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  • Image upload and Manipulation in Django

    - by Saransh Mohapatra
    I am trying upload images and than create an thumbnail of it and than store both in S3. After the file has been uploaded i am first uploading it to S3 and than trying to create thumbnail but it doesn't work as than PIL is not able to recognise the image. And secondly if I create the thumbnail first than while uploading original image I get EOF. I think Django allows just once for the uploaded files to be used only once....Please kindly tell me a way to do so....Thanks in advance

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  • PyQt - QLabel inheriting

    - by Ockonal
    Hello, i wanna inherit QLabel to add there click event processing. I'm trying this code: class NewLabel(QtGui.QLabel): def __init__(self, parent): QtGui.QLabel.__init__(self, parent) def clickEvent(self, event): print 'Label clicked!' But after clicking I have no line 'Label clicked!' EDIT: Okay, now I'm using not 'clickEvent' but 'mousePressEvent'. And I still have a question. How can i know what exactly label was clicked? For example, i have 2 edit box and 2 labels. Labels content are pixmaps. So there aren't any text in labels, so i can't discern difference between labels. How can i do that? EDIT2: I made this code: class NewLabel(QtGui.QLabel): def __init__(self, firstLabel): QtGui.QLabel.__init__(self, firstLabel) def mousePressEvent(self, event): print 'Clicked' #myLabel = self.sender() # None =) self.emit(QtCore.SIGNAL('clicked()'), "Label pressed") In another class: self.FirstLang = NewLabel(Form) QtCore.QObject.connect(self.FirstLang, QtCore.SIGNAL('clicked()'), self.labelPressed) Slot in the same class: def labelPressed(self): print 'in labelPressed' print self.sender() But there isn't sender object in self. What i did wrong?

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  • Optimizing list comprehension to find pairs of co-prime numbers

    - by user3685422
    Given A,B print the number of pairs (a,b) such that GCD(a,b)=1 and 1<=a<=A and 1<=b<=B. Here is my answer: return len([(x,y) for x in range(1,A+1) for y in range(1,B+1) if gcd(x,y) == 1]) My answer works fine for small ranges but takes enough time if the range is increased. such as 1 <= A <= 10^5 1 <= B <= 10^5 is there a better way to write this or can this be optimized?

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  • How do I create a Django ModelForm, so that it's fields are sometimes required, sometimes not?

    - by Graf
    Ok, here is the question. Imagine I have a ModelForm which have only two fields. like this one: class ColorForm(forms.Form): color_by_name = forms.CharField() color = forms.IntegerField(widget = forms.Select(choices=COLOR_CHOICES)) So a user can either input a color name, a choose it from a list. Color is required, but that doesn't mean, that user should enter it manually. There do I put validation, so that my code checks if user selected color in dropdownlist and if not then he should write it manually?

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  • chatbot using twisted and wokkel

    - by dmitriy k.
    I am writing a chatbot using Twisted and wokkel and everything seems to be working except that bot periodically logs off. To temporarily fix that I set presence to available on every connection initialized. Does anyone know how to prevent going offline? (I assume if i keep sending available presence every minute or so bot wont go offline but that just seems too wasteful.) Suggestions anyone? Here is the presence code: class BotPresenceClientProtocol(PresenceClientProtocol): def connectionInitialized(self): PresenceClientProtocol.connectionInitialized(self) self.available(statuses={None: 'Here'}) def subscribeReceived(self, entity): self.subscribed(entity) self.available(statuses={None: 'Here'}) def unsubscribeReceived(self, entity): self.unsubscribed(entity) Thanks in advance.

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  • Can I get the amount of time for which a key is pressed on a keyboard

    - by Adi
    Dear all, I am working on a project in which I have to develop bio-passwords based on user's keystroke style. Suppose a user types a password for 20 times, his keystrokes are recorded, like holdtime : time for which a particular key is pressed. digraph time : time it takes to press a different key. suppose a user types a password " COMPUTER". I need to know the time for which every key is pressed. something like : holdtime for the above password is C-- 200ms O-- 130ms M-- 150ms P-- 175ms U-- 320ms T-- 230ms E-- 120ms R-- 300ms The rational behind this is , every user will have a different holdtime. Say a old person is typing the password, he will take more time then a student. And it will be unique to a particular person. To do this project, I need to record the time for each key pressed. I would greatly appreciate if anyone can guide me in how to get these times. Editing from here.. Language is not important, but I would prefer it in C. I am more interested in getting the dataset.

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  • how to show the right word in my code, my code is : os.urandom(64)

    - by zjm1126
    My code is: print os.urandom(64) which outputs: > "D:\Python25\pythonw.exe" "D:\zjm_code\a.py" \xd0\xc8=<\xdbD' \xdf\xf0\xb3>\xfc\xf2\x99\x93 =S\xb2\xcd'\xdbD\x8d\xd0\\xbc{&YkD[\xdd\x8b\xbd\x82\x9e\xad\xd5\x90\x90\xdcD9\xbf9.\xeb\x9b>\xef#n\x84 which isn't readable, so I tried this: print os.urandom(64).decode("utf-8") but then I get: > "D:\Python25\pythonw.exe" "D:\zjm_code\a.py" Traceback (most recent call last): File "D:\zjm_code\a.py", line 17, in <module> print os.urandom(64).decode("utf-8") File "D:\Python25\lib\encodings\utf_8.py", line 16, in decode return codecs.utf_8_decode(input, errors, True) UnicodeDecodeError: 'utf8' codec can't decode bytes in position 0-3: invalid data What should I do to get human-readable output?

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  • changing order of items in tkinter listbox

    - by user1104854
    Is there an easier way to change the order of items in a tkinter listbox than deleting the values for specific key, then re-entering new info? For example, I want to be able to re-arrange items in a listbox. If I want to swap the position of two, this is what I've done. It works, but I just want to see if there's a quicker way to do this. def moveup(self,selection): value1 = int(selection[0]) - 1 #value to be moved down one position value2 = selection #value to be moved up one position nameAbove = self.fileListSorted.get(value1) #name to be moved down nameBelow = self.fileListSorted.get(value2) #name to be moved up self.fileListSorted.delete(value1,value1) self.fileListSorted.insert(value1,nameBelow) self.fileListSorted.delete(value2,value2) self.fileListSorted.insert(value2,nameAbove)

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  • Interrupting `while loop` with keyboard in Cython

    - by linello
    I want to be able to interrupt a long function with cython, using the usual CTRL+C interrupt command. My C++ long function is repeatedly called inside a while loop from Cython code, but I want to be able, during the loop, to send an "interrupt" and block the while loop. The interrupt also should wait the longFunction() to finish, so that no data are lost or kept in unknown status. This is one of my first implementation, which obviously doesn't work: computed=0; print "Computing long function..." while ( computed==0 ): try: computed = self.thisptr.aLongFunction() except (KeyboardInterrupt, SystemExit): computed=1 print '\n! Received keyboard interrupt.\n' break; (p.s. self.thisptr is the pointer to the current class which implements aLongFunction() )

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  • Testing for the existence of a field in a class

    - by Brett
    Hi, i have a quick question. I have a 2D array that stores an instance of a class. The elements of the array are assigned a particular class based on a text file that is read earlier in the program. Since i do not know without looking in the file what class is stored at a particular element i could refer to a field that doesn't exist at that index (referring to appearance when an instance of temp is stored in that index). i have come up with a method of testing this, but it is long winded and requires a second matrix. Is there a function to test for the existence of a field in a class? class temp(): name = "default" class temp1(): appearance = "@"

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  • Attribute Address getting displayed instead of Attribute Value

    - by Manish
    I am try to create the following. I want to have one drop down menu. Depending on the option selected in the first drop down menu, options in second drop down menu will be displayed. The options in 2nd drop down menu is supposed by dynamic, i.e., options change with the change of values in first menu. Here, instead of getting the drop down menus, I am getting the following Choose your Option1: Choose your Option2: Note: I strictly don't want to use javascript. home_form.py class HomeForm(forms.Form): def __init__(self, *args, **kwargs): var_filter_con = kwargs.pop('filter_con', None) super(HomeForm, self).__init__(*args, **kwargs) if var_filter_con == '***': var_empty_label = None else: var_empty_label = ' ' self.option2 = forms.ModelChoiceField(queryset = db_option2.objects.filter(option1_id = var_filter_con).order_by("name"), empty_label = var_empty_label, widget = forms.Select(attrs={"onChange":'this.form.submit();'}) ) self.option1 = forms.ModelChoiceField(queryset = db_option1.objects.all().order_by("name"), empty_label=None, widget=forms.Select(attrs={"onChange":'this.form.submit();'}) ) view.py def option_view(request): if request.method == 'POST': form = HomeForm(request.POST) if form.is_valid(): cd = form.cleaned_data if cd.has_key('option1'): f = HomeForm(filter_con = cd.get('option1')) return render_to_response('homepage.html', {'home_form':f,}, context_instance=RequestContext(request)) return render_to_response('invalid_data.html', {'form':form,}, context_instance=RequestContext(request)) else: f = HomeForm(filter_con = '***') return render_to_response('homepage.html', {'home_form':f,}, context_instance=RequestContext(request)) homepage.html <!DOCTYPE HTML> <head> <title>Nivaaran</title> </head> <body> <form method="post" name = 'choose_opt' action=""> {% csrf_token %} Choose your Option1: {{ home_form.option1 }} <br/> Choose your Option2: {{ home_form.option2 }} </form> </body>

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  • How do I restrict foreign keys choices to related objects only in django

    - by Jeff Mc
    I have a two way foreign relation similar to the following class Parent(models.Model): name = models.CharField(max_length=255) favoritechild = models.ForeignKey("Child", blank=True, null=True) class Child(models.Model): name = models.CharField(max_length=255) myparent = models.ForeignKey(Parent) How do I restrict the choices for Parent.favoritechild to only children whose parent is itself? I tried class Parent(models.Model): name = models.CharField(max_length=255) favoritechild = models.ForeignKey("Child", blank=True, null=True, limit_choices_to = {"myparent": "self"}) but that causes the admin interface to not list any children.

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  • Which is faster??

    - by kaki
    is opening a large file once reading it completely once to list faster (or) opening smaller files whose total sum of size is equal to large file and loading smaller file into list manupalating one by one faster? which is faster?? is the difference is time large enough to impact my program?? total time difference of lesser then of 30 sec is negligible for me

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  • I can't login to my Django app when debug is set to False

    - by Eric
    I have a very strange problem, and I don't know how to fix or debug it. Short Story: I get locked out of my Django app when Debug is set to False. Long story: Case 1 (the first time it happened): 1. I enter my login info, but It just redirects to the login page. 2. I restart the server, try to login, and it works fine, I get in. 3. a few hours later I come back, log out, try to log back in and I can't. It just redirects to the login page. Case 2 (I figure out how to provoke the login failure): 1. I restart the server and am able to login to the site. 2. I log in and log out several times, everything is fine. 3. I go to a non-existing page and get a server error. 4. I log out and try to log back in, and I can't, just get redirected back to the login page. Case 3 (I can't provoke the login failure with Debug set to True): 1. I restart the server and am able to login to the site. 2. I log in and log out several times, everything is fine. 3. I go to a non-existing page and get a traceback. 4. I log out and log back in, everything works. 5. I wait and play around with it and can't get the login to fail while in Debug mode. Please help!

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  • Is there a replacement for Paste.Template?

    - by Jorge Vargas
    I have grown tired of all the little issues with paste template, it's horrible to maintain the templates, it has no way of updating an old project and it's very hard to test. I'm wondering if someone knows of an alternative for quickstart generators as they have proven to be useful.

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  • Django admin page dropdowns

    - by zen
    I am building a high school team application using Django. Here is my working models file: class Directory(models.Model): school = models.CharField(max_length=60) website = models.URLField() district = models.SmallIntegerField() conference = models.ForeignKey(Conference) class Conference(models.Model): conference_name = models.CharField(max_length=50) url = models.URLField() class Meta: ordering = ['conference_name'] When I open my admin pages and go to edit a school's conference the drop down looks like this: <select> <option value="1">Conference Object</option> <option value="2">Conference Object</option> <select> How do I replace "Conference Object" with the conference_name?

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  • just-in-time list

    - by intuited
    I'd like to know if there is a class available, either in the standard library or in pypi, that fits this description. The constructor would take an iterator. It would implement the container protocol (ie _getitem_, _len_, etc), so that slices, length, etc., would work. In doing so, it would iterate and retain just enough values from its constructor argument to provide whatever information was requested. So if jitlist[6] was requested, it would call self.source.next() 7 times, save those elements in its list, and return the last one. This would allow downstream code to use it as a list, but avoid unnecessarily instantiating a list for cases where list functionality was not needed, and avoid allocating memory for the entire list if only a few members ended up being requested. It seems like a pretty easy one to write, but it also seems useful enough that it's likely that someone would have already made it available in a module.

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