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  • how to use @ in python.. and the @property

    - by zjm1126
    this is my code: def a(): print 'sss' @a() def b(): print 'aaa' b() and the Traceback is: sss Traceback (most recent call last): File "D:\zjm_code\a.py", line 8, in <module> @a() TypeError: 'NoneType' object is not callable so how to use the '@' thanks updated class a: @property def b(x): print 'sss' aa=a() print aa.b it print : sss None how to use @property thanks

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  • importing classes python

    - by Richard
    Just wondering why import sys exit(0) gives me this error: Traceback (most recent call last): File "<pyshell#1>", line 1, in ? exit(0) TypeError: 'str' object is not callable but from sys import exit exit(0) works fine?

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  • Convert a GTK python script to C

    - by Jessica
    The following script will take a screenshot on a Gnome desktop. import gtk.gdk w = gtk.gdk.get_default_root_window() sz = w.get_size() pb = gtk.gdk.Pixbuf(gtk.gdk.COLORSPACE_RGB,False, 8, sz[0], sz[1]) pb = pb.get_from_drawable(w, w.get_colormap(), 0, 0, 0, 0, sz[0], sz[1]) if (pb != None): pb.save("screenshot.png", "png") print "Screenshot saved to screenshot.png." else: print "Unable to get the screenshot." Now, I've been trying to convert this to C and use it in one of the apps I am writing but so far i've been unsuccessful. Is there any what to do this in C (on Linux)? Thanks! Jess.

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  • How to find links and modify an Html using BeautifulSoup in Python

    - by systempuntoout
    Starting from an Html input like this: <p> <a href="http://www.foo.com">this if foo</a> <a href="http://www.bar.com">this if bar</a> </p> using BeautifulSoup, i would like to change this Html in: <p> <a href="http://www.foo.com">this if foo[1]</a> <a href="http://www.bar.com">this if bar[2]</a> </p> saving parsed links in a dictionary with a result like this: links_dict = {"1":"http://www.foo.com","2":"http://www.bar.com"} Is it possible to do this using BeautifulSoup? Any valid alternative?

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  • python function that returns a function from list of functions

    - by thkang
    I want to make following function: 1)input is a number. 2)functions are indexed, return a function whose index matches given number here's what I came up with: def foo_selector(whatfoo): def foo1(): return def foo2(): return def foo3(): return ... def foo999(): return #something like return foo[whatfoo] the problem is, how can I index the functions (foo#)? I can see functions foo1 to foo999 by dir(). however, dir() returns name of such functions, not the functions themselves. In the example, those foo-functions aren't doing anything. However in my program they perform different tasks and I can't automatically generate them. I write them myself, and have to return them by their name.

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  • [Python/Tkinter] Grid within a frame?

    - by Sam
    Is it possible to place a grid of buttons in Tkinter inside another frame? I'm wanting to create a tic-tac-toe like game and want to use the grid feature to put gamesquares (that will be buttons). However, I'd like to have other stuff in the GUI other than just the game board so it's not ideal to just have everything in the one grid. To illustrate: O | X | X | ---------- | O | O | X | Player 2 wins! ---------- | X | O | X | The tic tac toe board is in a grid that is made up of all buttons and the 'player 2 wins' is a label inside a frame. This is an oversimplification of what I'm trying to do so bear with me, for the way I've designed the program so far (the board is dynamically created) a grid makes the most sense.

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  • How to replace only part of the match with python re.sub

    - by Arty
    I need to match two cases by one reg expression and do replacement 'long.file.name.jpg' - 'long.file.name_suff.jpg' 'long.file.name_a.jpg' - 'long.file.name_suff.jpg' I'm trying to do the following re.sub('(\_a)?\.[^\.]*$' , '_suff.',"long.file.name.jpg") But this is cut the extension '.jpg' and I'm getting long.file.name_suff. instead of long.file.name_suff.jpg I understand that this is because of [^.]*$ part, but I can't exclude it, because I have to find last occurance of '_a' to replace or last '.' Is there a way to replace only part of the match?

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  • using Python reduce over a list of pairs

    - by user248237
    I'm trying to pair together a bunch of elements in a list to create a final object, in a way that's analogous to making a sum of objects. I'm trying to use a simple variation on reduce where you consider a list of pairs rather than a flat list to do this. I want to do something along the lines of: nums = [1, 2, 3] reduce(lambda x, y: x + y, nums) except I'd like to add additional information to the sum that is specific to each element in the list of numbers nums. For example for each pair (a,b) in the list, run the sum as (a+b): nums = [(1, 0), (2, 5), (3, 10)] reduce(lambda x, y: (x[0]+x[1]) + (y[0]+y[1]), nums) This does not work: >>> reduce(lambda x, y: (x[0]+x[1]) + (y[0]+y[1]), nums) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 1, in <lambda> TypeError: 'int' object is unsubscriptable Why does it not work? I know I can encode nums as a flat list - that is not the point - I just want to be able to create a reduce operation that can iterate over a list of pairs, or over two lists of the same length simultaneously and pool information from both lists. thanks.

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  • I need to make a multithreading program (python)

    - by Andreawu98
    import multiprocessing import time from itertools import product out_file = open("test.txt", 'w') P = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p','q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',] N = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] M = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'] c = int(input("Insert the number of digits you want: ")) n = int(input("If you need number press 1: ")) m = int(input("If you need upper letters press 1: ")) i = [] if n == 1: P = P + N if m == 1: P = P + M then = time.time() def worker(): for i in product(P, repeat=c): #check every possibilities k = '' for z in range(0, c): # k = k + str(i[z]) # print each possibility in a txt without parentesis or comma out_file.write( k + '\n') # out_file.close() now = time.time() diff = str(now - then) # To see how long does it take print(diff) worker() time.sleep(10) # just to check console The code check every single possibility and print it out in a test.txt file. It works but I really can't understand how can I speed it up. I saw it use 1 core out of my quad core CPU so I thought Multi-threading might work even though I don't know how. Please help me. Sorry for my English, I am from Italy.

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  • encapsulation in python list (want to use " instead of ')

    - by Codehai
    I have a list of users users["pirates"] and they're stored in the format ['pirate1','pirate2']. If I hand the list over to a def and query for it in MongoDB, it returns data based on the first index (e.g. pirate1) only. If I hand over a list in the format ["pirate1","pirate"], it returns data based on all the elements in the list. So I think there's something wrong with the encapsulation of the elements in the list. My question: can I change the encapsulation from ' to " without replacing every ' on every element with a loop manually? Short Example: aList = list() # get pirate Stuff # users["pirates"] is a list returned by a former query # so e.g. users["pirates"][0] may be peter without any quotes for pirate in users["pirates"]: aList.append(pirate) aVar = pirateDef(aList) print(aVar) the definition: def pirateDef(inputList = list()): # prepare query col = mongoConnect().MYCOL # query for pirates Arrrr pirates = col.find({ "_id" : {"$in" : inputList}} ).sort("_id",1).limit(50) # loop over users userList = list() for person in pirates: # do stuff that has nothing to do with the problem # append user to userlist userList.append(person) return userList If the given list has ' encapsulation it returns: 'pirates': [{'pirate': 'Arrr', '_id': 'blabla'}] If capsulated with " it returns: 'pirates' : [{'_id': 'blabla', 'pirate' : 'Arrr'}, {'_id': 'blabla2', 'pirate' : 'cheers'}] EDIT: I tried figuring out, that the problem has to be in the MongoDB query. The list is handed over to the Def correctly, but after querying pirates only consists of 1 element... Thanks for helping me Codehai

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  • building a pairwise matrix in scipy/numpy in Python from dictionaries

    - by user248237
    I have a dictionary whose keys are strings and values are numpy arrays, e.g.: data = {'a': array([1,2,3]), 'b': array([4,5,6]), 'c': array([7,8,9])} I want to compute a statistic between all pairs of values in 'data' and build an n by x matrix that stores the result. Assume that I know the order of the keys, i.e. I have a list of "labels": labels = ['a', 'b', 'c'] What's the most efficient way to compute this matrix? I can compute the statistic for all pairs like this: result = [] for elt1, elt2 in itertools.product(labels, labels): result.append(compute_statistic(data[elt1], data[elt2])) But I want result to be a n by n matrix, corresponding to "labels" by "labels". How can I record the results as this matrix? thanks.

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  • Python method to remove iterability

    - by Debilski
    Suppose I have a function which can either take an iterable/iterator or a non-iterable as an argument. Iterability is checked with try: iter(arg). Depending whether the input is an iterable or not, the outcome of the method will be different. Not when I want to pass a non-iterable as iterable input, it is easy to do: I’ll just wrap it with a tuple. What do I do when I want to pass an iterable (a string for example) but want the function to take it as if it’s non-iterable? E.g. make that iter(str) fails.

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  • Using Memcached in Python/Django - questions.

    - by Thomas
    I am starting use Memcached to make my website faster. For constant data in my database I use this: from django.core.cache import cache cache_key = 'regions' regions = cache.get(cache_key) if result is None: """Not Found in Cache""" regions = Regions.objects.all() cache.set(cache_key, regions, 2592000) #(2592000sekund = 30 dni) return regions For seldom changed data I use signals: from django.core.cache import cache from django.db.models import signals def nuke_social_network_cache(self, instance, **kwargs): cache_key = 'networks_for_%s' % (self.instance.user_id,) cache.delete(cache_key) signals.post_save.connect(nuke_social_network_cache, sender=SocialNetworkProfile) signals.post_delete.connect(nuke_social_network_cache, sender=SocialNetworkProfile) Is it correct way? I installed django-memcached-0.1.2, which show me: Memcached Server Stats Server Keys Hits Gets Hit_Rate Traffic_In Traffic_Out Usage Uptime 127.0.0.1 15 220 276 79% 83.1 KB 364.1 KB 18.4 KB 22:21:25 Can sombody explain what columns means? And last question. I have templates where I am getting much records from a few table (relationships). So in my view I get records from one table and in templates show it and related info from others. Generating page last a few seconds for very small table (<100records). Is it some easy way to cache queries from templates? Have I to do some big structure in my view (with all related tables), cache it and send to template?

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  • python destructuring-bind dictionary contents

    - by Stephen
    Hi, I am trying to 'destructure' a dictionary and associate values with variables names after its keys. Something like params = {'a':1,'b':2} a,b = params.values() but since dictionaries are not ordered, there is no guarantee that params.values() will return values in the order of (a,b). Is there a nice way to do this? Thanks

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  • Optimizing python link matching regular expression

    - by Matt
    I have a regular expression, links = re.compile('<a(.+?)href=(?:"|\')?((?:https?://|/)[^\'"]+)(?:"|\')?(.*?)>(.+?)</a>',re.I).findall(data) to find links in some html, it is taking a long time on certain html, any optimization advice? One that it chokes on is http://freeyourmindonline.net/Blog/

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  • Plotting a cumulative graph of python datetimes

    - by ventolin
    Say I have a list of datetimes, and we know each datetime to be the recorded time of an event happening. Is it possible in matplotlib to graph the frequency of this event occuring over time, showing this data in a cumulative graph (so that each point is greater or equal to all of the points that went before it), without preprocessing this list? (e.g. passing datetime objects directly to some wonderful matplotlib function) Or do I need to turn this list of datetimes into a list of dictionary items, such as: {"year": 1998, "month": 12, "date": 15, "events": 92} and then generate a graph from this list? Sorry if this seems like a silly question - I'm not all too familiar with matplotlib, and would like to save myself the effort of doing this the latter way if matplotlib can already deal with datetime objects itself.

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  • Extract anything that looks like links from large amount of data in python

    - by Riz
    Hi, I have around 5 GB of html data which I want to process to find links to a set of websites and perform some additional filtering. Right now I use simple regexp for each site and iterate over them, searching for matches. In my case links can be outside of "a" tags and be not well formed in many ways(like "\n" in the middle of link) so I try to grab as much "links" as I can and check them later in other scripts(so no BeatifulSoup\lxml\etc). The problem is that my script is pretty slow, so I am thinking about any ways to speed it up. I am writing a set of test to check different approaches, but hope to get some advices :) Right now I am thinking about getting all links without filtering first(maybe using C module or standalone app, which doesn't use regexp but simple search to get start and end of every link) and then using regexp to match ones I need.

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  • Easy Python input question

    - by Josh K
    I'd like to have something similar to the following pseudo code: while input is not None and timer < 5: input = getChar() timer = time.time() - start if timer >= 5: print "took too long" else: print input Anyway to do this without threading? I would like an input method that returns whatever has been entered since the last time it was called, or None (null) if nothing was entered.

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