Search Results

Search found 70655 results on 2827 pages for 'python time'.

Page 257/2827 | < Previous Page | 253 254 255 256 257 258 259 260 261 262 263 264  | Next Page >

  • Why are Back In Time snapshots so large?

    - by Chethan S.
    I just backed up the contents of my home partition onto my external hard drive using Back In Time. I browsed to the backed up contents in the external drive and under properties it showed me the size as 9.6 GB. As I read that in next snapshots I create, Back In Time does not backup everything but creates hard links for older contents and saves newer contents, I wanted to test it. So I copied two small files into my home partition and ran 'Take Snapshot' again. The operation completed within a minute - first it checked previous snapshot, assessed the changes, detected two new files and synced them. After this when I browsed to the backed up contents, I was surprised to see the newer and older backup taking up 9.6 GB each. Isn't this a waste of hard drive space? Or did I interpret something wrongly?

    Read the article

  • How much time it will take to learn 3ds Max

    - by Mirror51
    I am not a 3d developer but i want to lean 3ds max just for simple house building with 2-3 rooms. Actually i don't want to develop from scratch . What i really want to do is get the existing models of homes , rooms , hotels from the internet and add my name there or my photo there , just for fun . SO i want to know that how much time do u think it will take me to that sort of stuff. Its not my career but just hobby . If its going to take longer time , then i don't want to waste but i can get going in one week or so that will go good but i want to ask from experience developers thanks

    Read the article

  • Contiguous Time Periods

    It is always more efficient to maintain referential integrity by using constraints rather than triggers. Sometimes it isn't obvious how to do this. Until a recent idea by Alex Kuznetsov, the history table presented problems for checking data that were difficult to solve with constraints. Joe Celko explains. Free trial of SQL Backup™“SQL Backup was able to cut down my backup time significantly AND achieved a 90% compression at the same time!” Joe Cheng. Download a free trial now.

    Read the article

  • Ubuntu 12.04 won't boot at all for the first time

    - by user76280
    Using http://www.ubuntu.com/download/desktop/windows-installer So I tried installing Ubuntu for the first time today, and it installed all well, but then it froze at the 'Preparing to run Ubuntu for the first time...' screen. I then proceeded to restart my computer and load Ubuntu from the OS selection screen. My screen completely froze and the picture didn't even come up fully as if my video driver was not installed correctly. Are there any fixes to this problem? Would greatly appreciate it. Possible useful information: Using http://www.ubuntu.com/download/desktop/windows-installer Dual-booting with Windows

    Read the article

  • IonMonkey : Mozilla renforce les performances JavaScript de Firefox, avec l'optimisation de la compilation Just in Time

    IonMonkey : Mozilla renforce les performances JavaScript de Firefox Avec l'optimisation de la compilation Just in Time Avec la complexité grandissante des applications Web interactives, les navigateurs web se doivent d'être toujours plus performants. C'est ce que fait Mozilla en travaillant sur une nouvelle architecture de compilation JavaScript. IonMonkey est le nouveau compilateur « Just In Time » en test du navigateur Firefox pour le langage JavaScript. Il verra le jour en tant que composant à part entière de Firefox 18 en début 2013. IonMonkey se distingue du courant JIT JägerMonkey avec l'étape d'optimisation. Il est destiné aux applications JavaScript qui fonct...

    Read the article

  • Game Asset Size Over Time

    - by jterrace
    The size (in bytes) of games have been growing over time. There are probably many factors contributing to this: trailer/cut scene videos being bundled with the game, more and higher-quality audio, multiple levels of detail being used, etc. What I'd really like to know is how the size of 3D models and textures that games ship with have changed over time. For example, if one were to look at the size of meshes and textures for Quake I (1996), Quake II (1997), Quake III: Arena (1999), Quake 4 (2005), and Enemy Territory: Quake Wars (2007), I'd imagine a steady increase in file size. Does anyone know of a data source for numbers like this?

    Read the article

  • Number of iterations to real time

    - by Ivansek
    I have an animation of traffic. I have 20 cars in road network, each car have a starting node and end node. Each car know how much distance does it need to travel in order to reach the end node. I move cars each 20 ms for 10 px. To move all cars from their start node to end node I need 60 iterations. That is 60*20ms = 1200ms. Now I want to convert this time, or use data that I have, to a real time where car move 50km/h. How can I do that? Any idea?

    Read the article

  • Part-time Programming Job London

    - by Bluechip Solutions
    I am a student at Middlesex Universtity, London studying Information Technology. I really love software development and I have taught myself how to write HTML + CSS, JavaScript (I use jQuery and AngularJS) and Java (I learnt this in school). I have developed few apps (a desktop app in Java and a mobile app with AngularJS and PhoneGap) I am looking at applying for a part-time programming job to develop myself. Are there part time jobs available for someone like me and are my skill set enough to get me a job? I understand this topic may not be ideal here but this is the only place I know can provide me answers. Thank you!

    Read the article

  • Python - calculate multinomial probability density functions on large dataset?

    - by Seafoid
    Hi, I originally intended to use MATLAB to tackle this problem but the inbuilt functions has limitations that do not suit my goal. The same limitation occurs in NumPy. I have two tab-delimited files. The first is a file showing amino acid residue, frequency and count for an in-house database of protein structures, i.e. A 0.25 1 S 0.25 1 T 0.25 1 P 0.25 1 The second file consists of quadruplets of amino acids and the number of times they occur, i.e. ASTP 1 Note, there are 8,000 such quadruplets. Based on the background frequency of occurence of each amino acid and the count of quadruplets, I aim to calculate the multinomial probability density function for each quadruplet and subsequently use it as the expected value in a maximum likelihood calculation. The multinomial distribution is as follows: f(x|n, p) = n!/(x1!*x2!*...*xk!)*((p1^x1)*(p2^x2)*...*(pk^xk)) where x is the number of each of k outcomes in n trials with fixed probabilities p. n is 4 four in all cases in my calculation. I have created three functions to calculate this distribution. # functions for multinomial distribution def expected_quadruplets(x, y): expected = x*y return expected # calculates the probabilities of occurence raised to the number of occurrences def prod_prob(p1, a, p2, b, p3, c, p4, d): prob_prod = (pow(p1, a))*(pow(p2, b))*(pow(p3, c))*(pow(p4, d)) return prob_prod # factorial() and multinomial_coefficient() work in tandem to calculate C, the multinomial coefficient def factorial(n): if n <= 1: return 1 return n*factorial(n-1) def multinomial_coefficient(a, b, c, d): n = 24.0 multi_coeff = (n/(factorial(a) * factorial(b) * factorial(c) * factorial(d))) return multi_coeff The problem is how best to structure the data in order to tackle the calculation most efficiently, in a manner that I can read (you guys write some cryptic code :-)) and that will not create an overflow or runtime error. To data my data is represented as nested lists. amino_acids = [['A', '0.25', '1'], ['S', '0.25', '1'], ['T', '0.25', '1'], ['P', '0.25', '1']] quadruplets = [['ASTP', '1']] I initially intended calling these functions within a nested for loop but this resulted in runtime errors or overfloe errors. I know that I can reset the recursion limit but I would rather do this more elegantly. I had the following: for i in quadruplets: quad = i[0].split(' ') for j in amino_acids: for k in quadruplets: for v in k: if j[0] == v: multinomial_coefficient(int(j[2]), int(j[2]), int(j[2]), int(j[2])) I haven'te really gotten to how to incorporate the other functions yet. I think that my current nested list arrangement is sub optimal. I wish to compare the each letter within the string 'ASTP' with the first component of each sub list in amino_acids. Where a match exists, I wish to pass the appropriate numeric values to the functions using indices. Is their a better way? Can I append the appropriate numbers for each amino acid and quadruplet to a temporary data structure within a loop, pass this to the functions and clear it for the next iteration? Thanks, S :-)

    Read the article

  • How to use Python list comprehension (or such) for retrieving rows when using MySQLdb?

    - by Erik Nygren
    Hey all, I use MySQLdb a lot when dealing with my webserver. I often find myself repeating the lines: row = cursor.fetchone() while row: do_processing(row) row = cursor.fetchone() Somehow this strikes me as somewhat un-pythonic. Is there a better, one-line way to accomplish the same thing, along the lines of inline assignment in C: while (row = do_fetch()) { do_processing(row); } I've tried figuring out the syntax using list comprehensions, but I can't seem to figure it out. Any recommendations? Thanks, Erik

    Read the article

  • Programmatically specifying Django model attributes

    - by mojbro
    Hi! I would like to add attributes to a Django models programmatically, at run time. For instance, lets say I have a Car model class and want to add one price attribute (database column) per currency, given a list of currencies. What is the best way to do this? I had an approach that I thought would work, but it didn't exactly. This is how I tried doing it, using the car example above: from django.db import models class Car(models.Model): name = models.CharField(max_length=50) currencies = ['EUR', 'USD'] for currency in currencies: Car.add_to_class('price_%s' % currency.lower(), models.IntegerField()) This does seem to work pretty well at first sight: $ ./manage.py syncdb Creating table shop_car $ ./manage.py dbshell shop=# \d shop_car Table "public.shop_car" Column | Type | Modifiers -----------+-----------------------+------------------------------------------------------- id | integer | not null default nextval('shop_car_id_seq'::regclass) name | character varying(50) | not null price_eur | integer | not null price_usd | integer | not null Indexes: "shop_car_pkey" PRIMARY KEY, btree (id) But when I try to create a new Car, it doesn't really work anymore: >>> from shop.models import Car >>> mycar = Car(name='VW Jetta', price_eur=100, price_usd=130) >>> mycar <Car: Car object> >>> mycar.save() Traceback (most recent call last): File "<console>", line 1, in <module> File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/base.py", line 410, in save self.save_base(force_insert=force_insert, force_update=force_update) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/base.py", line 495, in save_base result = manager._insert(values, return_id=update_pk) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/manager.py", line 177, in _insert return insert_query(self.model, values, **kwargs) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/query.py", line 1087, in insert_query return query.execute_sql(return_id) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/sql/subqueries.py", line 320, in execute_sql cursor = super(InsertQuery, self).execute_sql(None) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/models/sql/query.py", line 2369, in execute_sql cursor.execute(sql, params) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/django/db/backends/util.py", line 19, in execute return self.cursor.execute(sql, params) ProgrammingError: column "price_eur" specified more than once LINE 1: ...NTO "shop_car" ("name", "price_eur", "price_usd", "price_eur... ^

    Read the article

  • How to write the Visitor Pattern for Abstract Syntax Tree in Python?

    - by bodacydo
    My collegue suggested me to write a visitor pattern to navigate the AST. Can anyone tell me more how would I start writing it? As far as I understand, each Node in AST would have visit() method (?) that would somehow get called (from where?). That about concludes my understanding. To simplify everything, suppose I have nodes Root, Expression, Number, Op and the tree looks like this: Root | Op(+) / \ / \ Number(5) \ Op(*) / \ / \ / \ Number(2) Number(444) Can anyone think of how the visitor pattern would visit this tree to produce output: 5 + 2 * 444 Thanks, Boda Cydo.

    Read the article

  • Python: speed up removal of every n-th element from list.

    - by ChristopheD
    I'm trying to solve this programming riddle and althought the solution (see code below) works correct, it is too slow for succesful submission. Any pointers as how to make this run faster? (removal of every n-th element from a list)? Or suggestions for a better algorithm to calculate the same; seems I can't think of anything else then brute-force for now... Basically the task at hand is: GIVEN: L = [2,3,4,5,6,7,8,9,10,11,........] 1. Take the first remaining item in list L (in the general case 'n'). Move it to the 'lucky number list'. Then drop every 'n-th' item from the list. 2. Repeat 1 TASK: Calculate the n-th number from the 'lucky number list' ( 1 <= n <= 3000) My current code (it calculates the 3000 first lucky numbers in about a second on my machine - but unfortunately too slow): """ SPOJ Problem Set (classical) 1798. Assistance Required URL: http://www.spoj.pl/problems/ASSIST/ """ sieve = range(3, 33900, 2) luckynumbers = [2] while True: wanted_n = input() if wanted_n == 0: break while len(luckynumbers) < wanted_n: item = sieve[0] luckynumbers.append(item) items_to_delete = set(sieve[::item]) sieve = filter(lambda x: x not in items_to_delete, sieve) print luckynumbers[wanted_n-1]

    Read the article

  • Find new messages added to an imap mailbox since I last checked with python libimap2?

    - by vy32
    I am trying to write a program that monitors an IMAP mailbox and automatically copies every new incoming message into an "Archive" folder. I'm using imaplib2 which implements the IDLE command. Here's my basic program: M = imaplib2.IMAP4("mail.me.com") M.login(username,password) lst = M.list() assert lst[0]=='OK' for mbx in lst[1]: print "Mailboxes:",mbx def process(m): print "m=",m res = M.recent() print res M.select('INBOX') M.examine(mailbox='INBOX',callback=process) while True: print "Calling idle..." M.idle() print "back from idle" M.close() M.logout() It prints the mailboxes properly and runs process() when the first change happens to the mailbox. But the response from recent() doesn't make sense to me, and after the first message I never get any other notifications. Anyone know how to do this?

    Read the article

  • Python 3: timestamp to datetime: where does this additional hour come from?

    - by Beau Martínez
    I'm using the following functions: # The epoch used in the datetime API. EPOCH = datetime.datetime.fromtimestamp(0) def timedelta_to_seconds(delta): seconds = (delta.microseconds * 1e6) + delta.seconds + (delta.days * 86400) seconds = abs(seconds) return seconds def datetime_to_timestamp(date, epoch=EPOCH): # Ensure we deal with `datetime`s. date = datetime.datetime.fromordinal(date.toordinal()) epoch = datetime.datetime.fromordinal(epoch.toordinal()) timedelta = date - epoch timestamp = timedelta_to_seconds(timedelta) return timestamp def timestamp_to_datetime(timestamp, epoch=EPOCH): # Ensure we deal with a `datetime`. epoch = datetime.datetime.fromordinal(epoch.toordinal()) epoch_difference = timedelta_to_seconds(epoch - EPOCH) adjusted_timestamp = timestamp - epoch_difference date = datetime.datetime.fromtimestamp(adjusted_timestamp) return date And using them with the passed code: twenty = datetime.datetime(2010, 4, 4) print(twenty) print(datetime_to_timestamp(twenty)) print(timestamp_to_datetime(datetime_to_timestamp(twenty))) And getting the following results: 2010-04-04 00:00:00 1270339200.0 2010-04-04 01:00:00 For some reason, I'm getting an additional hour added in the last call, despite my code having, as far as I can see, no flaws. Where is this additional hour coming from?

    Read the article

< Previous Page | 253 254 255 256 257 258 259 260 261 262 263 264  | Next Page >