Search Results

Search found 13683 results on 548 pages for 'python sphinx'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 410/548 | < Previous Page | 406 407 408 409 410 411 412 413 414 415 416 417  | Next Page >

  • 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

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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!")

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • I'm getting the following error ''expected an indented block'' Where is the failing code?

    - by user1833814
    import math def area(base, height): '''(number,number) -> number Return the area of a wirh given base and height. >>>area(10,40) 200.0 ''' return base * height / 2 def perimeter(side1, side2, side3): '''(number,number,number) -> number Return the perimeter of the triangle with sides of length side1,side2 and side3. >>>perimeter(3,4,5) 12 >>>perimeter(10.5,6,9.3) 25.8 ''' return (side1 + side2 + side3) def semiperimeter(side1, side2, side3): return perimeter(side1, side2, side3) / 2 def area_hero(side1, side2, side3): semi = semiperimeter(side1, side2, side3) area = math.sqrt((semi * (semi - side1) * (semi - side2) * (semi - side3)) return area

    Read the article

  • efficiently convert string (or tuple) to ctypes array

    - by Mu Mind
    I've got code that takes a PIL image and converts it to a ctypes array to pass out to a C function: w_px, h_px = img.size pixels = struct.unpack('%dI'%(w_px*h_px), img.convert('RGBA').tostring()) pixels_array = (ctypes.c_int * len(pixels))(*pixels) But I'm dealing with big images, and unpacking that many items into function arguments seems to be noticeably slow. What's the simplest thing I can do to get a reasonable speedup? I'm only converting to a tuple as an intermediate step, so if it's unnecessary, all the better.

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • Best practice: How to persist simple data without a database in django?

    - by Infinity
    I'm building a website that doesn't require a database because a REST API "is the database". (Except you don't want to be putting site-specific things in there, since the API is used by mostly mobile clients) However there's a few things that normally would be put in a database, for example the "jobs" page. You have master list view, and the detail views for each job, and it should be easy to add new job entries. (not necessarily via a CMS, but that would be awesome) e.g. example.com/careers/ and example.com/careers/77/ I could just hardcode this stuff in templates, but that's no DRY- you have to update the master template and the detail template every time. What do you guys think? Maybe a YAML file? Or any better ideas? Thx

    Read the article

  • Change web service url for a suds client on runtime (keeping the wsdl)

    - by patanpatan
    Hi. First of all, my question is similar to this one But it's a little bit different. What we have is a series of environments, with the same set of services. For some environments (the local ones) we can get access to the wsdl, and thus generating the suds client. For external environment, we cannot access the wsdl. But being the same, I was hoping I can change just the URL without regenerating the client. I've tried cloning the client, but it doesn't work.

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 406 407 408 409 410 411 412 413 414 415 416 417  | Next Page >