Search Results

Search found 13693 results on 548 pages for 'python metaprogramming'.

Page 373/548 | < Previous Page | 369 370 371 372 373 374 375 376 377 378 379 380  | Next Page >

  • Attribute Error in django

    - by itsandy
    Hi all, I am having an attribute error while working with django-registration it says 'NoneType' object has no attribute 'strip' I dropped my db table and created again but the error doesnt go..can anyone help..

    Read the article

  • Obtaining references to function objects on the execution stack from the frame object?

    - by Marcin
    Given the output of inspect.stack(), is it possible to get the function objects from anywhere from the stack frame and call these? If so, how? (I already know how to get the names of the functions.) Here is what I'm getting at: Let's say I'm a function and I'm trying to determine if my caller is a generator or a regular function? I need to call inspect.isgeneratorfunction() on the function object. And how do you figure out who called you? inspect.stack(), right? So if I can somehow put those together, I'll have the answer to my question. Perhaps there is an easier way to do this?

    Read the article

  • matplotlib.pyplot, preserve aspect ratio of the plot

    - by Headcrab
    Assuming we have a polygon coordinates as polygon = [(x1, y1), (x2, y2), ...], the following code displays the polygon: import matplotlib.pyplot as plt plt.fill(*zip(*polygon)) plt.show() By default it is trying to adjust the aspect ratio so that the polygon (or whatever other diagram) fits inside the window, and automatically changing it so that it fits even after resizing. Which is great in many cases, except when you are trying to estimate visually if the image is distorted. How to fix the aspect ratio to be strictly 1:1? (Not sure if "aspect ratio" is the right term here, so in case it is not - I need both X and Y axes to have 1:1 scale, so that (0, 1) on both X and Y takes an exact same amount of screen space. And I need to keep it 1:1 no matter how I resize the window.)

    Read the article

  • socket.accept error 24: To many open files

    - by Creotiv
    I have a problem with open files under my Ubuntu 9.10 when running server in Python2.6 And main problem is that, that i don't know why it so.. I have set ulimit -n = 999999 net.core.somaxconn = 999999 fs.file-max = 999999 and lsof gives me about 12000 open files when server is running. And also i'm using epoll. But after some time it's start giving exeption: File "/usr/lib/python2.6/socket.py", line 195, in accept error: [Errno 24] Too many open files And i don't know how it can reach file limit when it isn't reached. Thanks for help)

    Read the article

  • Performing non-blocking requests? - Django

    - by RadiantHex
    Hi folks, I have been playing with other frameworks, such as NodeJS, lately. I love the possibility to return a response, and still being able to do further operations. e.g. def view(request): do_something() return HttpResponse() do_more_stuff() #not possible!!! Maybe Django already offers a way to perform operations after returning a request, if that is the case that would be great. Help would be very much appreciated! =D

    Read the article

  • how to speed up the code??

    - by kaushik
    i have very huge code about 600 lines plus. cant post the whole thing here. but a particular code snippet is taking so much time,leading to problems. here i post that part of code please tell me what to do speed up the processing.. please suggest the part which may be the reason and measure to improve them if this small part of code is understandable. using_data={} def join_cost(a , b): global using_data #print a #print b save_a=[] save_b=[] print 1 #for i in range(len(m)): #if str(m[i][0])==str(a): save_a=database_index[a] #for i in range(len(m)): # if str(m[i][0])==str(b): #print 'save_a',save_a #print 'save_b',save_b print 2 save_b=database_index[b] using_data[save_a[0]]=save_a s=str(save_a[1]).replace('phone','text') s=str(s)+'.pm' p=os.path.join("c:/begpython/wavnk/",s) x=open(p , 'r') print 3 for i in range(6): x.readline() k2='a' j=0 o=[] while k2 is not '': k2=x.readline() k2=k2.rstrip('\n') oj=k2.split(' ') o=o+[oj] #print o[j] j=j+1 #print j #print o[2][0] temp=long(1232332) end_time=save_a[4] #print end_time k=(j-1) for i in range(k): diff=float(o[i][0])-float(end_time) if diff<0: diff=diff*(-1) if temp>diff: temp=diff pm_row=i #print pm_row #print temp #print o[pm_row] #pm_row=3 q=[] print 4 l=str(p).replace('.pm','.mcep') z=open(l ,'r') for i in range(pm_row): z.readline() k3=z.readline() k3=k3.rstrip('\n') q=k3.split(' ') #print q print 5 s=str(save_b[1]).replace('phone','text') s=str(s)+'.pm' p=os.path.join("c:/begpython/wavnk/",s) x=open(p , 'r') for i in range(6): x.readline() k2='a' j=0 o=[] while k2 is not '': k2=x.readline() k2=k2.rstrip('\n') oj=k2.split(' ') o=o+[oj] #print o[j] j=j+1 #print j #print o[2][0] temp=long(1232332) strt_time=save_b[3] #print strt_time k=(j-1) for i in range(k): diff=float(o[i][0])-float(strt_time) if diff<0: diff=diff*(-1) if temp>diff: temp=diff pm_row=i #print pm_row #print temp #print o[pm_row] #pm_row=3 w=[] l=str(p).replace('.pm','.mcep') z=open(l ,'r') for i in range(pm_row): z.readline() k3=z.readline() k3=k3.rstrip('\n') w=k3.split(' ') #print w cost=0 for i in range(12): #print q[i] #print w[i] h=float(q[i])-float(w[i]) cost=cost+math.pow(h,2) j_cost=math.sqrt(cost) #print cost return j_cost def target_cost(a , b): a=(b+1)*3 b=(a+1)*2 t_cost=(a+b)*5/2 return t_cost r1='shht:ra_77' r2='grx_18' g=[] nodes=[] nodes=nodes+[[r1]] for i in range(len(y_in_db_format)): g=y_in_db_format[i] #print g #print g[0] g.remove(str(g[0])) nodes=nodes+[g] nodes=nodes+[[r2]] print nodes print "lenght of nodes",len(nodes) lists=[] #lists=lists+[r1] for i in range(len(nodes)): for j in range(len(nodes[i])): lists=lists+[nodes[i][j]] #lists=lists+[r2] print lists distance={} for i in range(len(lists)): if i==0: distance[str(lists[i])]=0 else: distance[str(lists[i])]=long(123231223) #print distance group_dist=[] infinity=long(123232323) for i in range(len(nodes)): distances=[] for j in range(len(nodes[i])): #distances=[] if i==0: distances=distances+[[nodes[i][j], 0]] else: distances=distances+[[nodes[i][j],infinity]] group_dist=group_dist+[distances] #print distances print "group_distances",group_dist #print "check",group_dist[0][0][1] #costs={} #for i in range(len(lists)): #if i==0: # costs[str(lists[i])]=1 #else: # costs[str(lists[i])]=get_selfcost(lists[i]) path=[] for i in range(len(nodes)): mini=[] if i!=(len(nodes)-1): #temp=long(123234324) #Now calculate the cost between the current node and each of its neighbour for k in range(len(nodes[(i+1)])): for j in range(len(nodes[i])): current=nodes[i][j] #print "current_node",current j_distance=join_cost( current , nodes[i+1][k]) #t_distance=target_cost( current , nodes[i+1][k]) t_distance=34 #print distance #print "distance between current and neighbours",distance total_distance=(.5*(float(group_dist[i][j][1])+float(j_distance))+.5*(float(t_distance))) #print "total distance between the intial_nodes and current neighbour",total_distance if int(group_dist[i+1][k][1]) > int(total_distance): group_dist[i+1][k][1]=total_distance #print "updated distance",group_dist[i+1][k][1] a=current #print "the neighbour",nodes[i+1][k],"updated the value",a mini=mini+[[str(nodes[i+1][k]),a]] print mini

    Read the article

  • how to speed up the code??

    - by kaushik
    in my program i have a method which requires about 4 files to be open each time it is called,as i require to take some data.all this data from the file i have been storing in list for manupalation. I approximatily need to call this method about 10,000 times.which is making my program very slow? any method for handling this files in a better ways and is storing the whole data in list time consuming what is better alternatives for list? I can give some code,but my previous question was closed as that only confused everyone as it is a part of big program and need to be explained completely to understand,so i am not giving any code,please suggest ways thinking this as a general question... thanks in advance

    Read the article

  • Error handling in the RequestHandler without embedding in URI

    - by hyn
    When a user sends a filled form, I want to print an error message in case there is an input error. One of the GAE sample codes does this by embedding the error message in the URI. Inside the form handler (get): self.redirect('/compose?error_message=%s' % message) and in the handler (get) of redirected URI, gets the message from request: values = { 'error_message': self.request.get('error_message'), ... Is there a way to accomplish the same without embedding the message in the URI?

    Read the article

  • Deterministic key serialization

    - by Mike Boers
    I'm writing a mapping class which uses SQLite as the storage backend. I am currently allowing only basestring keys but it would be nice if I could use a couple more types hopefully up to anything that is hashable (ie. same requirements as the builtin dict). To that end I would like to derive a deterministic serialization scheme. Ideally, I would like to know if any implementation/protocol combination of pickle is deterministic for hashable objects (e.g. can only use cPickle with protocol 0). I noticed that pickle and cPickle do not match: >>> import pickle >>> import cPickle >>> def dumps(x): ... print repr(pickle.dumps(x)) ... print repr(cPickle.dumps(x)) ... >>> dumps(1) 'I1\n.' 'I1\n.' >>> dumps('hello') "S'hello'\np0\n." "S'hello'\np1\n." >>> dumps((1, 2, 'hello')) "(I1\nI2\nS'hello'\np0\ntp1\n." "(I1\nI2\nS'hello'\np1\ntp2\n." Another option is to use repr to dump and ast.literal_eval to load. This would only be valid for builtin hashable types. I have written a function to determine if a given key would survive this process (it is rather conservative on the types it allows): def is_reprable_key(key): return type(key) in (int, str, unicode) or (type(key) == tuple and all( is_reprable_key(x) for x in key)) The question for this method is if repr itself is deterministic for the types that I have allowed here. I believe this would not survive the 2/3 version barrier due to the change in str/unicode literals. This also would not work for integers where 2**32 - 1 < x < 2**64 jumping between 32 and 64 bit platforms. Are there any other conditions (ie. do strings serialize differently under different conditions)? (If this all fails miserably then I can store the hash of the key along with the pickle of both the key and value, then iterate across rows that have a matching hash looking for one that unpickles to the expected key, but that really does complicate a few other things and I would rather not do it.) Any insights?

    Read the article

  • Form (or Formset?) to handle multiple table rows in Django

    - by Ben
    Hi, I'm working on my first Django application. In short, what it needs to do is to display a list of film titles, and allow users to give a rating (out of 10) to each film. I've been able to use the {{ form }} and {{ formset }} syntax in a template to produce a form which lets you rate one film at a time, which corresponds to one row in a MySQL table, but how do I produce a form that iterates over all the movie titles in the database and produces a form that lets you rate lots of them at once? At first, I thought this was what formsets were for, but I can't see any way to automatically iterate over the contents of a database table to produce items to go in the form, if you see what I mean. Currently, my views.py has this code: def survey(request): ScoreFormSet = formset_factory(ScoreForm) if request.method == 'POST': formset = ScoreFormSet(request.POST, request.FILES) if formset.is_valid(): return HttpResponseRedirect('/') else: formset = ScoreFormSet() return render_to_response('cf/survey.html', { 'formset':formset, }) And my survey.html has this: <form action="/survey/" method="POST"> <table> {{ formset }} </table> <input type = "submit" value = "Submit"> </form> Oh, and the definition of ScoreForm and Score from models.py are: class Score(models.Model): movie = models.ForeignKey(Movie) score = models.IntegerField() user = models.ForeignKey(User) class ScoreForm(ModelForm): class Meta: model = Score So, in case the above is not clear, what I'm aiming to produce is a form which has one row per movie, and each row shows a title, and has a box to allow the user to enter their score. If anyone can point me at the right sort of approach to this, I'd be most grateful. Thanks, Ben

    Read the article

  • csrf error in django

    - by niklasfi
    Hello, I want to realize a login for my site. I basically copied and pasted the following bits from the Django Book together. However I still get an error (CSRF verification failed. Request aborted.), when submitting my registration form. Can somebody tell my what raised this error and how to fix it? Here is my code: views.py: # Create your views here. from django import forms from django.contrib.auth.forms import UserCreationForm from django.http import HttpResponseRedirect from django.shortcuts import render_to_response def register(request): if request.method == 'POST': form = UserCreationForm(request.POST) if form.is_valid(): new_user = form.save() return HttpResponseRedirect("/books/") else: form = UserCreationForm() return render_to_response("registration/register.html", { 'form': form, }) register.html: <html> <body> {% block title %}Create an account{% endblock %} {% block content %} <h1>Create an account</h1> <form action="" method="post">{% csrf_token %} {{ form.as_p }} <input type="submit" value="Create the account"> </form> {% endblock %} </body> </html>

    Read the article

  • Scrape zipcode table for different urls based on county

    - by Dr.Venkman
    I used lxml and ran into a wall as my new computer wont install lxml and the code doesnt work. I know this is simple - maybe some one can help with a beautiful soup script. this is my code: import codecs import lxml as lh from selenium import webdriver import time import re results = [] city = [ 'amador'] state = [ 'CA'] for state in states: for city in citys: browser = webdriver.Firefox() link2 = 'http://www.getzips.com/cgi-bin/ziplook.exe?What=3&County='+ city +'&State=' + state + '&Submit=Look+It+Up' browser.get(link2) bcontent = browser.page_source zipcode = bcontent[bcontent.find('<td width="15%"'):bcontent.find('<p>')+0] if len(zipcode) > 0: print zipcode else: print 'none' browser.quit() Thanks for the help

    Read the article

  • django-uni-form helpers and CSRF tags over POST

    - by linked
    Hi, I'm using django-uni-forms to display my fields, with a rather rudimentary example straight out of their book. When I render the form fields using <form>{%csrf_tag%} {%form|as_uni_form%}</form>, everything works as expected. However, django-uni-form Helpers allow you to generate the form tag (and other helper-related content) using the following syntax -- {% with form.helper as helper %}{% uni_form form helper%}{%endwith%} -- This creates the <form> tag for me, so there's nowhere to embed my own CSRF_token. When I try to use this syntax, the form renders perfectly, but without a CSRF token, and so submitting the form fails every time. Does anyone have experience with this? Is there an established way to add the token? I much prefer the second syntax, for re-use reasons. Thanks!

    Read the article

  • In Django, why is user.is_authenticated a method and not a member variable like is_staff

    - by luc
    Hello all, I've lost some time with a bug in my app due to user authentication. I think that it's a bit confusing but maybe someone can explain the reason and it will appear to me very logical. The user.is_staff is a member variable while user.is_authenticated is a method. However is_authenticated only returns True or False depending if the class is User or AnonymousUser (see http://docs.djangoproject.com/en/dev/topics/auth/) Is there a reason for that? Why user.is_authenticated is a method? Thanks in advance

    Read the article

  • How to accept localized date format (e.g dd/mm/yy) in a DateField on an admin form ?

    - by tomjerry
    Is it possible to customize a django application to have accept localized date format (e.g dd/mm/yy) in a DateField on an admin form ? I have a model class : class MyModel(models.Model): date = models.DateField("Date") And associated admin class class MyModelAdmin(admin.ModelAdmin): pass On django administration interface, I would like to be able to input a date in following format : dd/mm/yyyy. However, the date field in the admin form expects yyyy-mm-dd. How can I customize things ? Nota bene : I have already specified my custom language code (fr-FR) in settings.py, but it seems to have no effect on this date input matter. Thanks in advance for your answer

    Read the article

  • how to login in google account with app engine webproxy

    - by user313446
    hi,a webproxy on app engine oncyberspace.appspot.com , save cookie in the database, when i try to login in the google with my account, it redirect to google.com . how to solve these problem ? and another problem , when i this the above web to login in twitter,it works !but i can not use it to update my tweet. i don't know why, may be i can't pass oauth . how to solve this ?

    Read the article

  • Infinite loop when adding a row to a list in a class in python3

    - by Margaret
    I have a script which contains two classes. (I'm obviously deleting a lot of stuff that I don't believe is relevant to the error I'm dealing with.) The eventual task is to create a decision tree, as I mentioned in this question. Unfortunately, I'm getting an infinite loop, and I'm having difficulty identifying why. I've identified the line of code that's going haywire, but I would have thought the iterator and the list I'm adding to would be different objects. Is there some side effect of list's .append functionality that I'm not aware of? Or am I making some other blindingly obvious mistake? class Dataset: individuals = [] #Becomes a list of dictionaries, in which each dictionary is a row from the CSV with the headers as keys def field_set(self): #Returns a list of the fields in individuals[] that can be used to split the data (i.e. have more than one value amongst the individuals def classified(self, predicted_value): #Returns True if all the individuals have the same value for predicted_value def fields_exhausted(self, predicted_value): #Returns True if all the individuals are identical except for predicted_value def lowest_entropy_value(self, predicted_value): #Returns the field that will reduce <a href="http://en.wikipedia.org/wiki/Entropy_%28information_theory%29">entropy</a> the most def __init__(self, individuals=[]): and class Node: ds = Dataset() #The data that is associated with this Node links = [] #List of Nodes, the offspring Nodes of this node level = 0 #Tree depth of this Node split_value = '' #Field used to split out this Node from the parent node node_value = '' #Value used to split out this Node from the parent Node def split_dataset(self, split_value): fields = [] #List of options for split_value amongst the individuals datasets = {} #Dictionary of Datasets, each one with a value from fields[] as its key for field in self.ds.field_set()[split_value]: #Populates the keys of fields[] fields.append(field) datasets[field] = Dataset() for i in self.ds.individuals: #Adds individuals to the datasets.dataset that matches their result for split_value datasets[i[split_value]].individuals.append(i) #<---Causes an infinite loop on the second hit for field in fields: #Creates subnodes from each of the datasets.Dataset options self.add_subnode(datasets[field],split_value,field) def add_subnode(self, dataset, split_value='', node_value=''): def __init__(self, level, dataset=Dataset()): My initialisation code is currently: if __name__ == '__main__': filename = (sys.argv[1]) #Takes in a CSV file predicted_value = "# class" #Identifies the field from the CSV file that should be predicted base_dataset = parse_csv(filename) #Turns the CSV file into a list of lists parsed_dataset = individual_list(base_dataset) #Turns the list of lists into a list of dictionaries root = Node(0, Dataset(parsed_dataset)) #Creates a root node, passing it the full dataset root.split_dataset(root.ds.lowest_entropy_value(predicted_value)) #Performs the first split, creating multiple subnodes n = root.links[0] n.split_dataset(n.ds.lowest_entropy_value(predicted_value)) #Attempts to split the first subnode.

    Read the article

  • Programmatically sync the db in Django

    - by Attila Oláh
    I'm trying to sync my db from a view, something like this: from django import http from django.core import management def syncdb(request): management.call_command('syncdb') return http.HttpResponse('Database synced.') The issue is, it will block the dev server by asking for user input from the terminal. How can I pass it the '--noinput' option to prevent asking me anything? I have other ways of marking users as super-user, so there's no need for the user input, but I really need to call syncdb (and flush) programmatically, without logging on to the server via ssh. Any help is appreciated.

    Read the article

  • how to fill a part of a circle using PIL?

    - by valya
    hello. I'm trying to use PIL for a task but the result is very dirty. What I'm doing is trying to fill a part of a piece of a circle, as you can see on the image. Here is my code: def gen_image(values): side = 568 margin = 47 image = Image.open(settings.MEDIA_ROOT + "/i/promo_circle.jpg") draw = ImageDraw.Draw(image) draw.ellipse((margin, margin, side-margin, side-margin), outline="white") center = side/2 r = side/2 - margin cnt = len(values) for n in xrange(cnt): angle = n*(360.0/cnt) - 90 next_angle = (n+1)*(360.0/cnt) - 90 nr = (r * values[n] / 5) max_r = r min_r = nr for cr in xrange(min_r*10, max_r*10): cr = cr/10.0 draw.arc((side/2-cr, side/2-cr, side/2+cr, side/2+cr), angle, next_angle, fill="white") return image

    Read the article

  • List Directories and get the name of the Directory

    - by chrissygormley
    Hello, I am trying to get the code to list all the directories in a folder, change directory into that folder and get the name of the current folder. The code I have so far is below and isn't working at the minute. I seem to be getting the parent folder name. import os for directories in os.listdir(os.getcwd()): dir = os.path.join('/home/user/workspace', directories) os.chdir(dir) current = os.path.dirname(dir) new = str(current).split("-")[0] print new I also have other files in the folder but I do not want to list them. I have tried the below code but I haven't got it working yet either. for directories in os.path.isdir(os.listdir(os.getcwd())): Can anyone see where I am going wrong? Thanks

    Read the article

< Previous Page | 369 370 371 372 373 374 375 376 377 378 379 380  | Next Page >