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  • Django: Determining if a user has voted or not

    - by TheLizardKing
    I have a long list of links that I spit out using the below code, total votes, submitted by, the usual stuff but I am not 100% on how to determine if the currently logged in user has voted on a link or not. I know how to do this from within my view but do I need to alter my below view code or can I make use of the way templates work to determine it? I have read http://stackoverflow.com/questions/1528583/django-vote-up-down-method but I don't quite understand what's going on ( and don't need any ofjavascriptery). Models (snippet): class Link(models.Model): category = models.ForeignKey(Category, blank=False, default=1) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add=True) modified = models.DateTimeField(auto_now=True) url = models.URLField(max_length=1024, unique=True, verify_exists=True) name = models.CharField(max_length=512) def __unicode__(self): return u'%s (%s)' % (self.name, self.url) class Vote(models.Model): link = models.ForeignKey(Link) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add=True) def __unicode__(self): return u'%s vote for %s' % (self.user, self.link) Views (snippet): links = Link.objects.select_related().annotate(votes=Count('vote')).order_by('-created')

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  • Create a color generator in matplotlib

    - by Brendan
    I have a series of lines that each need to be plotted with a separate colour. Each line is actually made up of several data sets (positive, negative regions etc.) and so I'd like to be able to create a generator that will feed one colour at a time across a spectrum, for example the gist_rainbow map shown here. I have found the following works but it seems very complicated and more importantly difficult to remember, from pylab import * NUM_COLORS = 22 mp = cm.datad['gist_rainbow'] get_color = matplotlib.colors.LinearSegmentedColormap.from_list(mp, colors=['r', 'b'], N=NUM_COLORS) ... # Then in a for loop this_color = get_color(float(i)/NUM_COLORS) Moreover, it does not cover the range of colours in the gist_rainbow map, I have to redefine a map. Maybe a generator is not the best way to do this, if so what is the accepted way?

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  • Django Querysets -- need a less expensive way to do this..

    - by rh0dium
    Hi all, I have a problem with some code and I believe it is because of the expense of the queryset. I am looking for a much less expensive (in terms of time) way to to this.. log.info("Getting Users") employees = Employee.objects.filter(is_active = True) log.info("Have Users") if opt.supervisor: if opt.hierarchical: people = getSubs(employees, " ".join(args)) else: people = employees.filter(supervisor__name__icontains = " ".join(args)) else: log.info("Filtering Users") people = employees.filter(name__icontains = " ".join(args)) | \ employees.filter(unix_accounts__username__icontains = " ".join(args)) log.info("Filtered Users") log.info("Processing data") np = [] for person in people: unix, p4, bugz = "No", "No", "No" if len(person.unix_accounts.all()): unix = "Yes" if len(person.perforce_accounts.all()): p4 = "Yes" if len(person.bugzilla_accounts.all()): bugz = "Yes" if person.cell_phone != "": exphone = fixphone(person.cell_phone) elif person.other_phone != "": exphone = fixphone(person.other_phone) else: exphone = "" np.append({ 'name':person.name, 'office_phone': fixphone(person.office_phone), 'position': person.position, 'location': person.location.description, 'email': person.email, 'functional_area': person.functional_area.name, 'department': person.department.name, 'supervisor': person.supervisor.name, 'unix': unix, 'perforce': p4, 'bugzilla':bugz, 'cell_phone': fixphone(exphone), 'fax': fixphone(person.fax), 'last_update': person.last_update.ctime() }) log.info("Have data") Now this results in a log which looks like this.. 19:00:55 INFO phone phone Getting Users 19:00:57 INFO phone phone Have Users 19:00:57 INFO phone phone Processing data 19:01:30 INFO phone phone Have data As you can see it's taking over 30 seconds to simply iterate over the data. That is way too expensive. Can someone clue me into a more efficient way to do this. I thought that if I did the first filter that would make things easier but seems to have no effect. I'm at a loss on this one. Thanks To be clear this is about 1500 employees -- Not too many!!

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  • Polynomial fitting with log log plot

    - by viral parekh
    I have a simple problem to fit a straight line on log-log scale. My code is, data=loadtxt(filename) xdata=data[:,0] ydata=data[:,1] polycoeffs = scipy.polyfit(xdata, ydata, 1) yfit = scipy.polyval(polycoeffs, xdata) pylab.plot(xdata, ydata, 'k.') pylab.plot(xdata, yfit, 'r-') Now I need to plot fit line on log scale so I just change x and y axis, ax.set_yscale('log') ax.set_xscale('log') then its not plotting correct fit line. So how can I change fit function (in log scale) so that it can plot fit line on log-log scale? Thanks -Viral

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  • need help in site classification

    - by goh
    hi guys, I have to crawl the contents of several blogs. The problem is that I need to classify whether the blogs the authors are from a specific school and is talking about the school's stuff. May i know what's the best approach in doing the crawling or how should i go about the classification?

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  • a simple smtp server

    - by fixxxer
    Could you please suggest a simple SMTP server with the very basic APIs(by very basic I mean, to read,write,delete email) that could be run on a linux box? I just need to convert the crux of the email into XML format and FTP it to another machine.

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  • Dropdown sorting in django-admin

    - by Andrey
    I'd like to know how can I sort values in the Django admin dropdowns. For example, I have a model called Article with a foreign key pointing to the Users model, smth like: class Article(models.Model): title = models.CharField(_('Title'), max_length=200) slug = models.SlugField(_('Slug'), unique_for_date='publish') author = models.ForeignKey(User) body = models.TextField(_('Body')) status = models.IntegerField(_('Status')) categories = models.ManyToManyField(Category, blank=True) publish = models.DateTimeField(_('Publish date')) I edit this model in django admin: class ArticleAdmin(admin.ModelAdmin): list_display = ('title', 'publish', 'status') list_filter = ('publish', 'categories', 'status') search_fields = ('title', 'body') prepopulated_fields = {'slug': ('title',)} admin.site.register(Article, ArticleAdmin) and of course it makes the nice user select dropdown for me, but it's not sorted and it takes a lot of time to find a user by username.

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  • TypeError: 'NoneType' object does not support item assignment

    - by R S John
    I am trying to do some mathematical calculation according to the values at particular index of a NumPy array with the following code X = np.arange(9).reshape(3,3) temp = X.copy().fill(5.446361E-01) ind = np.where(X < 4.0) temp[ind] = 0.5*X[ind]**2 - 1.0 ind = np.where(X >= 4.0 and X < 9.0) temp[ind] = (5.699327E-1*(X[ind]-1)**4)/(X[ind]**4) print temp But I am getting the following error Traceback (most recent call last): File "test.py", line 7, in <module> temp[ind] = 0.5*X[ind]**2 - 1.0 TypeError: 'NoneType' object does not support item assignment Would you please help me in solving this? Thanks

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  • Django: Get remote IP address inside settings.py

    - by Silver Light
    Hello! I want to enable debug (DEBUG = True) For my Django project only if it runs on localhost. How can I get user IP address inside settings.py? I would like something like this to work: #Debugging only on localhost if user_ip = '127.0.0.1': DEBUG = True else: DEBUG = False How do I put user IP address in user_ip variable inside settings.py file?

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  • Content-Length header not returned from Pylons response

    - by Evgeny
    I'm still struggling to Stream a file to the HTTP response in Pylons. In addition to the original problem, I'm finding that I cannot return the Content-Length header, so that for large files the client cannot estimate how long the download will take. I've tried response.content_length = 12345 and I've tried response.headers['Content-Length'] = 12345 In both cases the HTTP response (viewed in Fiddler) simply does not contain the Content-Length header. How do I get Pylons to return this header? (Oh, and if you have any ideas on making it stream the file please reply to the original question - I'm all out of ideas there.)

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  • Scraping paginated items from a website using scrapy

    - by Mridang Agarwalla
    I'm using scrapy to scrape items from a site. I'm not being able to implement this scraping pattern. The site I'm trying to scrape is a forum and I scrape the site once a day. Each page has a table containing posts. New posts are added to the top of the table and as more and more posts are posted to the site, the older posts go further into the pages due to pagination. This is a very simple scenario and we will assume that the order of the posts never change. I would like to scrape this site and scrape all the "new" records until the last scraped post from yesterday is encountered. I have configured my spider to paginate endlessly and when it encounters yesterday's last scraped post, it should stop. How can implement this? (My Scrapy installation works with my Django installation using django-dynamic-scraper )

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  • SQLAlchemy - select for update example

    - by Mark
    I'm looking for a complete example of using select for update in SQLAlchemy, but haven't found one googling. I need to lock a single row and update a column, the following code doesn't work (blocks forever): s = table.select(table.c.user=="test",for_update=True) u = table.update().where(table.c.user=="test") u.execute(email="foo") Do I need a commit? How do I do that? As far as I know you need to: begin transaction select ... for update update commit

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  • Prepopulate drop-box according to another drop-box choice in Django Admin

    - by onorua
    I have models like this: class User(models.Model): Switch = models.ForeignKey(Switch, related_name='SwitchUsers') Port = models.ForeignKey(Port) class Switch(models.Model): Name = models.CharField(max_length=50) class Port(models.Model): PortNum = models.PositiveIntegerField() Switch = models.ForeignKey(Switch, related_name = "Ports") When I'm in Admin interface and choose Switch from Switches available, I would like to have Port prepopulated accordingly with Ports from the related Switch. As far as I understand I need to create some JS script to prepopulate it. Unfortunately I don't have this experience, and I would like to keep things simple as it possible and don't rewrite all Django admin interface. Just add this functionality for one Field. Could you please help me with my problem? Thank you.

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  • Django does not load internal .css files

    - by Rubén Jiménez
    I have created a Django project in local which runs without any kind of problem. But, after an annoying and difficult Cherokee + uWSGI installation on Amazon AWS, my project does not show Django .css internal files. http://f.cl.ly/items/2Q2W3I3R0X1n2X3v0q2P/django_error.jpg <-- /Admin/ looks like The image is a screen of my /admin/, which should have a different style, but .css files are not loaded. [pid: 23206|app: 0|req: 19/19] 83.49.10.217 () {56 vars in 1121 bytes} [Sun Apr 15 05:50:24 2012] GET /static/admin/css/base.css = generated 2896 bytes in 6 msecs (HTTP/1.1 404) 1 headers in 51 bytes (1 switches on core 0) [pid: 23206|app: 0|req: 20/20] 83.49.10.217 () {56 vars in 1125 bytes} [Sun Apr 15 05:50:24 2012] GET /static/admin/css/login.css = generated 2899 bytes in 5 msecs (HTTP/1.1 404) 1 headers in 51 bytes (1 switches on core 0) This is a log from Cherokee. I don't understand why it is looking for the .css files in that path. Cherokee should be searching the files in Django original directory so i didn't change .css files in my project. Any advice? Thanks a lot.

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  • Locating file path from a <InMemoryUploadedFile> Djnago object

    - by PirosB3
    Hi all I have a Django app which, submitting a package, should return values that are inside it.. Submitted the form to a view called "insert": request.FILES['file'] returns the file objects, but it is of kind < InMemoryUploadedFile. What i need is a way to get the absolute path of the uploaded file, so that i can feed it to a method that will return the values needed Anyone know how i can accomplish this? Thanks

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  • Sqlalchemy complex in_ clause

    - by lostlogic
    I'm trying to find a way to cause sqlalchemy to generate sql of the following form: select * from t where (a,b) in ((a1,b1),(a2,b2)); Is this possible? If not, any suggestions on a way to emulate it? Thanks kindly!

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  • How do I code this relationship in SQLAlchemy?

    - by Martin Del Vecchio
    I am new to SQLAlchemy (and SQL, for that matter). I can't figure out how to code the idea I have in my head. I am creating a database of performance-test results. A test run consists of a test type and a number (this is class TestRun below) A test suite consists the version string of the software being tested, and one or more TestRun objects (this is class TestSuite below). A test version consists of all test suites with the given version name. Here is my code, as simple as I can make it: from sqlalchemy import * from sqlalchemy.ext.declarative import declarative_base from sqlalchemy.orm import relationship, backref, sessionmaker Base = declarative_base() class TestVersion (Base): __tablename__ = 'versions' id = Column (Integer, primary_key=True) version_name = Column (String) def __init__ (self, version_name): self.version_name = version_name class TestRun (Base): __tablename__ = 'runs' id = Column (Integer, primary_key=True) suite_directory = Column (String, ForeignKey ('suites.directory')) suite = relationship ('TestSuite', backref=backref ('runs', order_by=id)) test_type = Column (String) rate = Column (Integer) def __init__ (self, test_type, rate): self.test_type = test_type self.rate = rate class TestSuite (Base): __tablename__ = 'suites' directory = Column (String, primary_key=True) version_id = Column (Integer, ForeignKey ('versions.id')) version_ref = relationship ('TestVersion', backref=backref ('suites', order_by=directory)) version_name = Column (String) def __init__ (self, directory, version_name): self.directory = directory self.version_name = version_name # Create a v1.0 suite suite1 = TestSuite ('dir1', 'v1.0') suite1.runs.append (TestRun ('test1', 100)) suite1.runs.append (TestRun ('test2', 200)) # Create a another v1.0 suite suite2 = TestSuite ('dir2', 'v1.0') suite2.runs.append (TestRun ('test1', 101)) suite2.runs.append (TestRun ('test2', 201)) # Create another suite suite3 = TestSuite ('dir3', 'v2.0') suite3.runs.append (TestRun ('test1', 102)) suite3.runs.append (TestRun ('test2', 202)) # Create the in-memory database engine = create_engine ('sqlite://') Session = sessionmaker (bind=engine) session = Session() Base.metadata.create_all (engine) # Add the suites in version1 = TestVersion (suite1.version_name) version1.suites.append (suite1) session.add (suite1) version2 = TestVersion (suite2.version_name) version2.suites.append (suite2) session.add (suite2) version3 = TestVersion (suite3.version_name) version3.suites.append (suite3) session.add (suite3) session.commit() # Query the suites for suite in session.query (TestSuite).order_by (TestSuite.directory): print "\nSuite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs)) for run in suite.runs: print " Test '%s', result %d" % (run.test_type, run.rate) # Query the versions for version in session.query (TestVersion).order_by (TestVersion.version_name): print "\nVersion %s has %d test suites:" % (version.version_name, len (version.suites)) for suite in version.suites: print " Suite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs)) for run in suite.runs: print " Test '%s', result %d" % (run.test_type, run.rate) The output of this program: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 Version v1.0 has 1 test suites: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Version v1.0 has 1 test suites: Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Version v2.0 has 1 test suites: Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 This is not correct, since there are two TestVersion objects with the name 'v1.0'. I hacked my way around this by adding a private list of TestVersion objects, and a function to find a matching one: versions = [] def find_or_create_version (version_name): # Find existing for version in versions: if version.version_name == version_name: return (version) # Create new version = TestVersion (version_name) versions.append (version) return (version) Then I modified my code that adds the records to use it: # Add the suites in version1 = find_or_create_version (suite1.version_name) version1.suites.append (suite1) session.add (suite1) version2 = find_or_create_version (suite2.version_name) version2.suites.append (suite2) session.add (suite2) version3 = find_or_create_version (suite3.version_name) version3.suites.append (suite3) session.add (suite3) Now the output is what I want: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 Version v1.0 has 2 test suites: Suite directory dir1, version v1.0 has 2 test runs: Test 'test1', result 100 Test 'test2', result 200 Suite directory dir2, version v1.0 has 2 test runs: Test 'test1', result 101 Test 'test2', result 201 Version v2.0 has 1 test suites: Suite directory dir3, version v2.0 has 2 test runs: Test 'test1', result 102 Test 'test2', result 202 This feels wrong to me; it doesn't feel right that I am manually keeping track of the unique version names, and manually adding the suites to the appropriate TestVersion objects. Is this code even close to being correct? And what happens when I'm not building the entire database from scratch, as in this example. If the database already exists, do I have to query the database's TestVersion table to discover the unique version names? Thanks in advance. I know this is a lot of code to wade through, and I appreciate the help.

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  • Django and json request

    - by Hulk
    In a template i have the following code <script> var url="/mypjt/my_timer" $.post(url, paramarr, function callbackHandler(dict) { alert('got response back'); if (dict.flag == 2) { alert('1'); $.jGrowl("Data could not be saved"); } else if(dict.ret_status == 1) { alert('2'); $.jGrowl("Data saved successfully"); window.location = "/mypjt/display/" + dict.rid; } }, "json" ); </script> In views i have the following code, def my_timer(request): dict={} try: a= timer.objects.get(pk=1) dict({'flag':1}) return HttpResponse(simplejson.dumps(dict), mimetype='application/javascript') except: dict({'flag':1}) return HttpResponse(simplejson.dumps(dict), mimetype='application/javascript') My question is since we are making a json request and in the try block ,after setting the flag ,cant we return a page directly as return render_to_response('mypjt/display.html',context_instance=RequestContext(request,{'dict': dict})) instead of sending the response, because on success again in the html page we redirect the code Also if there is a exception then only can we return the json request. My only concern is that the interaction between client and server should be minimal. Thanks..

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  • Is django orm & templates thread safe?

    - by Piotr Czapla
    I'm using django orm and templates to create a background service that is ran as management command. Do you know if django is thread safe? I'd like to use threads to speed up processing. The processing is blocked by I/O not CPU so I don't care about performance hit caused by GIL.

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  • SQLAlchemy unsupported type error - and table design issues?

    - by Az
    Hi there, back again with some more SQLAlchemy shenanigans. Let me step through this. My table is now set up as so: engine = create_engine('sqlite:///:memory:', echo=False) metadata = MetaData() students_table = Table('studs', metadata, Column('sid', Integer, primary_key=True), Column('name', String), Column('preferences', Integer), Column('allocated_rank', Integer), Column('allocated_project', Integer) ) metadata.create_all(engine) mapper(Student, students_table) Fairly simple, and for the most part I've been enjoying the ability to query almost any bit of information I want provided I avoid the error cases below. The class it is mapped from is: class Student(object): def __init__(self, sid, name): self.sid = sid self.name = name self.preferences = collections.defaultdict(set) self.allocated_project = None self.allocated_rank = 0 def __repr__(self): return str(self) def __str__(self): return "%s %s" %(self.sid, self.name) Explanation: preferences is basically a set of all the projects the student would prefer to be assigned. When the allocation algorithm kicks in, a student's allocated_project emerges from this preference set. Now if I try to do this: for student in students.itervalues(): session.add(student) session.commit() It throws two errors, one for the allocated_project column (seen below) and a similar error for the preferences column: sqlalchemy.exc.InterfaceError: (InterfaceError) Error binding parameter 4 - probably unsupported type. u'INSERT INTO studs (sid, name, allocated_rank, allocated_project) VALUES (?, ?, ?, ?, ?, ?, ?)' [1101, 'Muffett,M.', 1, 888 Human-spider relationships (Supervisor id: 123)] If I go back into my code I find that, when I'm copying the preferences from the given text files, it actually refers to the Project class which is mapped to a dictionary, using the unique project id's (pid) as keys. Thus, as I iterate through each student via their rank and to the preferences set, it adds not a project id, but the reference to the project id from the projects dictionary. students[sid].preferences[int(rank)].add(projects[int(pid)]) Now this is very useful to me since I can find out all I want to about a student's preferred projects without having to run another check to pull up information about the project id. The form you see in the error has the object print information passed as: return "%s %s (Supervisor id: %s)" %(self.proj_id, self.proj_name, self.proj_sup) My questions are: I'm trying to store an object in a database field aren't I? Would the correct way then, be copying the project information (project id, name, etc) into its own table, referenced by the unique project id? That way I can just have the project id field for one of the student tables just be an integer id and when I need more information, just join the tables? So and so forth for other tables? If the above makes sense, then how does one maintain the relationship with a column of information in one table which is a key index on another table? Does this boil down into a database design problem? Are there any other elegant ways of accomplishing this? Apologies if this is a very long-winded question. It's rather crucial for me to solve this, so I've tried to explain as much as I can, whilst attempting to show that I'm trying (key word here sadly) to understand what could be going wrong.

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  • the error "invalid literal for int() with base 10:" keeps coming up

    - by ratce003
    I'm trying to write a very simple program, I want to print out the sum of all the multiples of 3 and 5 below 100, but, an error keeps accuring, saying "invalid literal for int() with base 10:" my program is as follows: sum = "" sum_int = int(sum) for i in range(1, 101): if i % 5 == 0: sum += i elif i % 3 == 0: sum += i else: sum += "" print sum Any help would be much appreciated.

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