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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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  • Display additional data while iterating over a Django formset

    - by Jannis
    Hi, I have a list of soccer matches for which I'd like to display forms. The list comes from a remote source. matches = ["A vs. B", "C vs. D", "E vs, F"] matchFormset = formset_factory(MatchForm,extra=len(matches)) formset = MatchFormset() On the template side, I would like to display the formset with the according title (i.e. "A vs. B"). {% for form in formset.forms %} <fieldset> <legend>{{TITLE}}</legend> {{form.team1}} : {{form.team2}} </fieldset> {% endfor %} Now how do I get TITLE to contain the right title for the current form? Or asked in a different way: how do I iterate over matches with the same index as the iteration over formset.forms? Thanks for your input!

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  • stdout and stderr anomalies

    - by momo
    from the interactive prompt: >>> import sys >>> sys.stdout.write('is the') is the6 what is '6' doing there? another example: >>> for i in range(3): ... sys.stderr.write('new black') ... 9 9 9 new blacknew blacknew black where are the numbers coming from?

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  • wx Menu disappears from frame when shown as a popup

    - by Adam Fraser
    I'm trying to create a wx.Menu that will be shared between a popup (called on right-click), and a sub menu accessible from the frame menubar. The following code demonstrates the problem. If you open the "MENUsubmenu" from the menubar the item "asdf" is visible. If you right click on the frame content area, "asdf" will be visible from there as well... however, returning to the menubar, you will find that "MENUsubmenu" is vacant. Why is this happening and how can I fix it? import wx app = wx.PySimpleApp() m = wx.Menu() m.Append(-1, 'asdf') def show_popup(evt): ''' R-click callback ''' f.PopupMenu(m, (evt.X, evt.Y)) f = wx.Frame(None) f.SetMenuBar(wx.MenuBar()) frame_menu = wx.Menu() f.MenuBar.Append(frame_menu, 'MENU') frame_menu.AppendMenu(-1,'submenu', m) f.Show() f.Bind(wx.EVT_RIGHT_DOWN, show_popup) app.MainLoop() Interestingly, appending the menu to MenuBar works, but is not the behavior I want: import wx app = wx.PySimpleApp() m = wx.Menu() m.Append(-1, 'asdf') def show_popup(evt): f.PopupMenu(m, (evt.X, evt.Y)) f = wx.Frame(None) f.SetMenuBar(wx.MenuBar()) f.MenuBar.Append(m, 'MENU') f.Show() f.Bind(wx.EVT_RIGHT_DOWN, show_popup) app.MainLoop()

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  • JSON serialization of Google App Engine models

    - by user111677
    I've been search for quite a while with no success. My project isn't using Django, is there a simple way to serialize App Engine models (google.appengine.ext.db.Model) into JSON or do I need to write my own serializer? My model class is fairly simple. For instance: class Photo(db.Model): filename = db.StringProperty() title = db.StringProperty() description = db.StringProperty(multiline=True) date_taken = db.DateTimeProperty() date_uploaded = db.DateTimeProperty(auto_now_add=True) album = db.ReferenceProperty(Album, collection_name='photo') Thanks in advance.

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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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  • VS2010 patce: why it's take so muce time to install it?

    - by Mendy
    Visual Studio 2010 RC has a few of patches release. For more information about them take a look here. What I'm expect from patch program, is to replace a few dll's of the program to a new fixed version of them. But when I run each of this 3 patches, they take a lot of time (5 minutes each), and you think that the program was frozen because the progress bar stay on the begging. This is question may not be so important, but it really interesting me to know, why this happens? It's really confusing to see that each VS2010 (or Microsoft in general) is frozen to 4-5 minutes.

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  • How to call Twiter's Streaming/Filter Feed with urllib2/httplib?

    - by Simon
    Update: I switched this back from answered as I tried the solution posed in cogent Nick's answer and switched to Google's urlfetch: logging.debug("starting urlfetch for http://%s%s" % (self.host, self.url)) result = urlfetch.fetch("http://%s%s" % (self.host, self.url), payload=self.body, method="POST", headers=self.headers, allow_truncated=True, deadline=5) logging.debug("finished urlfetch") but unfortunately finished urlfetch is never printed - I see the timeout happen in the logs (it returns 200 after 5 seconds), but execution doesn't seem tor return. Hi All- I'm attempting to play around with Twitter's Streaming (aka firehose) API with Google App Engine (I'm aware this probably isn't a great long term play as you can't keep the connection perpetually open with GAE), but so far I haven't had any luck getting my program to actually parse the results returned by Twitter. Some code: logging.debug("firing up urllib2") req = urllib2.Request(url="http://%s%s" % (self.host, self.url), data=self.body, headers=self.headers) logging.debug("called urlopen for %s %s, about to call urlopen" % (self.host, self.url)) fobj = urllib2.urlopen(req) logging.debug("called urlopen") When this executes, unfortunately, my debug output never shows the called urlopen line printed. I suspect what's happening is that Twitter keeps the connection open and urllib2 doesn't return because the server doesn't terminate the connection. Wireshark shows the request being sent properly and a response returned with results. I tried adding Connection: close to my request header, but that didn't yield a successful result. Any ideas on how to get this to work? thanks -Simon

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  • socket.shutdown vs socket.close

    - by Jason Baker
    I recently saw a bit of code that looked like this (with sock being a socket object of course): sock.shutdown(socket.SHUT_RDWR) sock.close() What exactly is the purpose of calling shutdown on the socket and then closing it? If it makes a difference, this socket is being used for non-blocking IO.

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  • cannot output a json encoded dict containing accents (noob inside)

    - by user296546
    Hi all, here is a fairly simple example wich is driving me nuts since a couple of days. Considering the following script: # -*- coding: utf-8 -* from json import dumps as json_dumps machaine = u"une personne émérite" print(machaine) output = {} output[1] = machaine jsonoutput = json_dumps(output) print(jsonoutput) The result of this from cli: une personne émérite {"1": "une personne \u00e9m\u00e9rite"} I don't understand why their such a difference between the two strings. i have been trying all sorts of encode, decode etc but i can't seem to be able to find the right way to do it. Does anybody has an idea ? Thanks in advance. Matthieu

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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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  • Added tagging to existing model, now how does its admin work?

    - by Oli
    I wanted to add a StackOverflow-style tag input to a blog model of mine. This is a model that has a lot of data already in it. class BlogPost(models.Model): # my blog fields try: tagging.register(BlogPost) except tagging.AlreadyRegistered: pass I thought that was all I needed so I went through my old database of blog posts (this is a newly ported blog) and copied the tags in. It worked and I could display tags and filter by tag. However, I just wrote a new BlogPost and realise there's no tag field there. Reading the documentation (coincidentally, dry enough to be used as an antiperspirant), I found the TagField. Thinking this would just be a manager-style layer over the existing tagging register, I added it. It complained about there not being a Tag column. I'd rather not denormalise on tags just to satisfy create an interface for inputting them. Is there a TagManager class that I can just set on the model? tags = TagManager() # or somesuch

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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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  • Preserve time stamp when shrinking an image

    - by Ckhrysze
    My digital camera takes pictures with a very high resolution, and I have a PIL script to shrink them to 800x600 (or 600x800). However, it would be nice for the resultant file to retain the original timestamp. I noticed in the docs that I can use a File object instead of a name in PIL's image save method, but I don't know if that will help or not. My code is basically name, ext = os.path.splitext(filename) # open an image file (.bmp,.jpg,.png,.gif) you have in the working folder image = Image.open(filename) width = 800 height = 600 w, h = image.size if h > w: width = 600 height = 800 name = name + ".jpg" shunken = image.resize((width, height), Image.ANTIALIAS) shunken.save(name) Thank you for any help you can give!

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  • Perceptron Classification and Model Training

    - by jake pinedo
    I'm having an issue with understanding how the Perceptron algorithm works and implementing it. cLabel = 0 #class label: corresponds directly with featureVectors and tweets for m in range(miters): for point in featureVectors: margin = answers[cLabel] * self.dot_product(point, w) if margin <= 0: modifier = float(lrate) * float(answers[cLabel]) modifiedPoint = point for x in modifiedPoint: if x != 0: x *= modifier newWeight = [modifiedPoint[i] + w[i] for i in range(len(w))] w = newWeight self._learnedWeight = w This is what I've implemented so far, where I have a list of class labels in answers and a learning rate (lrate) and a list of feature vectors. I run it for the numbers of iterations in miter and then get the final weight at the end. However, I'm not sure what to do with this weight. I've trained the perceptron and now I have to classify a set of tweets, but I don't know how to do that. EDIT: Specifically, what I do in my classify method is I go through and create a feature vector for the data I'm given, which isn't a problem at all, and then I take the self._learnedWeight that I get from the earlier training code and compute the dot-product of the vector and the weight. My weight and feature vectors include a bias in the 0th term of the list so I'm including that. I then check to see if the dotproduct is less than or equal to 0: if so, then I classify it as -1. Otherwise, it's 1. However, this doesn't seem to be working correctly.

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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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  • do I need to use partial?

    - by wiso
    I've a general function, for example (only a simplified example): def do_operation(operation, a, b, name): print name do_something_more(a,b,name, operation(a,b)) def operation_x(a,b): return a**2 + b def operation_y(a,b): return a**10 - b/2. and some data: data = {"first": {"name": "first summation", "a": 10, "b": 20, "operation": operation_x}, "second": {"name": "second summation", "a": 20, "b": 50, "operation": operation_y}, "third": {"name": "third summation", "a": 20, "b": 50, "operation": operation_x}, # <-- operation_x again } now I can do: what_to_do = ("first", "third") # this comes from command line for sum_id in what_to_do: do_operation(data["operation"], data["a"], data["b"], data["name"]) or maybe it's better if I use functools.partial? from functools import partial do_operation_one = do_operation(name=data["first"]["name"], operation=data["first"]["operation"], a=data["first"]["a"], b=data["first"]["b"]) do_operation_two = do_operation(name=data["second"]["name"], operation=data["second"]["operation"] a=data["second"]["a"], b=data["second"]["b"]) do_operation_three = do_operation(name=data["third"]["name"], operation=data["third"]["operation"] a=data["third"]["a"], b=data["third"]["b"]) do_dictionary = { "first": do_operation_one, "second": do_operation_two, "third": do_operation_three } for what in what_to_do: do_dictionary[what]()

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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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