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  • command line arg?

    - by kaushik
    This is a module named XYZ. def func(x) ..... ..... if __name__=="__main__": print func(sys.argv[1]) Now I have imported this module in another code and want to use the func. How can i use it? import XYZ After this, where to give the argument, and syntax on how to call it, please?

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  • Sqlalchemy: Many to Many relationship error

    - by 1001010101
    Dear everyone, I am following the Many to many relationship described on http://www.sqlalchemy.org/docs/mappers.html#many-to-many #This is actually a VIEW tb_mapping_uGroups_uProducts = Table( 'mapping_uGroups_uProducts', metadata, Column('upID', Integer, ForeignKey('uProductsInfo.upID')), Column('ugID', Integer, ForeignKey('uGroupsInfo.ugID')) ) tb_uProducts = Table( 'uProductsInfo', metadata, Column('upID', Integer, primary_key=True) ) mapper( UnifiedProduct, tb_uProducts) tb_uGroupsInfo = Table( 'uGroupsInfo', metadata, Column('ugID', Integer, primary_key=True) ) mapper( UnifiedGroup, tb_uGroupsInfo, properties={ 'unifiedProducts': relation(UnifiedProduct, secondary=tb_mapping_uGroups_uProducts, backref="unifiedGroups") }) where the relationship between uProduct and uGroup are N:M. When I run the following sess.query(UnifiedProduct).join(UnifiedGroup).distinct()[:10] I am getting the error: sqlalchemy.exc.ArgumentError: Can't find any foreign key relationships between 'uProductsInfo' and 'uGroupsInfo' What am I doing wrong?

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  • delete common dictionaries in list based on a value

    - by pythoonatic
    How would I delete all corresponding dictionaries in a list of dictionaries based on one of the dictionaries having a character in it. data = [ { 'x' : 'a', 'y' : '1' }, { 'x' : 'a', 'y' : '1/1' }, { 'x' : 'a', 'y' : '2' }, { 'x' : 'b', 'y' : '1' }, { 'x' : 'b', 'y' : '1' }, { 'x' : 'b', 'y' : '1' }, ] For example, how would I delete all of the x = a due to one of the y in the x=a having a / in it? Based on the example data above, here is where I would like to get to: cleaneddata = [ { 'x' : 'b', 'y' : '1' }, { 'x' : 'b', 'y' : '1' }, { 'x' : 'b', 'y' : '1' }, ]

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  • How to merge or copy anonymous session data into user data when user logs in?

    - by benhoyt
    This is a general question, or perhaps a request for pointers to other open source projects to look at: I'm wondering how people merge an anonymous user's session data into the authenticated user data when a user logs in. For example, someone is browsing around your websites saving various items as favourites. He's not logged in, so they're saved to an anonymous user's data. Then he logs in, and we need to merge all that data into their (possibly existing) user data. Is this done different ways in an ad-hoc fashion for different applications? Or are there some best practices or other projects people can direct me to?

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  • does webapp has 'elseif' or 'elif' in template tags..

    - by zjm1126
    my code is : Hello!~~~ {% if user %} <p>Logged in as {{ user.first_name }} {{ user.last_name }}.</p> {% elif openid_user%} <p>Hello, {{openid_user.nickname}}! Do you want to <a href="{{openid_logout_url}}">Log out?</p> {% else %} <p><a href="/login?redirect={{ current_url }}">google Log in</a>.</p> <p><a href="/twitter">twitter Log in</a>.</p> <p><a href="/facebook">facebook Log in</a>.</p> <p><a href="{{openid_login_url}}">openid Log in</a>.</p> <iframe src="/_openid/login?continue=/"></iframe> {% endif %} the error is : TemplateSyntaxError: Invalid block tag: 'elif' does not webapp has a 'else if ' ? thanks

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  • How to compare two lists with duplicated items in one list?

    - by eladc
    I need to compare list_a against many others. my problem starts when there's a duplicated item in the other lists (two k's in other_b). my goal is to filter out all the lists with the same items (up to three matching items). list_a = ['j','k','a','7'] other_b = ['k', 'j', 'k', 'q'] other_c = ['k','k','9','k'] >>>filter(lambda x: not x in list_a,other_b) ['q'] I need a way that would return ['k', 'q'], because 'k' appears only once in list_a. comparing list_a and other_c with set() isn't good for my purpose since it will return only one element: k. while I need ['k','9','k'] I hope I was clear enough. Thank you

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  • Aggregation over a few models - Django

    - by RadiantHex
    Hi folks, I'm trying to compute the average of a field over various subsets of a queryset. Player.objects.order_by('-score').filter(sex='male').aggregate(Avg('level')) This works perfectly! But... if I try to compute it for the top 50 players it does not work. Player.objects.order_by('-score').filter(sex='male')[:50].aggregate(Avg('level')) This last one returns the exact same result as the query above it, which is wrong. What am I doing wrong? Help would be very much appreciated!

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  • muti user dungeon help

    - by mudman
    ive created a single user dungeon which i would like to create into a multi user dungoen so at least two plays can play how would i do that what code do i need to add can anyone help? i would show coding but if i do then everyone would see it and all my work will be copied as i know other students do use this site to so plz understand my situation and yes this is a homework/assignment work.

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  • nested list comprehension using intermediate result

    - by KentH
    I am trying to grok the output of a function which doesn't have the courtesy of setting a result code. I can tell it failed by the "error:" string which is mixed into the stderr stream, often in the middle of a different conversion status message. I have the following list comprehension which works, but scans for the "error:" string twice. Since it is only rescanning the actual error lines, it works fine, but it annoys me I can't figure out how to use a single scan. Here's the working code: errors = [e[e.find('error:'):] for e in err.splitlines() if 'error:' in e] The obvious (and wrong) way to simplify is to save the "find" result errors = [e[i:] for i in e.find('error:') if i != -1 for e in err.splitlines()] However, I get "UnboundLocalError: local variable 'e' referenced before assignment". Blindly reversing the 'for's in the comprehension also fails. How is this done? THanks. Kent

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  • How to attach a line to a moving object?

    - by snow-spur
    Hello i have designed a maze and i want to draw a path between the cells as the 'person' moves from one cell to the next. So each time i move the cell a line is drawn I have done this so far but do not want to show my full code However i get an error saying Circle has no attribute center my circle which is my cell center = Point(15, 15) c = Circle(center, 12) c.setFill('blue') c.setOutline('yellow') c.draw(win) p1 = Point(c.center().getx(), c.center().gety()) this bit is in my loop p2 = Point(getx(), gety()) line = graphics.Line(p1, p2)

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  • a function that returns a random number that is a multiple of 3 between 0 and the function's non-negative integer parameter n

    - by martin
    I need to write a function called multipleOf3 that returns a random number that is a multiple of 3 between 0 and the function's non-negative integer parameter n and here is the result i want [Note: No number returned can be greater than the value of the parameter n] Examples: multipleOf3(0) -- 0 multipleOf3(1) -- 0 multipleOf3(2) -- 0 multipleOf3(3) -- 0 or 3 multipleOf3(20) -- 0 or 3 or 6 or 9 or 12 or 15 or 18

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  • Using classes for the first time,help in debugging

    - by kaushik
    here is post my code:this is no the entire code but enough to explain my doubt.please discard any code line which u find irrelavent enter code here saving_tree={} isLeaf=False class tree: global saving_tree rootNode=None lispTree=None def __init__(self,x): file=x string=file.readlines() #print string self.lispTree=S_expression(string) self.rootNode=BinaryDecisionNode(0,'Root',self.lispTree) class BinaryDecisionNode: global saving_tree def __init__(self,ind,name,lispTree,parent=None): self.parent=parent nodes=lispTree.getNodes(ind) print nodes self.isLeaf=(nodes[0]==1) nodes=nodes[1]#Nodes are stored self.name=name self.children=[] if self.isLeaf: #Leaf Node print nodes #Set the leaf data self.attribute=nodes print "LeafNode is ",nodes else: #Set the question self.attribute=lispTree.getString(nodes[0]) self.attribute=self.attribute.split() print "Question: ",self.attribute,self.name tree={} tree={str(self.name):self.attribute} saving_tree=tree #Add the children for i in range(1,len(nodes)):#Since node 0 is a question # print "Adding child ",nodes[i]," who has ",len(nodes)-1," siblings" self.children.append(BinaryDecisionNode(nodes[i],self.name+str(i),lispTree,self)) print saving_tree i wanted to save some data in saving_tree{},which i have declared previously and want to use that saving tree in the another function outside the class.when i asked to print saving_tree it printing but,only for that instance.i want the saving_tree{} to have the data to store data of all instance and access it outside. when i asked for print saving_tree outside the class it prints empty{}.. please tell me the required modification to get my required output and use saving_tree{} outside the class..

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  • Crossed import in django

    - by Kuhtraphalji
    On example, i have 2 apps: alpha and beta in alpha/models.py import of model from beta.models and in beta/models.py import of model from alpha.models manage.py validate says that ImportError: cannot import name ModelName how to solve this problem?

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  • Save JSON outputed from a URL to a file

    - by Aidan
    Hey Guys, How would I save JSON outputed by an URL to a file? e.g from the Twitter search API (this http://search.twitter.com/search.json?q=hi) Language isn't important. Thanks! edit // How would I then append further updates to EOF?

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  • list in loop, Nonetype errors

    - by user2926755
    Here is my Code def printList(stringlist): empty = [] if stringlist is None: print empty else: print stringlist def add (stringlist, string): string = [] if string is None else string if stringlist is not None: stringlist.insert(0, string) else: stringlist.append(1) it somehow appears "AttributeError: 'NoneType' object has no attribute 'append'" I was originally looking for the code to be run like this: >>> myList = None >>> printList(myList) [] >>> for word in ['laundry','homework','cooking','cleaning']: myList = add(myList, word) printList(myList) [laundry] [homework, laundry] [cooking, homework, laundry] [cleaning, cooking, homework, laundry]

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  • Display graph without saving using pydot

    - by user506710
    Hello all I am trying to display a simple graph using pydot. My question is that is there any way to display the graph without writing it to a file as currently I use write function to first draw and then have to use the Image module to show the files. However is there any way that the graph directly gets printed on the screen without being saved ?? Also as an update I would like to ask in this same question that I observe that while the image gets saved very quickly when I use the show command of the Image module it takes noticeable time for the image to be seen .... Also sometimes I get the error that the image could'nt be opened because it was either deleted or saved in unavailable location which is not correct as I am saving it at my Desktop..... Does anyone know what's happening and is there a faster way to get the image loaded..... Thanks a lot....

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  • Are there viable alternatives for Web 2.0 apps besides lots of Javascript?

    - by djembe
    If you say find C-style syntax to be in the axis of evil are you just hopelessly condemned to suck it up and deal with it if you want to provide your users with cool web 2.0 applications - for example stuff that's generally done using JQuery and Ajax etc? Are there no other choices out there? We're currently building intranet apps using pylons and a bunch of JavaScript along with a bit of Evoque. So obviously for us the world would be a better place if instead something equivalent existed written in like PythonScript. But I've yet to seen anything approaching that aside from the Android system's ASE - but obviously that's something rather unrelated. Still - if browsers could support other scripting languages....

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