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  • reading specific lines from a file

    - by MacUsers
    What's the best way of reading only the specific lines (based on matching text) from a file? This is what I'm doing now: match_txt = "lhcb" for inFile in os.listdir('.'): readFile = open(inFile, 'r') lines = readFile.readlines() readFile.close() for line in lines: if line.find(match_txt)==0: #< do stuff here > i.e. I'm reading the lines, only with "lhcb" in it, from all the files in the present directory one by one. Is it the best way of doing that? Can it be done without loading the whole file in the memory in the first place?

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  • Many-to-one relationship in SQLAlchemy

    - by Arrieta
    This is a beginner-level question. I have a catalog of mtypes: mtype_id name 1 'mtype1' 2 'mtype2' [etc] and a catalog of Objects, which must have an associated mtype: obj_id mtype_id name 1 1 'obj1' 2 1 'obj2' 3 2 'obj3' [etc] I am trying to do this in SQLAlchemy by creating the following schemas: mtypes_table = Table('mtypes', metadata, Column('mtype_id', Integer, primary_key=True), Column('name', String(50), nullable=False, unique=True), ) objs_table = Table('objects', metadata, Column('obj_id', Integer, primary_key=True), Column('mtype_id', None, ForeignKey('mtypes.mtype_id')), Column('name', String(50), nullable=False, unique=True), ) mapper(MType, mtypes_table) mapper(MyObject, objs_table, properties={'mtype':Relationship(MType, backref='objs', cascade="all, delete-orphan")} ) When I try to add a simple element like: mtype1 = MType('mtype1') obj1 = MyObject('obj1') obj1.mtype=mtype1 session.add(obj1) I get the error: AttributeError: 'NoneType' object has no attribute 'cascade_iterator' Any ideas?

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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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  • GQL Request BadArgument Error. How to get around with my case?

    - by awegawef
    My query is essentially the following: entries=Entry.all().order("-votes").order("-date").filter("votes >", VOTE_FILTER).fetch(PAGE_SIZE+1, page* PAGE_SIZE) I want to grab N of the latest entries that have a voting score above some benchmark (VOTE_FILTER). Google currently says that I cannot filter on 'votes' because I order by 'date.' I don't see a way that I can do this the way I want to, so I'd appreciate any advice.

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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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  • How do I do a semijoin using SQLAlchemy?

    - by Jason Baker
    http://en.wikipedia.org/wiki/Relational_algebra#Semijoin Let's say that I have two tables: A and B. I want to make a query that would work similarly to the following SQL statement using the SQLAlchemy orm: SELECT A.* FROM A, B WHERE A.id = B.id AND B.type = 'some type'; The thing is that I'm trying to separate out A and B's logic into different places. So I'd like to make two queries that I can define in separate places: one where A uses B as a subquery, but only returns rows from A. I'm sure this is fairly easy to do, but an example would be nice if someone could show me.

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  • Connect to a DB with an encrypted password with Django?

    - by Liam
    My place of employment requires that all passwords must be encrypted, including the ones used to connect to a database. What's the best way of handling this? I'm using the development version of Django with MySQL at the moment, but I will be eventually migrating to Oracle. Is this a job for Django, or the database? Edit: The encrypted password should be stored in the settings.py file, or somewhere else in the filesystem. This is the password that will be used to connect to the database.

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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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  • UDP security and identifying incoming data.

    - by Charles
    I have been creating an application using UDP for transmitting and receiving information. The problem I am running into is security. Right now I am using the IP/socketid in determining what data belongs to whom. However, I have been reading about how people could simply spoof their IP, then just send data as a specific IP. So this seems to be the wrong way to do it (insecure). So how else am I suppose to identify what data belongs to what users? For instance you have 10 users connected, all have specific data. The server would need to match the user data to this data we received. The only way I can see to do this is to use some sort of client/server key system and encrypt the data. I am curious as to how other applications (or games, since that's what this application is) make sure their data is genuine. Also there is the fact that encryption takes much longer to process than unencrypted. Although I am not sure by how much it will affect performance. Any information would be appreciated. Thanks.

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  • Flask Admin didn't show all fields

    - by twoface88
    I have model like this: class User(db.Model): __tablename__ = 'users' __table_args__ = {'mysql_engine' : 'InnoDB', 'mysql_charset' : 'utf8'} id = db.Column(db.Integer, primary_key=True) username = db.Column(db.String(80), unique=True) email = db.Column(db.String(120), unique=True) _password = db.Column('password', db.String(80)) def __init__(self, username = None, email = None, password = None): self.username = username self.email = email self._set_password(password) def _set_password(self, password): self._password = generate_password_hash(password) def _get_password(self): return self._password def check_password(self, password): return check_password_hash(self._password, password) password = db.synonym("_password", descriptor=property(_get_password, _set_password)) def __repr__(self): return '<User %r>' % self.username I have ModelView: class UserAdmin(sqlamodel.ModelView): searchable_columns = ('username', 'email') excluded_list_columns = ['password'] list_columns = ('username', 'email') form_columns = ('username', 'email', 'password') But no matter what i do, flask admin didn't show password field when i'm editing user info. Is there any way ? Even just to edit hash code. UPDATE: https://github.com/mrjoes/flask-admin/issues/78

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  • Django shell command to change a value in json data

    - by crozzfire
    I am a django newbie and i was playing around in django's manage.py shell. Here is something i am trying in the shell: >>> data [{'primary_program': False, 'id': 3684}, {'primary_program': True, 'id': 3685}] >>> data[0] {'primary_program': False, 'id': 3684} >>> data[1] {'primary_program': True, 'id': 3685} >>> data[0].values() [False, 3684] >>> data[1].values() [True, 3685] >>> How should i give a command here to update the value of primary_program in data[1] to False and keep the rest of the json the same?

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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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  • Pygame single push event

    - by Miller92Time
    in Pygame i am trying to translate an image by 10% in each direction using each arrow key. right now the code i am using moves the image as long as the key is pushed down, what I want is for it to move only once regardless if the key is still pushed down or not. if event.type == KEYDOWN: if (event.key == K_RIGHT): DISPLAYSURF.fill((255,255,255)) #Clears the screen translation_x(100) draw(1) if (event.key == K_LEFT): DISPLAYSURF.fill((255,255,255)) #Clears the screen translation_x(-100) draw(2) if (event.key == K_UP): DISPLAYSURF.fill((255,255,255)) #Clears the screen translation_y(100) draw(3) if (event.key == K_DOWN): DISPLAYSURF.fill((255,255,255)) #Clears the screen translation_y(-100) draw(4) is there a simpler way of implementing this besides using time.sleep

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  • How to make zebra table with Sphinx? Or how to use jQuery with Sphinx?

    - by prosseek
    I think the table generated from Sphinx is not pretty, as it produces the following HTML code for table. <table border="1" class="docutils"> <colgroup> <col width="43%" /> <col width="29%" /> <col width="29%" /> </colgroup> <thead valign="bottom"> <tr><th class="head">Graph</th> <th class="head">HIR</th> <th class="head">AIR</th> </tr> </thead> <tbody valign="top"> <tr><td>Graph</td> <td>Circuit</td> <td>System</td> </tr> </tbody> </table> How can I turn in into pretty one, for example, zebra table? The HTML generated html has the jQuery, and according to this site, it's just one line of code to have a zebra table, but I'm not sure how to use jQuery to make a zebra table. $("tr:nth-child(odd)").addClass("odd"); Q: How to use jQuery with Sphinx? Q: Is there any other way to have a zebra table with Sphinx?

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  • Infinite recursion trying to check all elements of a TreeCtrl

    - by mavnn
    I have a TreeCtrl in which more than one Item can be assigned the same object as PyData. When the object is updated, I want to update all of the items in the tree which have that object as their PyData. I thought the following code would solve the problem quite neatly, but for some reason the logical test (current != self.GetFirstVisibleItem()) always returns true leading to infinite recursion. Can anyone explain why? def RefreshNodes(self, obj, current=None): print "Entered refresh" current = current or self.GetFirstVisibleItem() if current.IsOk(): print self.GetPyData(current).name if self.GetPyData(current) == obj: self.RefreshNode(current) current = self.GetNextVisible(current) if current != self.GetFirstVisibleItem(): self.RefreshNodes(obj, current) Edit: the above is obviously part of a class based on wx.TreeCtrl

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  • Recursion function not working properly

    - by jakecar
    I'm having quite a hard time figuring out what's going wrong here: class iterate(): def init(self): self.length=1 def iterated(self, n): if n==1: return self.length elif n%2==0: self.length+=1 self.iterated(n/2) elif n!=1: self.length+=1 self.iterated(3*n+1) For example, x=iterate() x.iterated(5) outputs None. It should output 6 because the length would look like this: 5 -- 16 -- 8 -- 4 -- 2 -- 1 After doing some debugging, I see that the self.length is returned properly but something goes wrong in the recursion. I'm not really sure. Thanks for any help.

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  • How do I delete in Django? (mysql transactions)

    - by alex
    If you are familiar with Django, you know that they have a Authentication system with User model. Of course, I have many other tables that have a Foreign Key to this User model. If I want to delete this user, how do I architect a script (or through mysql itself) to delete every table that is related to this user? My only worry is that I can do this manually...but if I add a table , but I forget to add that table to my DELETE operation...then I have a row that links to a deleted, non-existing User.

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  • Processing forms that generate many rows in DB

    - by Zack
    I'm wondering what the best approach to take here is. I've got a form that people use to register for a class and a lot of times the manager of a company will register multiple people for the class at the same time. Presently, they'd have to go through the registration process multiple times and resubmit the form once for every person they want to register. What I want to do is give the user a form that has a single <input/> for one person to register with, along with all the other fields they'll need to fill out (Email, phone number, etc); if they want to add more people, they'll be able to press a button and a new <input/> will be generated. This part I know how to do, but I'm including it to best describe what I'm aiming to do. The part I don't know how to approach is processing that data the form submits, I need some way of making a new row in the Registrant table for every <input/> that's added and include the same contact information (phone, email, etc) as the first row with that row. For the record, I'm using the Django framework for my back-end code. What's the best approach here? Should it just POST the form x times for x people, or is there a less "brute force" way of handling this?

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  • string problems, tuple strings.

    - by suresh
    a tuple representing starting points for the first substring, a tuple representing starting points for the second substring, and the length of the first substring. The function should return a tuple of all members (call it n) of the first tuple for which there is an element in the second tuple n+m+1 = k, where m is the length of the first substring. Complete the definition def constrainedMatchPair(firstMatch,secondMatch,length):

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