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  • How detect length of a numpy array with only one element?

    - by mishaF
    I am reading in a file using numpy.genfromtxt which brings in columns of both strings and numeric values. One thing I need to do is detect the length of the input. This is all fine provided there are more than one value read into each array. But...if there is only one element in the resulting array, the logic fails. I can recreate an example here: import numpy as np a = np.array(2.3) len(a) returns an error saying: TypeError: len() of unsized object however, If a has 2 or more elements, len() behaves as one would expect. import numpy as np a = np.array([2.3,3.6]) len(a) returns 2 My concern here is, if I use some strange exception handling, I can't distinguish between a being empty and a having length = 1.

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  • Django & custom auth backend (web service) + no database. How to save stuff in session?

    - by Infinity
    I've been searching here and there, and based on this answer I've put together what you see below. It works, but I need to put some stuff in the user's session, right there inside authenticate. How would I store acme_token in the user's session, so that it will get cleared if they logged out? class AcmeUserBackend(object): # Create a User object if not already in the database? create_unknown_user = False def get_user(self, username): return AcmeUser(id=username) def authenticate(self, username=None, password=None): """ Check the username/password and return an AcmeUser. """ acme_token = ask_another_site_about_creds(username, password) if acme_token: return AcmeUser(id=username) return None ################## from django.contrib.auth.models import User class AcmeUser(User): objects = None # we cannot really use this w/o local DB def save(self): """saving to DB disabled""" pass def get_group_permissions(self): """If you don't make your own permissions module, the default also will use the DB. Throw it away""" return [] # likewise with the other permission defs def get_and_delete_messages(self): """Messages are stored in the DB. Darn!""" return []

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  • How to turn this simple 10 digit hex number back into 8 digits?

    - by Babil
    The algorithm to convert input 8 digit hex number into 10 digit are following: Given that the 8 digit number is: '12 34 56 78' x1 = 1 * 16^8 * 2^3 x2 = 2 * 16^7 * 2^2 x3 = 3 * 16^6 * 2^1 x4 = 4 * 16^4 * 2^4 x5 = 5 * 16^3 * 2^3 x6 = 6 * 16^2 * 2^2 x7 = 7 * 16^1 * 2^1 x8 = 8 * 16^0 * 2^0 Final 10 digit hex is: = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = '08 86 42 98 E8' The problem is - how to go back to 8 digit hex from a given 10 digit hex (for example: 08 86 42 98 E8 to 12 34 56 78) Some sample input and output are following: input output 11 11 11 11 08 42 10 84 21 22 22 33 33 10 84 21 8C 63 AB CD 12 34 52 D8 D0 88 64 45 78 96 32 21 4E 84 98 62 FF FF FF FF 7B DE F7 BD EF

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  • Error 400 in urllib2 when using cookies

    - by Hempage
    I've had some success using a different method to load cookies, but now I'm wanting to use the cookielib.MozillaCookieJar method to open a cookies.txt file. Here's the snippet of code that does this. cookieJar=MozillaCookieJar() cookieJar.load(argv[2]) After creating an HTTPCookieProcessor opener, and installing it, whenever I use urlopen, I get an HTTP 400 error. However, if I don't use a CookieJar, the urlopen method succeeds (though the response doesn't contain the data I need). I'm not sure whether the cookies.txt file is malformed, or whether there's something else going on.

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  • muti user dungeon help

    - by mudman
    ive created a single user dungeon which i would like to create into a multi user dungoen so at least two plays can play how would i do that what code do i need to add can anyone help? i would show coding but if i do then everyone would see it and all my work will be copied as i know other students do use this site to so plz understand my situation and yes this is a homework/assignment work.

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  • jEdit+JythonInterpreter: how to import java class?

    - by JChao
    Hi, I'm running jEdit with the JythonInterprete and I have a .jar file called JavaTest.jar. JavaTest has a class called SampleJavaClass which has a method printerCount. From my .py file, I want to do: from javatest import SampleJavaClass class SampleClass(SampleJavaClass): def pymain(self): SampleJavaClass.printerCount(4) Java code: package javatest; public class SampleJavaClass { public static void printerCount(int i){ for(int j=0; j< i; j++){ System.out.println("hello world"); } } (etc...) In the JythonInterpreter, I have already tried clicking "Edit Jython Path" and adding the .jar file then running the interpreter again, but it still gives me ImportError: cannot import name SampleJavaClass

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  • rearranging a list of months

    - by MacUsers
    How can I list the numbers 01 to 12 (one for each of the 12 months) in such a way so that the current month always comes last where the oldest one is first. In other words, if the number is grater than the current month, it's from the previous year. e.g. 02 is Feb, 2011 (the current month right now), 03 is March, 2010 and 09 is Sep, 2010 but 01 is Jan, 2011. In this case, I'd like to have [09, 03, 01, 02]. This is what I'm doing to determine the year: for inFile in os.listdir('.'): if inFile.isdigit(): month = months[int(inFile)] if int(inFile) <= int(strftime("%m")): year = strftime("%Y") else: year = int(strftime("%Y"))-1 mnYear = month + ", " + str(year) I don't have a clue what to do next. What should I do here?

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  • indexing for faster search of lists in a file??

    - by kaushik
    i have a file having around 1 lakh lists and have a another file with again a list of around an average of 50.. I want to compare 2nd item of list in second file with the 2nd element of 1st file and repeat this for each of the 50 lists in 2nd file and get the result of all the matching element. I have written the code for all this,but this is taking a lot of time as it need to check the whole the 1lakh list some 50 times..i want to improve the speed... please tell me how can i do this.... i cant not post my code as it is part of big code and will be difficult to infer anything from that... please tell what can be done to improve the speed?? thank u,

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  • Get Username from a Cookie

    - by craphunter
    Hi, I use the backend solution from django. I just want to get a username from the cookie or the session_key to get to know the user. How I can do it? from django.contrib.auth.models import User from django.contrib.sessions.models import Session def start(request, template_name="registration/my_account.html"): user_id = request.session.get('session_key') if user_id: name = request.user.username return render_to_response(template_name, locals()) else: return render_to_response('account/noauth.html') Only else is coming up. What am I doing wrong? Am I right then that authenticated means he is logged in? -- Okay this I got! Firstly, if you have some clarification to a question, update the question, don't post an answer or (even worse) another question, as you have done. Secondly, if the user is logged out, by definition he doesn't have a username. I mean the advantage of Cookies is to identify a user again. I just want to place his name on the webpage. Even if he is logged out. Or isnt't it possible?

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  • List comprehension, map, and numpy.vectorize performance

    - by mcstrother
    I have a function foo(i) that takes an integer and takes a significant amount of time to execute. Will there be a significant performance difference between any of the following ways of initializing a: a = [foo(i) for i in xrange(100)] a = map(foo, range(100)) vfoo = numpy.vectorize(foo) a = vfoo(range(100)) (I don't care whether the output is a list or a numpy array.) Is there a better way?

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  • Using adaptive step sizes with scipy.integrate.ode

    - by Mike
    The (brief) documentation for scipy.integrate.ode says that two methods (dopri5 and dop853) have stepsize control and dense output. Looking at the examples and the code itself, I can only see a very simple way to get output from an integrator. Namely, it looks like you just step the integrator forward by some fixed dt, get the function value(s) at that time, and repeat. My problem has pretty variable timescales, so I'd like to just get the values at whatever time steps it needs to evaluate to achieve the required tolerances. That is, early on, things are changing slowly, so the output time steps can be big. But as things get interesting, the output time steps have to be smaller. I don't actually want dense output at equal intervals, I just want the time steps the adaptive function uses.

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  • text overlay for tray icon

    - by AnC
    I have a simple tray icon using PyGTK's gtk.StatusIcon: import pygtk pygtk.require('2.0') import gtk statusIcon = gtk.StatusIcon() statusIcon.set_from_stock(gtk.STOCK_EDIT) statusIcon.set_tooltip('Hello World') statusIcon.set_visible(True) gtk.main() How can I add a text label (one or two characters; basically, unread count) to the tooltip - without creating separate images for set_from_file?

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  • Yet another list comprehension question

    - by relima
    I had this: if Setting["Language"] == "en": f.m_radioBox3.SetSelection(0) elif Setting["Language"] == "pt": f.m_radioBox3.SetSelection(1) elif Setting["Language"] == "fr": f.m_radioBox3.SetSelection(2) elif Setting["Language"] == "es": f.m_radioBox3.SetSelection(3) Then I did this: Linguas = ["en","pt","fr","es"] a = 0 for i in Linguas: if i == Setting["Language"]: f.m_radioBox3.SetSelection(a) a += 1 Is it possible to further simplify this and make it into a one-liner?

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  • get the list and input from one function and run them in different function

    - by rookie
    i have a programm that generate the list and then i ask them to tell me what they want to do from the menu and this is where my problem start i was able to get the input form the user to different function but when i try to use the if else condition it doesn't check, below are my code def menu(x,l): print (x) if x == 1: return make_table(l) if x == 2: y= input("enter a row (as a number) or a column (as an uppercase letter") if y in [ "1",'2','3']: print("Minmum is:",minimum(y,l)) if x== 3: print ('bye') def main(): bad_filename = True l =[] while bad_filename == True: try: filename = input("Enter the filename: ") fp = open(filename, "r") for f_line in fp: f_str=f_line.strip() f_str=f_str.split(',') for unit_str in f_str: unit=float(unit_str) l.append(unit) bad_filename = False except IOError: print("Error: The file was not found: ", filename) #print(l) condition=True while condition==True: print('1- open\n','2- maximum') x=input("Enter the choice") menu(x,l) main() from the bottom function i can get list and i can get the user input and i can get the data and move it in second function but it wont work after that.thank you

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  • List filtering: list comprehension vs. lambda + filter

    - by Agos
    I happened to find myself having a basic filtering need: I have a list and I have to filter it by an attribute of the items. My code looked like this: list = [i for i in list if i.attribute == value] But then i thought, wouldn't it be better to write it like this? filter(lambda x: x.attribute == value, list) It's more readable, and if needed for performance the lambda could be taken out to gain something. Question is: are there any caveats in using the second way? Any performance difference? Am I missing the Pythonic Way™ entirely and should do it in yet another way (such as using itemgetter instead of the lambda)? Thanks in advance

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  • does webapp has 'elseif' or 'elif' in template tags..

    - by zjm1126
    my code is : Hello!~~~ {% if user %} <p>Logged in as {{ user.first_name }} {{ user.last_name }}.</p> {% elif openid_user%} <p>Hello, {{openid_user.nickname}}! Do you want to <a href="{{openid_logout_url}}">Log out?</p> {% else %} <p><a href="/login?redirect={{ current_url }}">google Log in</a>.</p> <p><a href="/twitter">twitter Log in</a>.</p> <p><a href="/facebook">facebook Log in</a>.</p> <p><a href="{{openid_login_url}}">openid Log in</a>.</p> <iframe src="/_openid/login?continue=/"></iframe> {% endif %} the error is : TemplateSyntaxError: Invalid block tag: 'elif' does not webapp has a 'else if ' ? thanks

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  • urllib open - how to control the number of retries

    - by user1641071
    how can i control the number of retries of the "opener.open"? for example, in the following code, it will send about 6 "GET" HTTP requests (i saw it in the Wireshark sniffer) before it goes to the " except urllib.error.URLError" success/no-success lines. password_mgr = urllib.request.HTTPPasswordMgrWithDefaultRealm() password_mgr.add_password(None,url, username, password) handler = urllib.request.HTTPBasicAuthHandler(password_mgr) opener = urllib.request.build_opener(handler) try: resp = opener.open(url,None,1) except urllib.error.URLError as e: print ("no success") else: print ("success!")

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  • Django - Expression based model constraints

    - by rtmie
    Is it possible to set an expression based constraint on a django model object, e.g. If I want to impose a constraint where an owner can have only one widget of a given type that is not in an expired state, but can have as many others as long as they are expired. Obviously I can do this by overriding the save method, but I am wondering if it can be done by setting constraints, e.g. some derivative of the unique_together constraint WIDGET_STATE_CHOICES = ( ('NEW', 'NEW'), ('ACTIVE', 'ACTIVE'), ('EXPIRED', 'EXPIRED') ) class MyWidget(models.Model): owner = models.CharField(max_length=64) widget_type = models.CharField(max_length = 10) widget_state = models.CharField(max_length = 10, choices = WIDGET_STATE_CHOICES) #I'd like to be able to do something like class Meta: unique_together = (("owner","widget_type","widget_state" != 'EXPIRED')

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  • Updating a module level shared dictionary

    - by Vishal
    Hi, A module level dictionary 'd' and is accessed by different threads/requests in a django web application. I need to update 'd' every minute with a new data and the process takes about 5 seconds. What could be best solution where I want the users to get either the old value or the new value of d and nothing in between. I can think of a solution where a temp dictionary is constructed with a new data and assigned to 'd' but not sure how this works! Appreciate your ideas. Thanks

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