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  • MongoDB lists with paginations?

    - by Timmy
    for documents with lists with pagination, is it better to embed or use reference? im reading the custom type "SONManipulator" and it appears to transform every thing on retrieval, even the sub docs. i want to keep the list in the document sorted, should this impact anything?

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  • How to insert several thousand columns into sqlite3?

    - by user291071
    Similar to my last question, but I ran into problem lets say I have a simple dictionary like below but its Big, when I try inserting a big dictionary using the methods below I get operational error for the c.execute(schema) for too many columns so what should be my alternate method to populate an sql databases columns? Using the alter table command and add each one individually? import sqlite3 con = sqlite3.connect('simple.db') c = con.cursor() dic = { 'x1':{'y1':1.0,'y2':0.0}, 'x2':{'y1':0.0,'y2':2.0,'joe bla':1.5}, 'x3':{'y2':2.0,'y3 45 etc':1.5} } # 1. Find the unique column names. columns = set() for _, cols in dic.items(): for key, _ in cols.items(): columns.add(key) # 2. Create the schema. col_defs = [ # Start with the column for our key name '"row_name" VARCHAR(2) NOT NULL PRIMARY KEY' ] for column in columns: col_defs.append('"%s" REAL NULL' % column) schema = "CREATE TABLE simple (%s);" % ",".join(col_defs) c.execute(schema) # 3. Loop through each row for row_name, cols in dic.items(): # Compile the data we have for this row. col_names = cols.keys() col_values = [str(val) for val in cols.values()] # Insert it. sql = 'INSERT INTO simple ("row_name", "%s") VALUES ("%s", "%s");' % ( '","'.join(col_names), row_name, '","'.join(col_values) )

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  • Django: automatically import MEDIA_URL in context

    - by pistacchio
    Hi, like exposed here, one can set a MEDIA_URL in settings.py (for example i'm pointing to Amazon S3) and serve the files in the view via {{ MEDIA_URL }}. Since MEDIA_URL is not automatically in the context, one have to manually add it to the context, so, for example, the following works: #views.py from django.shortcuts import render_to_response from django.template import RequestContext def test(request): return render_to_response('test.html', {}, context_instance=RequestContext(request)) This means that in each view.py file i have to add from django.template import RequestContext and in each response i have to explicitly specify context_instance=RequestContext(request). Is there a way to automatically (DRY) add MEDIA_URL to the default context? Thanks in advance.

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  • Replace ",**" with a linebreak using RegEx (or something else)

    - by John
    I'm getting started with RegEx and I was wondering if anyone could help me craft a statement to convert coordinates as follows: 145.00694,-37.80421,9 145.00686,-37.80382,9 145.00595,-37.8035,16 145.00586,-37.80301,16 to 145.00694,-37.80421 145.00686,-37.80382 145.00595,-37.8035 145.00586,-37.80301 (Strip off the last comma and value and turn it into a line break.) I can't figure out how to use wildcards to do something like that. Any help would be greatly appreciated! Thanks.

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  • Sending object C from class A to class B

    - by user278618
    Hi, I can't figure out how to design classes in my system. In classA I create object selenium (it simulates user actions at website). In this ClassA I create another objects like SearchScreen, Payment_Screen and Summary_Screen. # -*- coding: utf-8 -*- from selenium import selenium import unittest, time, re class OurSiteTestCases(unittest.TestCase): def setUp(self): self.verificationErrors = [] self.selenium = selenium("localhost", 5555, "*chrome", "http://www.someaddress.com/") time.sleep(5) self.selenium.start() def test_buy_coffee(self): sel = self.selenium sel.open('/') sel.window_maximize() search_screen=SearchScreen(self.selenium) search_screen.choose('lavazza') payment_screen=PaymentScreen(self.selenium) payment_screen.fill_test_data() summary_screen=SummaryScreen(selenium) summary_screen.accept() def tearDown(self): self.selenium.stop() self.assertEqual([], self.verificationErrors) if __name__ == "__main__": unittest.main() It's example SearchScreen module: class SearchScreen: def __init__(self,selenium): self.selenium=selenium def search(self): self.selenium.click('css=button.search') I want to know if there is anything ok with a design of those classes?

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  • App Engine Bulkloader

    - by gurkan
    Hi all, I am trying to use Bulkloader of google app engine but unfortunately could not understand what to do from documentation. It says add this part to app.yaml builtins: - remote_api: on ok i have added. Then says that i have to execute this command appcfg.py update but i don't have any appcfg.py file. And also what is the command which executes this line? Please somebody tell me what i am missing I use AppEngineLauncher to upload my project to server.. I have naver used a command to update or upload it. Thanks in advance..

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  • Trie Backtracking in Recursion

    - by Darksky
    I am building a tree for a spell checker with suggestions. Each node contains a key (a letter) and a value (array of letters down that path). So assume the following sub-trie in my big trie: W / \ a e | | k k | | is word--> e e | ... This is just a subpath of a sub-trie. W is a node and a and e are two nodes in its value array etc... At each node, I check if the next letter in the word is a value of the node. I am trying to support mistyped vowels for now. So 'weke' will yield 'wake' as a suggestion. Here's my searchWord function in my trie: def searchWord(self, word, path=""): if len(word) > 0: key = word[0] word = word[1:] if self.values.has_key(key): path = path + key nextNode = self.values[key] return nextNode.searchWord(word, path) else: # check here if key is a vowel. If it is, check for other vowel substitutes else: if self.isWord: return path # this is the word found else: return None Given 'weke', at the end when word is of length zero and path is 'weke', my code will hit the second big else block. weke is not marked as a word and so it will return with None. This will return out of searchWord with None. To avoid this, at each stack unwind or recursion backtrack, I need to check if a letter is a vowel and if it is, do the checking again. I changed the if self.values.has_key(key) loop to the following: if self.values.has_key(key): path = path + key nextNode = self.values[key] ret = nextNode.searchWord(word, path) if ret == None: # check if key == vowel and replace path # return nextNode.searchWord(... return ret What am I doing wrong here? What can I do when backtracking to achieve what I'm trying to do?

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  • PyGTK "assertion GTK_IS_WINDOW failed

    - by iAndr0idOs
    I'm trying to build a web browser using PyGTK and PyWebKit However, I'm pretty sure my question only concerns PyGTK I have a custom gtk.Notebook class, with an "add tab" button as the last tab. When I click it, it gives me the error /home/ruiqimao/workspace/PyBrowser/src/browser/__init__.py:161: GtkWarning: IA__gdk_window_get_cursor: assertion `GDK_IS_WINDOW (window)' failed gtk.main() twice. And then, my new tab won't show up. I have no idea what is going on, so here is my whole code If any of you could help me, that would be great! Thanks! [EDIT]: Just found out that the problem lies in the w.show_all() line What could this mean?

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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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  • Encrypt file using M2Crypto

    - by Bear
    It is known that I can read the whole file content in memory and encrypt it using the following code. contents = fin.read() cipher = M2Crypto.EVP.Cipher(alg="aes_128_cbc", key = aes_key, iv = aes_iv, op = 1) encryptedContents = cipher.update(contents) encryptedContents += cipher.final() But what if the file size is large, is there a way for me to pass the input stream to M2Crypto instead of reading the whole file first?

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  • Twisted Spread: How to authenticate each RPC with digital signature

    - by kronat
    I have remote objects which talk each others with RPCs, using Twisted Spread. I want that objects authenticate messages, before using them, with digital signatures, but I don't know where to start to implement this. In my head, the Root object must have a public/private key pair, and the Client too. When a message is sent, a digital signature of the hash is added, and when it is received, the signature is checked. Is the Protocol part where I need to add these adds and checks? Thank you

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  • Best practice- How to team-split a django project while still allowing code reusal

    - by Infinity
    I know this sounds kind of vague, but please let me explain- I'm starting work on a brand new project, it will have two main components: "ACME PRODUCT" (think Gmail, Meebo, etc), and "THE SITE" (help, information, marketing stuff, promotional landing pages, etc lots of marketing-induced cruft). So basically the url /acme/* will load stuff in the uber cool ajaxy application, and every other URI will load stuff in the other site. Problem: "THE SITE" component is out of my hands, and will be handled by a consultants team that will work closely with marketing, And I and my team will work solely on the ACME PRODUCT. Question: How to set up the django project in such a way that we can have: Seperate releases. (They can push new marketing pages and functionality without having to worry about the state of our code. Maybe even separate Subversion "projects") Minimize impact (on our product) of whatever flying-unicorns-hocus-pocus the other team codes into the site. Still allow some code reusal. My main concern is that the ACME product needs to be rock solid, and therefore needs to be somewhat isolated of whatever mistakes/code bloopers the consultants make in their marketing side of the site. How have you handled this? Any ideas? Thanks!

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  • How to sort a Pandas DataFrame according to multiple criteria?

    - by user1715271
    I have the following DataFrame containing song names, their peak chart positions and the number of weeks they spent at position no 1: Song Peak Weeks 76 Paperback Writer 1 16 117 Lady Madonna 1 9 118 Hey Jude 1 27 22 Can't Buy Me Love 1 17 29 A Hard Day's Night 1 14 48 Ticket To Ride 1 14 56 Help! 1 17 109 All You Need Is Love 1 16 173 The Ballad Of John And Yoko 1 13 85 Eleanor Rigby 1 14 87 Yellow Submarine 1 14 20 I Want To Hold Your Hand 1 24 45 I Feel Fine 1 15 60 Day Tripper 1 12 61 We Can Work It Out 1 12 10 She Loves You 1 36 155 Get Back 1 6 8 From Me To You 1 7 115 Hello Goodbye 1 7 2 Please Please Me 2 20 92 Strawberry Fields Forever 2 12 93 Penny Lane 2 13 107 Magical Mystery Tour 2 16 176 Let It Be 2 14 0 Love Me Do 4 26 157 Something 4 9 166 Come Together 4 10 58 Yesterday 8 21 135 Back In The U.S.S.R. 19 3 164 Here Comes The Sun 58 19 96 Sgt. Pepper's Lonely Hearts Club Band 63 12 105 With A Little Help From My Friends 63 7 I'd like to rank these songs in order of popularity, so I'd like to sort them according to the following criteria: songs that reached the highest position come first, but if there is a tie, the songs that remained in the charts for the longest come first. I can't seem to figure out how to do this in Pandas.

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  • How can this code be made more Pythonic?

    - by usethedeathstar
    This next part of code does exactly what I want it to do. dem_rows and dem_cols contain float values for a number of things i can identify in an image, but i need to get the nearest pixel for each of them, and than to make sure I only get the unique points, and no duplicates. The problem is that this code is ugly and as far as I get it, as unpythonic as it gets. If there would be a pure-numpy-solution (without for-loops) that would be even better. # next part is to make sure that we get the rounding done correctly, and than to get the integer part out of it # without the annoying floatingpoint-error, and without duplicates fielddic={} for i in range(len(dem_rows)): # here comes the ugly part: abusing the fact that i overwrite dictionary keys if I get duplicates fielddic[int(round(dem_rows[i]) + 0.1), int(round(dem_cols[i]) + 0.1)] = None # also very ugly: to make two arrays of integers out of the first and second part of the keys field_rows = numpy.zeros((len(fielddic.keys())), int) field_cols = numpy.zeros((len(fielddic.keys())), int) for i, (r, c) in enumerate(fielddic.keys()): field_rows[i] = r field_cols[i] = c

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  • Can Django be used for web services?

    - by alex
    My friend said, "Pylons is so much better for web services." My other friend said, "You can modify Django in a way to do exactly whatever you like." In Django, what is necessary to be modified (urls.py? models classes? settings?) in order to do "web services" with APIs and REST and versioning, etc etc.?

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  • A faster alternative to Pandas `isin` function

    - by user3576212
    I have a very large data frame df that looks like: ID Value1 Value2 1345 3.2 332 1355 2.2 32 2346 1.0 11 3456 8.9 322 And I have a list that contains a subset of IDs ID_list. I need to have a subset of df for the ID contained in ID_list. Currently, I am using df_sub=df[df.ID.isin(ID_list)] to do it. But it takes a lot time. IDs contained in ID_list doesn't have any pattern, so it's not within certain range. (And I need to apply the same operation to many similar dataframes. I was wondering if there is any faster way to do this. Will it help a lot if make ID as the index? Thanks!

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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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  • Google App Engine - Is os.environ reset between requests?

    - by Ian Charnas
    Hello I can't think of a way to test this and was hoping someone here knew the answer... I'm storing some request-specific data in os.environ, and was wondering if that data was going to leak to other requests. Does anyone know? Yes I realize that it's normal to use request.environ for this, and usually I do, but I want to store the currently authorized user ID (I'm using custom auth, not GAE auth) inside os.environ so that the models know the currently logged in user (remember, they don't have access to request.environ) without me having to pass the request object to just about every single model method. any help would be greatly appreciated Ian

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  • programs hangs during socket interaction

    - by herrturtur
    I have two programs, sendfile.py and recvfile.py that are supposed to interact to send a file across the network. They communicate over TCP sockets. The communication is supposed to go something like this: sender =====filename=====> receiver sender <===== 'ok' ======= receiver or sender <===== 'no' ======= receiver if ok: sender ====== file ======> receiver I've got The sender and receiver code is here: Sender: import sys from jmm_sockets import * if len(sys.argv) != 4: print "Usage:", sys.argv[0], "<host> <port> <filename>" sys.exit(1) s = getClientSocket(sys.argv[1], int(sys.argv[2])) try: f = open(sys.argv[3]) except IOError, msg: print "couldn't open file" sys.exit(1) # send filename s.send(sys.argv[3]) # receive 'ok' buffer = None response = str() while 1: buffer = s.recv(1) if buffer == '': break else: response = response + buffer if response == 'ok': print 'receiver acknowledged receipt of filename' # send file s.send(f.read()) elif response == 'no': print "receiver doesn't want the file" # cleanup f.close() s.close() Receiver: from jmm_sockets import * s = getServerSocket(None, 16001) conn, addr = s.accept() buffer = None filename = str() # receive filename while 1: buffer = conn.recv(1) if buffer == '': break else: filename = filename + buffer print "sender wants to send", filename, "is that ok?" user_choice = raw_input("ok/no: ") if user_choice == 'ok': # send ok conn.send('ok') #receive file data = str() while 1: buffer = conn.recv(1) if buffer=='': break else: data = data + buffer print data else: conn.send('no') conn.close() I'm sure I'm missing something here in the sorts of a deadlock, but don't know what it is.

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