293, 1.8002049922943115, 1.721085958480835]
py> timeit.repeat("'a' * 1000 + 'ẞ'")
[1.7209150791168213, 1.7162668704986572, 1.7260780334472656]
In other words, if you stick to non-ASCII strings, Python 3.3 is no
slower than Python 3.2.
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On 03/31/2013 03:33 AM, Steven D'Aprano wrote: On Sat, 30 Mar 2013 23:56:46 -0700, morphex wrote: Hi. I was just doodling around with the python interpreter today, and here is the dump from the terminal: morphex@laptop:~$ python Python 2.7.3 (default, Sep 26 2012, 21:53:58) [GCC 4.7.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. 1**2 1 1**2**3 1 1**2**3**4 1L 1**2**3**4**5 Traceback (most recent call last): File "", line 1, in MemoryError Does anyone know why this raises a MemoryError? Doesn't make sense to me. Because exponentiation is right-associative, not left. 1**2**3**4**5 is calculated like this: 1**2**3**4**5 => 1**2**3**1024 => 1**2**373...481 # 489-digit number Oops, you're right, it's 489. I figured 488 but was wrong. -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: flaming vs accuracy [was Re: Performance of int/long in Python 3]
--
Neil Hodgson:
"The counter-problem is that a French document that needs to include
one mathematical symbol (or emoji) outside Latin-1 will double in size
as a Python string."
Serious developers/typographers/users know that you can not compose
a text in French with "latin-1". This is now also the case with
German (Germany).
---
Neil's comment is correct,
>>> sys.getsizeof('a' * 1000 + 'z')
1026
>>> sys.getsizeof('a' * 1000 + '€')
2040
This is not really the problem. "Serious users" may
notice sooner or later, Python and Unicode are walking in
opposite directions (technically and in spirit).
>>> timeit.repeat("'a' * 1000 + 'ẞ'")
[1.1088995672090292, 1.0842266613261913, 1.1010779011941594]
>>> timeit.repeat("'a' * 1000 + 'z'")
[0.6362570846925735, 0.6159128762502917, 0.6200501673623791]
(Just an opinion)
jmf
--
http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On Sat, 30 Mar 2013 23:56:46 -0700, morphex wrote: > Hi. > > I was just doodling around with the python interpreter today, and here > is the dump from the terminal: > > morphex@laptop:~$ python > Python 2.7.3 (default, Sep 26 2012, 21:53:58) [GCC 4.7.2] on linux2 > Type "help", "copyright", "credits" or "license" for more information. 1**2 > 1 1**2**3 > 1 1**2**3**4 > 1L 1**2**3**4**5 > Traceback (most recent call last): > File "", line 1, in > MemoryError > Does anyone know why this raises a MemoryError? Doesn't make sense to > me. Because exponentiation is right-associative, not left. 1**2**3**4**5 is calculated like this: 1**2**3**4**5 => 1**2**3**1024 => 1**2**373...481 # 489-digit number => 1**(something absolutely humongous) => 1 except of course you get a MemoryError in calculating the intermediate values. In other words, unlike you or me, Python is not smart enough to realise that 1**(...) is automatically 1, it tries to calculate the humongous intermediate result, and that's what fails. For what it's worth, that last intermediate result (two to the power of the 489-digit number) has approximately a billion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion digits. (American billion and trillion, 10**9 and 10**12 respectively.) -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Why does 1**2**3**4**5 raise a MemoryError?
On 03/31/2013 02:56 AM, morphex wrote: Hi. I was just doodling around with the python interpreter today, and here is the dump from the terminal: morphex@laptop:~$ python Python 2.7.3 (default, Sep 26 2012, 21:53:58) [GCC 4.7.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. 1**2 1 1**2**3 1 1**2**3**4 1L 1**2**3**4**5 Traceback (most recent call last): File "", line 1, in MemoryError Does anyone know why this raises a MemoryError? Doesn't make sense to me. Perhaps you didn't realize that the expression will be done from right to left. So first it calculates 4**5 which is 1024 Then it calculates 3**1024, which has 488 digits. then it calculates 2 to that power, which is a number large enough to boggle the mind. That number's storage needs makes a few gigabytes seem like a molecule in the ocean. Anyway, it never gets around to doing the 1** part. On the other hand, perhaps you wanted to do a different calculation: >>> 1**2)**3)**4)**5) 1 -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Why does 1**2**3**4**5 raise a MemoryError?
Hi. I was just doodling around with the python interpreter today, and here is the dump from the terminal: morphex@laptop:~$ python Python 2.7.3 (default, Sep 26 2012, 21:53:58) [GCC 4.7.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> 1**2 1 >>> 1**2**3 1 >>> 1**2**3**4 1L >>> 1**2**3**4**5 Traceback (most recent call last): File "", line 1, in MemoryError >>> Does anyone know why this raises a MemoryError? Doesn't make sense to me. Happy Easter! -Morten -- http://mail.python.org/mailman/listinfo/python-list
