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  • Javascript with Django?

    - by Rosarch
    I know this has been asked before, but I'm having a hard time setting up JS on my Django web app, even though I'm reading the documentation. I'm running the Django dev server. My file structure looks like this: mysite/ __init__.py MySiteDB manage.py settings.py urls.py myapp/ __init__.py admin.py models.py test.py views.py templates/ index.html Where do I want to put the Javascript and CSS? I've tried it in a bunch of places, including myapp/, templates/ and mysite/, but none seem to work. From index.html: <head> <title>Degree Planner</title> <script type="text/javascript" src="/scripts/JQuery.js"></script> <script type="text/javascript" src="/media/scripts/sprintf.js"></script> <script type="text/javascript" src="/media/scripts/clientside.js"></script> </head> From urls.py: (r'^admin/', include(admin.site.urls)), (r'^media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': 'media'}) (r'^.*', 'mysite.myapp.views.index'), I suspect that the serve() line is the cause of errors like: TypeError at /admin/auth/ 'tuple' object is not callable Just to round off the rampant flailing, I changed these settings in settings.py: MEDIA_ROOT = '/media/' MEDIA_URL = 'http://127.0.0.1:8000/media'

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  • Inside a decorator-class, access instance of the class which contains the decorated method

    - by ifischer
    I have the following decorator, which saves a configuration file after a method decorated with @saveconfig is called: class saveconfig(object): def __init__(self, f): self.f = f def __call__(self, *args): self.f(object, *args) # Here i want to access "cfg" defined in pbtools print "Saving configuration" I'm using this decorator inside the following class. After the method createkvm is called, the configuration object self.cfg should be saved inside the decorator: class pbtools() def __init__(self): self.configfile = open("pbt.properties", 'r+') # This variable should be available inside my decorator self.cfg = ConfigObj(infile = self.configfile) @saveconfig def createkvm(self): print "creating kvm" My problem is that i need to access the object variable self.cfg inside the decorator saveconfig. A first naive approach was to add a parameter to the decorator which holds the object, like @saveconfig(self), but this doesn't work. How can I access object variables of the method host inside the decorator? Do i have to define the decorator inside the same class to get access?

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  • Group Chat XMPP with Google App Engine

    - by David Shellabarger
    Google App Engine has a great XMPP service built in. One of the few limitations it has is that it doesn't support receiving messages from a group chat. That's the one thing I want to do with it. :( Can I run a 3rd party XMPP/Jabber server on App Engine that supports group chat? If so, which one?

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  • Reliable and fast way to convert a zillion ODT files in PDF?

    - by Marco Mariani
    I need to pre-produce a million or two PDF files from a simple template (a few pages and tables) with embedded fonts. Usually, I would stay low level in a case like this, and compose everything with a library like ReportLab, but I joined late in the project. Currently, I have a template.odt and use markers in the content.xml files to fill with data from a DB. I can smoothly create the ODT files, they always look rigth. For the ODT to PDF conversion, I'm using openoffice in server mode (and PyODConverter w/ named pipe), but it's not very reliable: in a batch of documents, there is eventually a point after which all the processed files are converted into garbage (wrong fonts and letters sprawled all over the page). Problem is not predictably reproducible (does not depend on the data), happens in OOo 2.3 and 3.2, in Ubuntu, XP, Server 2003 and Windows 7. My Heisenbug detector is ticking. I tried to reduce the size of batches and restarting OOo after each one; still, a small percentage of the documents are messed up. Of course I'll write about this on the Ooo mailing lists, but in the meanwhile, I have a delivery and lost too much time already. Where do I go? Completely avoid the ODT format and go for another template system. Suggestions? Anything that takes a few seconds to run is way too slow. OOo takes around a second and it sums to 15 days of processing time. I had to write a program for clustering the jobs over several clients. Keep the format but go for another tool/program for the conversion. Which one? There are many apps in the shareware or commercial repositories for windows, but trying each one is a daunting task. Some are too slow, some cannot be run in batch without buying it first, some cannot work from command line, etc. Open source tools tend not to reinvent the wheel and often depend on openoffice. Converting to an intermediate .DOC format could help to avoid the OOo bug, but it would double the processing time and complicate a task that is already too hairy. Try to produce the PDFs twice and compare them, discarding the whole batch if there's something wrong. Although the documents look equal, I know of no way to compare the binary content. Restart OOo after processing each document. it would take a lot more time to produce them it would lower the percentage of the wrong files, and make it very hard to identify them. Go for ReportLab and recreate the pages programmatically. This is the approach I'm going to try in a few minutes. Learn to properly format bulleted lists Thanks a lot.

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  • How to limit choice field options based on another choice field in django admin

    - by umnik700
    I have the following models: class Category(models.Model): name = models.CharField(max_length=40) class Item(models.Model): name = models.CharField(max_length=40) category = models.ForeignKey(Category) class Demo(models.Model): name = models.CharField(max_length=40) category = models.ForeignKey(Category) item = models.ForeignKey(Item) In the admin interface when creating a new Demo, after user picks category from the dropdown, I would like to limit the number of choices in the "items" drop-down. If user selects another category then the item choices should update accordingly. I would like to limit item choices right on the client, before it even hits the form validation on the server. This is for usability, because the list of items could be 1000+ being able to narrow it down by category would help to make it more manageable. Is there a "django-way" of doing it or is custom JavaScript the only option here?

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  • Django: How to create a model dynamically just for testing

    - by muhuk
    I have a Django app that requires a settings attribute in the form of: RELATED_MODELS = ('appname1.modelname1.attribute1', 'appname1.modelname2.attribute2', 'appname2.modelname3.attribute3', ...) Then hooks their post_save signal to update some other fixed model depending on the attributeN defined. I would like to test this behaviour and tests should work even if this app is the only one in the project (except for its own dependencies, no other wrapper app need to be installed). How can I create and attach/register/activate mock models just for the test database? (or is it possible at all?) Solutions that allow me to use test fixtures would be great.

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  • django admin site make CharField a PasswordInput

    - by Paul
    I have a Django site in which the site admin inputs their Twitter Username/Password in order to use the Twitter API. The Model is set up like this: class TwitterUser(models.Model): screen_name = models.CharField(max_length=100) password = models.CharField(max_length=255) def __unicode__(self): return self.screen_name I need the Admin site to display the password field as a password input, but can't seem to figure out how to do it. I have tried using a ModelAdmin class, a ModelAdmin with a ModelForm, but can't seem to figure out how to make django display that form as a password input...

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  • Keeping track of changes - Django

    - by RadiantHex
    Hi folks!! I have various models of which I would like to keep track and collect statistical data. The problem is how to store the changes throughout time. I thought of various alternative: Storing a log in a TextField, open it and update it every time the model is saved. Alternatively pickle a list and store it in a TextField. Save logs on hard drive. What are your suggestions?

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  • Stream a file to the HTTP response in Pylons

    - by Evgeny
    I have a Pylons controller action that needs to return a file to the client. (The file is outside the web root, so I can't just link directly to it.) The simplest way is, of course, this: with open(filepath, 'rb') as f: response.write(f.read()) That works, but it's obviously inefficient for large files. What's the best way to do this? I haven't been able to find any convenient methods in Pylons to stream the contents of the file. Do I really have to write the code to read a chunk at a time myself from scratch?

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  • Graphing a line and scatter points using Matplotlib?

    - by Patrick O'Doherty
    Hi guys I'm using matplotlib at the moment to try and visualise some data I am working on. I'm trying to plot around 6500 points and the line y = x on the same graph but am having some trouble in doing so. I can only seem to get the points to render and not the line itself. I know matplotlib doesn't plot equations as such rather just a set of points so I'm trying to use and identical set of points for x and y co-ordinates to produce the line. The following is my code from matplotlib import pyplot import numpy from pymongo import * class Store(object): """docstring for Store""" def __init__(self): super(Store, self).__init__() c = Connection() ucd = c.ucd self.tweets = ucd.tweets def fetch(self): x = [] y = [] for t in self.tweets.find(): x.append(t['positive']) y.append(t['negative']) return [x,y] if __name__ == '__main__': c = Store() array = c.fetch() t = numpy.arange(0., 0.03, 1) pyplot.plot(array[0], array[1], 'ro', t, t, 'b--') pyplot.show() Any suggestions would be appreciated, Patrick

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  • SQL Alchemy MVC and cross controller joins

    - by Khorkrak
    When using SQL Alchemy for abstracting your data access layer and using controllers as the way to access objects from that abstraction layer, how should joins be handled? So for example, say you have an Orders controller class that manages Order objects such that it provides getOrder, saveOrder, etc methods and likewise a similar controller for User objects. First of all do you even need these controllers? Should you instead just treat SQL Alchemy as "the" thing for handling data access. Why bother with object oriented controller stuff there when you instead have a clean declarative way to obtain and persist objects without having to write SQL directly either. Well one reason could be that perhaps you may want to replace SQL Alchemy with direct SQL or Storm or whatever else. So having controller classes there to act as an intermediate layer helps limit what would need to change then. Anyway - back to the main question - so assuming you have these two controllers, now lets say you want the list of orders for a certain set of users meeting some criteria. How do you go about doing this? Generally you don't want the controllers crossing domains - the Orders controllers knows only about Orders and the User controller just about Users - they don't mess with each other. You also don't want to go fetch all the Users that match and then feed a big list of user ids to the Orders controller to go find the matching Orders. What's needed is a join. Here's where I'm stuck - that seems to mean either the controllers must cross domains or perhaps they should be done away with altogether and you simply do the join via SQL Alchemy directly and get the resulting User and / or Order objects as needed. Thoughts?

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  • IOError: [Errno 13] Permission denied when trying to read an file in google app engine

    - by mahesh
    I want to read an XML file and parse it, for that I had used SAX parser which requires file as input to parse. For that I had stored my XML file in Entity called XMLDocs with following property XMLDocs Entity Name name : Property of string type content : property of blob type (will contain my xml file) Reason I had to store file like this as I had not yet provide my billing detail to google Now when I try to open this file in my I am getting error of permission denied.. Please help me, what I have to do... You can see that error by running my app at www.parsepython.appspot.com

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  • Pyplot connect to timer event?

    - by Baron Yugovich
    The same way I now have plt.connect('button_press_event', self.on_click) I would like to have something like plt.connect('each_five_seconds_event', self.on_timer) How can I achieve this in a way that's most similar to what I've shown above? EDIT: I tried fig = plt.subplot2grid((num_cols, num_rows), (col, row), rowspan=rowspan, colspan=colspan) timer = fig.canvas.new_timer(interval=100, callbacks=[(self.on_click)]) timer.start() And got AttributeError: 'AxesSubplot' object has no attribute 'canvas'

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  • Fastest way to generate delimited string from 1d numpy array

    - by Abiel
    I have a program which needs to turn many large one-dimensional numpy arrays of floats into delimited strings. I am finding this operation quite slow relative to the mathematical operations in my program and am wondering if there is a way to speed it up. For example, consider the following loop, which takes 100,000 random numbers in a numpy array and joins each array into a comma-delimited string. import numpy as np x = np.random.randn(100000) for i in range(100): ",".join(map(str, x)) This loop takes about 20 seconds to complete (total, not each cycle). In contrast, consider that 100 cycles of something like elementwise multiplication (x*x) would take than one 1/10 of a second to complete. Clearly the string join operation creates a large performance bottleneck; in my actual application it will dominate total runtime. This makes me wonder, is there a faster way than ",".join(map(str, x))? Since map() is where almost all the processing time occurs, this comes down to the question of whether there a faster to way convert a very large number of numbers to strings.

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  • Convert list to sequence of variables

    - by wtzolt
    I was wondering if this was possible... I have a sequence of variables that have to be assigned to a do.something (a, b) a and b variables accordingly. Something like this: # # Have a list of sequenced variables. list = 2:90 , 1:140 , 3:-40 , 4:60 # # "Template" on where to assign the variables from the list. do.something (a,b) # # Assign the variables from the list in a sequence with possibility of "in between" functions like print and time.sleep() added. do.something (2,90) time.sleep(1) print "Did something (%d,%d)" % (# # vars from list?) do.something (1,140) time.sleep(1) print "Did something (%d,%d)" % (# # vars from list?) do.something (3,-40) time.sleep(1) print "Did something (%d,%d)" % (# # vars from list?) do.something (4,60) time.sleep(1) print "Did something (%d,%d)" % (# # vars from list?) Any ideas?

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  • How to make socket.listen(1) work for some time and then continue rest of code???

    - by Rami Jarrar
    I'm making server that make a tcp socket and work over port range, with each port it will listen on that port for some time, then continue the rest of the code. like this:: import socket sck = socket.socket(socket.AF_INET, socket.SOCK_STREAM) sck.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1) msg ='' ports = [x for x in xrange(4000)] while True: try: for i in ports: sck.bind(('',i)) ## sck.listen(1) ## make it just for some time and then continue this ## if there a connection do this conn, addr = sck.accept() msg = conn.recv(2048) ## do something ##if no connection continue the for loop conn.close() except KeyboardInterrupt: exit() so how i could make sck.listen(1) work just for some time ??

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  • How to write program to do file transfer based on based omniORBpy

    - by cofthew7
    I'm now writing a Corba project to do file transfering between client and server. But I face trouble when I want to upload file from the client to the server. The IDL I defined is: interface SecretMessage { string send_file(in string file_name, in string file_obj); }; And I implemented the uploading function in the client code: f = open('SB.docx', 'rb') data = '' for piece in read_in_chunks(f): data += piece result = mo.send_file('2.docx', data) If the file is a plain txt file, there is no problem. But if the file is a, like jpg, doc, or others except txt, then it does work. It gives me the error: omniORB.CORBA.BAD_PARAM: CORBA.BAD_PARAM(omniORB.BAD_PARAM_WrongPythonType, CORBA.COMPLETED_NO) Where is the problem?

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  • Google App Engine + Form Validation

    - by Iwona
    Hi, I would like to do google app engine form validation but I dont know how to do it? I tried like this: from google.appengine.ext.db import djangoforms from django import newforms as forms class SurveyForm(forms.Form): occupations_choices = ( ('1', ""), ('2', "Undergraduate student"), ('3', "Postgraduate student (MSc)"), ('4', "Postgraduate student (PhD)"), ('5', "Lab assistant"), ('6', "Technician"), ('7', "Lecturer"), ('8', "Other" ) ) howreach_choices = ( ('1', ""), ('2', "Typed the URL directly"), ('3', "Site is bookmarked"), ('4', "A search engine"), ('5', "A link from another site"), ('6', "From a book"), ('7', "Other") ) boxes_choices = ( ("des", "Website Design"), ("svr", "Web Server Administration"), ("com", "Electronic Commerce"), ("mkt", "Web Marketing/Advertising"), ("edu", "Web-Related Education") ) name = forms.CharField(label='Name', max_length=100, required=True) email = forms.EmailField(label='Your Email Address:') occupations = forms.ChoiceField(choices=occupations_choices, label='What is your occupation?') howreach = forms.ChoiceField(choices=howreach_choices, label='How did you reach this site?') # radio buttons 1-5 rating = forms.ChoiceField(choices=range(1,6), label='What is your occupation?', widget=forms.RadioSelect) boxes = forms.ChoiceField(choices=boxes_choices, label='Are you involved in any of the following? (check all that apply):', widget=forms.CheckboxInput) comment = forms.CharField(widget=forms.Textarea, required=False) And I wanted to display it like this: template_values = { 'url' : url, 'url_linktext' : url_linktext, 'userName' : userName, 'item1' : SurveyForm() } And I have this error message: Traceback (most recent call last): File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp_init_.py", line 515, in call handler.get(*groups) File "C:\Program Files\Google\google_appengine\demos\b00213576\main.py", line 144, in get self.response.out.write(template.render(path, template_values)) File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 143, in render return t.render(Context(template_dict)) File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 183, in wrap_render return orig_render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 168, in render return self.nodelist.render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 705, in render bits.append(self.render_node(node, context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 718, in render_node return(node.render(context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template\defaulttags.py", line 209, in render return self.nodelist_true.render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 705, in render bits.append(self.render_node(node, context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 718, in render_node return(node.render(context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 768, in render return self.encode_output(output) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 757, in encode_output return str(output) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\util.py", line 26, in str return self.unicode().encode(settings.DEFAULT_CHARSET) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 73, in unicode return self.as_table() File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 144, in as_table return self._html_output(u'%(label)s%(errors)s%(field)s%(help_text)s', u'%s', '', u'%s', False) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 129, in _html_output output.append(normal_row % {'errors': bf_errors, 'label': label, 'field': unicode(bf), 'help_text': help_text}) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 232, in unicode value = value.str() File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\util.py", line 26, in str return self.unicode().encode(settings.DEFAULT_CHARSET) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 246, in unicode return u'\n%s\n' % u'\n'.join([u'%s' % w for w in self]) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 238, in iter yield RadioInput(self.name, self.value, self.attrs.copy(), choice, i) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 212, in init self.choice_value = smart_unicode(choice[0]) TypeError: 'int' object is unsubscriptable Do You have any idea how I can do this validation in different case? I have tried to do it using this kind of: class ItemUserAnswer(djangoforms.ModelForm): class Meta: model = UserAnswer But I dont know how to add extra labels to this form and it is displayed in one line. Do You have any suggestions? Thanks a lot as it making me crazy why it is still not working:/

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  • Using PyLab to create a 2D graph from two separate lists

    - by user324333
    Hey Guys, This seems like a basic problem with an easy answer but I simply cannot figure it out no matter how much I try. I am trying to create a line graph based on two lists. For my x-axis, I want my list to be a set of strings. x_axis_list = ["Jan-06","Jul-06","Jan-07","Jul-07","Jan-08"] y_axis_list = [5,7,6,8,9] Any suggestions on how to best graph these items?

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  • How to update the contents of a FigureCanvasTkAgg

    - by Copo
    I'm plotting some data in a Tkinter FigureCanvasTkagg using matplotlib. I need to clear the figure where i plot data and draw new data when a button is pressed. here is the plotting part of the code (there's an App class defined before..) self.fig = figure() self.ax = self.fig.add_subplot(111) self.ax.set_ylim( min(y), max(y) ) self.line, = self.ax.semilogx(x,y,'.-') #tuple of a single element self.canvas = FigureCanvasTkAgg(self.fig,master=master) self.ax.semilogx(x,y,'o-') self.canvas.show() self.canvas.get_tk_widget().pack(side='top', fill='both', expand=1) self.frame.pack() how do i update the contents of such a canvas? regards, Jacopo

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