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  • how to fetch more than 1000 entities NON keybased?

    - by user291071
    If I should be approaching this problem through a different method, please suggest so. I am creating an item based collaborative filter. I populate the db with the LinkRating2 class and for each link there are more than a 1000 users that I need to call and collect their ratings to perform calculations which I then use to create another table. So I need to call more than 1000 entities for a given link. For instance lets say there are over a 1000 users rated 'link1' there will be over a 1000 instances of this class for the given link property that I need to call. How would I complete this example? class LinkRating2(db.Model): user = db.StringProperty() link = db.StringProperty() rating2 = db.FloatProperty() query =LinkRating2.all() link1 = 'link string name' a = query.filter('link = ', link1) aa = a.fetch(1000)##how would i get more than 1000 for a given link1 as shown? ##keybased over 1000 in other post example i need method for a subset though not key class MyModel(db.Expando): @classmethod def count_all(cls): """ Count *all* of the rows (without maxing out at 1000) """ count = 0 query = cls.all().order('__key__') while count % 1000 == 0: current_count = query.count() if current_count == 0: break count += current_count if current_count == 1000: last_key = query.fetch(1, 999)[0].key() query = query.filter('__key__ > ', last_key) return count

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  • How can I identify an element from a list within another list

    - by Alex
    I have been trying to make a block of code that finds the index of the largest bid for each item. Then I was going to use the index as a way to identify the person who paid that much moneys name. However no matter what i try I can't link the person and what they have gained from the auction together. Here is the code I have been writing: It has to be able to work with any information inputted def sealedBids(): n = int(input('\nHow many people are in the group? ')) z = 0 g = [] s = [] b = [] f = [] w = []#goes by number of items q = [] while z < n: b.append([]) z = z + 1 z = 0 while z < n: g.append(input('Enter a bidders name: ')) z = z + 1 z = 0 i = int(input('How many items are being bid on?')) while z < i: s.append(input('Enter the name of an item: ')) w.append(z) z = z + 1 z = 0 for j in range(n):#specifies which persons bids your taking for k in range(i):#specifies which item is being bid on b[j].append(int(input('How much money has {0} bid on the {1}? '.format(g[j], s[k])))) print(' ') for j in range(n):#calculates fair share f.append(sum(b[j])/n) for j in range(i):#identifies which quantity of money was the largest for each item for k in range(n): if w[j] < b[k][j]: w[j] = b[k][j] q.append(k) any advice is much appreciated.

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  • Tkinter after that survives clock rewinding.

    - by Oren
    I noticed that in my version of Tkinter, the after() call does not survive system clock rewinding. If the after(x, func) was called, and the system clock was rewinded, func will be called only after the clock returned to its time before the rewind + x milliseconds. I assume this is because Tkinter uses the system-clock instead of the "time.clock" (the amount of time that the program is running). I tested it only on windows, and maybe its because I have an old version of Tkinter. I want my App to work on computers that synchronize their clock from the network... Does anyone have a simple solution?

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  • slicing arrays in numpy/scipy

    - by user248237
    I have an array like: a = array([[1,2,3],[3,4,5],[4,5,6]]) what's the most efficient way to slice out a 1x2 array out of this that has only the first two columns of "a"? I.e., array([[2,3],[4,5],[5,6]]) in this case. thanks.

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  • using dictionary to assign misspelled words to its line number

    - by jad
    This is the code I have so far d = {} counter = 0 for lines in words: counter += 1 for word in text1: if word not in words: d[word] = [counter] else: d[word].append(counter) print(word, d[counter]) words = my text file text1 is my misspelled words But this gives me an error. What I want to do is print the word and the line number e.g. togeher 5 7

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  • What does this `_time_independent_equals` mean?

    - by Satoru.Logic
    In the tornado.web module there is a function called _time_independent_equals: def _time_independent_equals(a, b): if len(a) != len(b): return False result = 0 for x, y in zip(a, b): result |= ord(x) ^ ord(y) return result == 0 It is used to compare secure cookie signatures, and thus the name. But regarding the implementation of this function, is it just a complex way to say a==b?

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  • Construct Numpy index given list of starting and ending positions

    - by Abiel
    I have two identically-sized numpy.array objects (both one-dimensional), one of which contains a list of starting index positions, and the other of which contains a list of ending index positions (alternatively you could say I have a list of starting positions and window lengths). In case it matters, the slices formed by the starting and ending positions are guaranteed to be non-overlapping. I am trying to figure out how to use these starting and ending positions to form an index for another array object, without having to use a loop. For example: import numpy as np start = np.array([1,7,20]) end = np.array([3,10,25]) Want to reference somearray[1,2,7,8,9,20,21,22,23,24])

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  • How to execute machine language from memory?

    - by Mike Curry
    I wrote a program to compile a simple text program to a compiled executable... Is it possible that I can load an executable to memory an some how point a pc counter to the memory space at will? Here is what I made that I would like to store the programs to memory for execution on demand... Kind of wanting to make a little web language like php but compile it... Just for learning. http://spiceycurry.blogspot.com/2010/05/simple-compilable-programming-language.html

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  • Does Google appengine cache external requests?

    - by Andy Hume
    I have a very simple application running on appengine that requests a web page every five minutes and parses for a specific piece of data. Everything works fine except that the response I get back from the external request (using urllib2) doesn't reflect the latest changes to the page. Sometimes it takes a few minutes to get the latest, sometimes over an hour. Is there a transparent layer of caching that appengine puts in place? Or is there something else I am missing here? I've looked at the caching headers of the requested page and there is no Expires or LastModified's sent. Update: Sometimes, it will get the new version of the page for a number of requests and then randomly later get an old out of date version.

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  • Django filter by hour

    - by DJPy
    I've found that link: http://code.djangoproject.com/attachment/ticket/8424/time_filters.diff and changed my django 1.2 files by adding taht what you can see there. But now, when I'm trying to write Entry.objects.filter(pub_date__hour = x) - the result is following error: Field has invalid lookup: hour What should I do else, to make it work? (sorry for my english)

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  • Calculating the null space of a matrix

    - by Ainsworth
    I'm attempting to solve a set of equations of the form Ax = 0. A is known 6x6 matrix and I've written the below code using SVD to get the vector x which works to a certain extent. The answer is approximately correct but not good enough to be useful to me, how can I improve the precision of the calculation? Lowering eps below 1.e-4 causes the function to fail. from numpy.linalg import * from numpy import * A = matrix([[0.624010149127497 ,0.020915658603923 ,0.838082638087629 ,62.0778180312547 ,-0.336 ,0], [0.669649399820597 ,0.344105317421833 ,0.0543868015800246 ,49.0194290212841 ,-0.267 ,0], [0.473153758252885 ,0.366893577716959 ,0.924972565581684 ,186.071352614705 ,-1 ,0], [0.0759305208803158 ,0.356365401030535 ,0.126682113674883 ,175.292109352674 ,0 ,-5.201], [0.91160934274653 ,0.32447818779582 ,0.741382053883291 ,0.11536775372698 ,0 ,-0.034], [0.480860406786873 ,0.903499596111067 ,0.542581424762866 ,32.782593418975 ,0 ,-1]]) def null(A, eps=1e-3): u,s,vh = svd(A,full_matrices=1,compute_uv=1) null_space = compress(s <= eps, vh, axis=0) return null_space.T NS = null(A) print "Null space equals ",NS,"\n" print dot(A,NS)

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  • Django: Get bound 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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  • how to change headers data

    - by Moayyad Yaghi
    hello i have the following class class AssetTableModel(QtCore.QAbstractTableModel): def init(self,filename=''): super(AssetTableModel,self).init() self.fileName=filename self.dirty = False self.assets = [] self.setHeaderData(0,QtCore.Qt.Horizontal,QtCore.QVariant('moayyad'),QtCore.Qt.EditRole) and i need to change the headers of the columns or the rows ,i used ( self.setHeaderdata()) but its not working ,i have a table that consistes of 2 columns and 2 rows only. is there any other function that changes headers ??. please help thanx in adnvance

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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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  • More compact layout

    - by Jesse Aldridge
    In the following code, I'd like to get rid of the margin around the buttons. I'd like to have the buttons stretch all the way to the edge of the frame. How can I do that? import sys from PyQt4.QtGui import * from PyQt4.QtCore import * app = QApplication(sys.argv) window = QWidget() layout = QVBoxLayout() layout.setSpacing(0) window.setLayout(layout) for i in range(2): layout.addWidget(QPushButton()) window.show() app.exec_()

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  • Extract files from zip folder and store these files in blobstore

    - by Eng_Engineer
    i want to upload zip folder from file input in form the i want to extract the contents of this uploaded zip folder,and store the contents (files)of this zip in the blobstore in order to download them after putting these files in one folder,but the problem is that i can't deal with the zip folder directly(to read it), i tried as this: form = cgi.FieldStorage() file_upload = form['file'] zip1=file_upload.filename zipstream=StringIO.StringIO(zip1.read()) But the problem still that i can't read the zip as previous,also i tried to read zip folder directly like this: z1=zipfile.ZipFile(zip1,"r") But there was an error in this way.Please can any one help me.Thanks in advance.

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  • make a tree based on the key of each element in list.

    - by cocobear
    >>> s [{'000000': [['apple', 'pear']]}, {'100000': ['good', 'bad']}, {'200000': ['yeah', 'ogg']}, {'300000': [['foo', 'foo']]}, {'310000': [['#'], ['#']]}, {'320000': ['$', ['1']]}, {'321000': [['abc', 'abc']]}, {'322000': [['#'], ['#']]}, {'400000': [['yeah', 'baby']]}] >>> for i in s: ... print i ... {'000000': [['apple', 'pear']]} {'100000': ['good', 'bad']} {'200000': ['yeah', 'ogg']} {'300000': [['foo', 'foo']]} {'310000': [['#'], ['#']]} {'320000': ['$', ['1']]} {'321000': [['abc', 'abc']]} {'322000': [['#'], ['#']]} {'400000': [['yeah', 'baby']]} i want to make a tree based on the key of each element in list. result in logic will be: {'000000': [['apple', 'pear']]} {'100000': ['good', 'bad']} {'200000': ['yeah', 'ogg']} {'300000': [['foo', 'foo']]} {'310000': [['#'], ['#']]} {'320000': ['$', ['1']]} {'321000': [['abc', 'abc']]} {'322000': [['#'], ['#']]} {'400000': [['yeah', 'baby']]} perhaps a nested list can implement this or I need a tree type?

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  • Saving data in a inherited django model

    - by aldeano
    I'm building an app to save data and some calculations made with those datas, the idea is keep the data in one model and the calculations in other. So, the models are like this: class FreshData(models.Model): name = models.CharField(max_length=20) one = models.IntegerField() two = models.IntegerField() def save(self, *args, **kwargs): Calculations() Calculations.three = self.one + self.two super(FreshData, self).save(*args, **kwargs) Calculations.save() class Calculations(FreshData): three = models.IntegerField() I've got a valueerror pointing out "self.one" and "self.two" as without value. I keep the idea in witch my design is wrong and django has a simpler way to store related data.

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  • Clear sqlalchemy reflection cache

    - by OrganicPanda
    Hi all, I'm using sqlalchemy's reflection tools to get a Table object. I do this because these tables are dynamic and tables/columns can change. Here's the code I'm using: def getTableByReflection(self, tableName, metadata, engine): return Table(tableName, metadata, autoload = True, autoload_with = engine) The problem is that when the above code is run twice it seems to return the same results regardless of whether or not the columns have changed. I have tried refreshing using the mysession.refresh(mytable) but that fails because the table is not attached to any metadata - which makes sense but then why am I seeing cached results? Is there any way to tell the metadata/engine/session to forget about this table and let me load it cleanly?

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