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  • Collectable<T> serialization, Root Namespaces on T in .xml files.

    - by Stacey
    I have a Repository Class with the following method... public T Single<T>(Predicate<T> expression) { using (var list = (Models.Collectable<T>)System.Xml.Serializer.Deserialize(typeof(Models.Collectable<T>), FileName)) { return list.Find(expression); } } Where Collectable is defined.. [Serializable] public class Collectable<T> : List<T>, IDisposable { public Collectable() { } public void Dispose() { } } And an Item that uses it is defined.. [Serializable] [System.Xml.Serialization.XmlRoot("Titles")] public partial class Titles : Collectable<Title> { } The problem is when I call the method, it expects "Collectable" to be the XmlRoot, but the XmlRoot is "Titles" (all of object Title). I have several classes that are collected in .xml files like this, but it seems pointless to rewrite the basic methods for loading each up when the generic accessors do it - but how can I enforce the proper root name for each file without hard coding methods for each one? The [System.Xml.Serialization.XmlRoot] seems to be ignored.

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  • Methodology for a Rails app

    - by Aaron Vegh
    I'm undertaking a rather large conversion from a legacy database-driven Windows app to a Rails app. Because of the large number of forms and database tables involved, I want to make sure I've got the right methodology before getting too far. My chief concern is minimizing the amount of code I have to write. There are many models that interact together, and I want to make sure I'm using them correctly. Here's a simplified set of models: class Patient < ActiveRecord::Base has_many :PatientAddresses has_many :PatientFileStatuses end class PatientAddress < ActiveRecord::Base belongs_to :Patient end class PatientFileStatus < ActiveRecord::Base belongs_to :Patient end The controller determines if there's a Patient selected; everything else is based on that. In the view, I will be needing data from each of these models. But it seems like I have to write an instance variable in my controller for every attribute that I want to use. So I start writing code like this: @patient = Patient.find(session[:patient]) @patient_addresses = @patient.PatientAddresses @patient_file_statuses = @patient.PatientFileStatuses @enrollment_received_when = @patient_file_statuses[0].EnrollmentReceivedWhen @consent_received = @patient_file_statuses[0].ConsentReceived @consent_received_when = @patient_file_statuses[0].ConsentReceivedWhen The first three lines grab the Patient model and its relations. The next three lines are examples of my providing values to the view from one of those relations. The view has a combination of text fields and select fields to show the data above. For example: <%= select("patientfilestatus", "ConsentReceived", {"val1"="val1", "val2"="val2", "Written"="Written"}, :include_blank=true )% <%= calendar_date_select_tag "patient_file_statuses[EnrollmentReceivedWhen]", @enrollment_complete_when, :popup=:force % (BTW, the select tag isn't really working; I think I have to use collection_select?) My questions are: Do I have to manually declare the value of every instance variable in the controller, or can/should I do it within the view? What is the proper technique for displaying a select tag for data that's not the primary model? When I go to save changes to this form, will I have to manually pick out the attributes for each model and save them individually? Or is there a way to name the fields such that ActiveRecord does the right thing? Thanks in advance, Aaron.

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  • Rails: Design Pattern to Store Order of Relations

    - by ChrisInCambo
    Hi, I have four models: Customer, QueueRed, QueueBlue, QueueGreen. The Queue models have a one to many relationship with customers A customer must always be in a queue A customer can only be in one queue at a time A customer can change queues We must be able to find out the customers current position in their respective queue In an object model the queues would just have an array property containing customers, but ActiveRecord doesn't have arrays. In a DB I would probably create some extra tables just to handle the order of the stories in the queue. My question is what it the best way to model the relationship in ActiveRecord? Obviously there are many ways this could be done, but what is the best or the most in line with how ActiveRecord should be used? Cheers, Chris

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  • CakePHP hasOne ineffeciency?

    - by Andre
    I was looking at examples on the CakePHP website, in particular hasOne used in linking models. http://book.cakephp.org/view/78/Associations-Linking-Models-Together My question is this, is CakePHP using two queries to build the array structure of data returned in a model that uses hasOne linkage? Taken from CakePHP: //Sample results from a $this-User-find() call. Array ( [User] => Array ( [id] => 121 [name] => Gwoo the Kungwoo [created] => 2007-05-01 10:31:01 ) [Profile] => Array ( [id] => 12 [user_id] => 121 [skill] => Baking Cakes [created] => 2007-05-01 10:31:01 ) ) Hope this all makes sense.

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  • What is the best way to mock a 3rd party object in ruby?

    - by spinlock
    I'm writing a test app using the twitter gem and I'd like to write an integration test but I can't figure out how to mock the objects in the Twitter namespace. Here's the function that I want to test: def build_twitter(omniauth) Twitter.configure do |config| config.consumer_key = TWITTER_KEY config.consumer_secret = TWITTER_SECRET config.oauth_token = omniauth['credentials']['token'] config.oauth_token_secret = omniauth['credentials']['secret'] end client = Twitter::Client.new user = client.current_user self.name = user.name end and here's the rspec test that I'm trying to write: feature 'testing oauth' do before(:each) do @twitter = double("Twitter") @twitter.stub!(:configure).and_return true @client = double("Twitter::Client") @client.stub!(:current_user).and_return(@user) @user = double("Twitter::User") @user.stub!(:name).and_return("Tester") end scenario 'twitter' do visit root_path login_with_oauth page.should have_content("Pages#home") end end But, I'm getting this error: 1) testing oauth twitter Failure/Error: login_with_oauth Twitter::Error::Unauthorized: GET https://api.twitter.com/1/account/verify_credentials.json: 401: Invalid / expired Token # ./app/models/user.rb:40:in `build_twitter' # ./app/models/user.rb:16:in `build_authentication' # ./app/controllers/authentications_controller.rb:47:in `create' # ./spec/support/integration_spec_helper.rb:3:in `login_with_oauth' # ./spec/integration/twit_test.rb:16:in `block (2 levels) in <top (required)>' The mocks above are using rspec but I'm open to trying mocha too. Any help would be greatly appreciated.

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  • How do I construct a Django form with model objects in a Select widget?

    - by Thierry Lam
    Let's say I'm using the Django Site model: class Site(models.Model): name = models.CharField(max_length=50) My Site values are (key, value): 1. Stackoverflow 2. Serverfault 3. Superuser I want to construct a form with an html select widget with the above values: <select> <option value="1">Stackoverflow</option> <option value="2">Serverfault</option> <option value="3">Superuser</option> </select> I'm thinking of starting with the following code but it's incomplete: class SiteForm(forms.Form): site = forms.IntegerField(widget=forms.Select()) Any ideas how I can achieve that with Django form?

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  • Does Python Django support custom SQL and denormalized databases with no Foreign Key relationships?

    - by Jay
    I've just started learning Python Django and have a lot of experience building high traffic websites using PHP and MySQL. What worries me so far is Python's overly optimistic approach that you will never need to write custom SQL and that it automatically creates all these Foreign Key relationships in your database. The one thing I've learned in the last few years of building Chess.com is that its impossible to NOT write custom SQL when you're dealing with something like MySQL that frequently needs to be told what indexes it should use (or avoid), and that Foreign Keys are a death sentence. Percona's strongest recommendation was for us to remove all FKs for optimal performance. Is there a way in Django to do this in the models file? create relationships without creating actual DB FKs? Or is there a way to start at the database level, design/create my database, and then have Django reverse engineer the models file?

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  • Doctrine 1.2 Column Naming Conventions for Many To Many Relationships

    - by Alan Storm
    I'm working with an existing database schema, and trying to setup two Doctrine models with a Many to Many relationship, as described in this document When creating tables from scratch, I have no trouble getting this working. However, the existing join tables use a different naming convention that what's described in the Doctrine document. Specifically Table 1 -------------------------------------------------- table_1_id ....other columns.... Table 2 -------------------------------------------------- table_2_id ....other columns.... Join Table -------------------------------------------------- fktable1_id fktable_2_id Basically, the previous developers prefaced all forign keys with an fk. From the examples I've seen and some brief experimenting with code, it appears that Doctrine 1.2 requires that the join table use the same column names as the tables it's joining in Is my assumption correct? If so, has the situation changed in Doctrine 2? If the answers to either of the above are true, how do you configure the models so that all the columns "line up"

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  • How to page multiple data sets in ASP.NET MVC

    - by REA_ANDREW
    On a single view I will have three sets of paged data. Which means for each model I will have The Objects The Page Index The Page Size My initial thought was for example: public class PagedModel<T> where T:class { public IList<T> Objects { get; set; } public int ModelPageIndex { get; set; } public int ModelPageSize { get; set; } } Then having a model which is to be supplied to the action as for example: public class TypesViewModel { public PagedModel<ObjectA> Types1 { get; set; } public PagedModel<ObjectB> Typed2 { get; set; } public PagedModel<ObjectC> Types3 { get; set; } } So if I then for example have the Index view inherit from the type: System.Web.Mvc.ViewPage<uk.co.andrewrea.forum.Web.Models.TypesViewModel> Now my initial aciton method for the index is simply: public ActionResult Index() { var forDisplayPurposes = new TypesViewModel(); return View(forDisplayPurposes); } If I then want to page, it is here where I am struggling to decide which action to take. Lets say that I select the next page of the Types2 PageModel. What should the action look like for this in order to return the new view showing the second page of the Types2 PageModel I was thinking possibly to duplicate the action but use it with POST [AcceptVerbs(HttpVerbs.Post)] public ActionResult Index(TypesViewModel model) { return View(model); } Is this a good way to approach it. I understand there is always Session, but I was just wondering how such a thing is achieved currently out there. If any best methods have been mutually accepted and things. So simply, one page with multiple paged models. How to persist the data for each using a wrapper model. Which way should you pass in the model and which way should you page the data, i.e. Form Post Lastly, I have seen the routes take this into account i.e. {controller}/{action}/{id}/{pageindex}/{pagesize} but this only accounts for one model and I do not really wwant to repeat the pagesize and pageindex values for the number of models I have inside the wrapper model. Thanks for your time!! Andrew

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  • Multi domain rails app. How to intelligently use MVC?

    - by denial
    Background: We have app a, b, and plan to add more apps into this same application. The apps are similar enough they could share many views, assets, and actions. Currently a,b live in a single rails app(2.3.10). c will be similar enough that it could also be in this rails app. The problem: As we continue to add more apps to this one app, there's going to be too much case logic that the app will soon become a nightmare to maintain. There will also be potential namespace issues. However, the apps are very similar in function and layout, it also makes sense to keep them in one app so that it's one app to maintain(since roughly 50% of site look/functionality will be shared). What we are trying to do is keep this as clean as possible so it's easy for multiple teams to work on and easy to maintain. Some things we've thought about/are trying: Engines. Make each app an engine. This would let us base routes on the domain. It also allows us to pull out controllers, models and views for the specific app. This solution does not seem ideal as we won't be reusing the apps any time soon. And explicitly stating the host in the routes doesn't seem right. Skinning/themes. The auth logic would be different between the apps. Each user model would be different. So it's not just a skinning problem. In app/view add folder sitea for sitea views, siteb for siteb views and so on. Do the same for controllers and models. This is still pretty messy and since it didn't follow naming conventions, it did not work with rails so nicely and made much of the code messier. Making another rails app. We just didn't want to maintain the same controller or view in 2 apps if they are identical. What we want to do is make the app intelligently use a controller based on the host. So there would be a sessions controller for each app, and perhaps some parent session controller for shared logic(not needed now). In each of these session controllers, it handles authentication for that specific app. So if the domain is a.mysite.com, it would use session controller for app a and know to use app a's views,models,controllers. And if the domain is b.mysite, it would use the session controller for b. And there would be a user model for a and user model for b, which also would be determined by the domain. Does anyone have any suggestions or experience with this situation? And ideally using rails 2.3.x as updating to rails 3 isn't an option right now.

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  • Accessing properties through Generic type parameter

    - by Veer
    I'm trying to create a generic repository for my models. Currently i've 3 different models which have no relationship between them. (Contacts, Notes, Reminders). class Repository<T> where T:class { public IQueryable<T> SearchExact(string keyword) { //Is there a way i can make the below line generic //return db.ContactModels.Where(i => i.Name == keyword) //I also tried db.GetTable<T>().Where(i => i.Name == keyword) //But the variable i doesn't have the Name property since it would know it only in the runtime //db also has a method ITable GetTable(Type modelType) but don't think if that would help me } } In MainViewModel, I call the Search method like this: Repository<ContactModel> _contactRepository = new Repository<ContactModel>(); public void Search(string keyword) { var filteredList = _contactRepository.SearchExact(keyword).ToList(); } I use Linq-To-Sql.

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  • Django ORM dealing with MySQL BIT(1) field

    - by Carles Barrobés
    In a Django application, I'm trying to access an existing MySQL database created with Hibernate (a Java ORM). I reverse engineered the model using: $ manage.py inspectdb > models.py This created a nice models file from the Database and many things were quite fine. But I can't find how to properly access boolean fields, which were mapped by Hibernate as columns of type BIT(1). The inspectdb script by default creates these fields in the model as TextField and adds a comment saying that it couldn't reliably obtain the field type. I changed these to BooleanField but it doesn't work (the model objects always fetch a value of true for these fields). Using IntegerField won't work as well (e.g. in the admin these fields show strange non-ascii characters). Any hints of doing this without changing the database? (I need the existing Hibernate mappings and Java application to still work with the database).

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  • Domain object validation vs view model validation

    - by Brendan Vogt
    I am using ASP.NET MVC 3 and I am using FluentValidation to validate my view models. I am just a little concerned that I might not be on the correct track. As far as what I know, model validation should be done on the domain object. Now with MVC you might have multiple view models that are similar that needs validation. What happens if a property from a domain object occurs in more than one view model? Now you are validating the same property twice, and they might not even be in sync. So if I have a User domain object then I would like to do validation on this object. Now what happens if I have UserAViewModel and UserBViewModel, so now it is multiple validations that needs to be done. The scenario above is just an example, so please don't critise on it.

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  • How do I specify (1) an order and (2) a meaninful string representation for users in my Django application?

    - by David Faux
    I have a Django application with users. I have a model called "Course" with a foreign key called "teacher" to the default User model that Django provides: class Course(models.Model): ... teacher = models.ForeignKey(User, related_name='courses_taught') When I create a model form to edit information for individual courses, the possible users for the teacher field appear in this long select menu of user names. These users are ordered by ID, which is of meager use to me. How can I order these users by their last names? change the string representation of the User class to be "Firstname Lastname (username)" instead of "username"?

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  • Model class is not found in zend framework project (quickstart) ,please help

    - by Dumindu
    The things I did is zf create project demo1 in command prompt add the lines to application.ini appnamespace = "Application" resources.layout.layoutPath = APPLICATION_PATH "/layouts/scripts" add a layout with header and footer using partial() (They are perfectly worked) create Data.php in models directory and add this simple class <?php class Application_Model_Data{ }//Application for appnamespace then I tried to load this class(by creating instance)from index controller index action $data = new Application_Model_Data(); but when I test it even in this level it gives an error Fatal error: Class 'Application_Model_Data' not found in C:\Zend\...\IndexController.php Question Do I want to add a autoloader to load models in the application( I'm not used modules) if not what was I missed to add please help I'm stuck in the beginning,Thank you

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  • django updating m2m field

    - by Marconi
    I have a model service and a ModelForm named Service which I use to add and update the service model. The model looks like this: class Service(models.Model): categories = models.ManyToManyField(Category) The categories field is displayed as a tag with that allows multiple selection. It works well when I'm adding a new record but when I'm updating it, only one service is showing up on the request.POST['categories'] even if I selected multiple categories. I tried dumping the request object and I can see that the categories is showing something like: u'categories': [u'3', u'4', u'2'] I tried calling the request._get_post() and it did return only 1 category, hence the request.POST['categories'] returns only 1. Anybody who knows what's happening and how to fix it?

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  • Play 2.0 javaToDo tutorial doesn't compile

    - by chsn
    I'm trying to follow the Play2.0 JavaToDO tutorial and for some reason it just doesn't want to work. Have looked through stackoverflow and other online resources, but haven't find an answer to this and it's driving me crazy. Attached code of the Application.java package controllers; import models.Task; import play.data.Form; import play.mvc.Controller; import play.mvc.Result; public class Application extends Controller { static Form<Task> taskForm = form(Task.class); public static Result index() { return redirect(routes.Application.tasks()); } public static Result tasks() { return ok( views.html.index.render(Task.all(), taskForm)); } public static Result newTask() { return TODO; } public static Result deleteTask(Long id) { return TODO; } } Attached code of the Task java package models; import java.util.List; import javax.persistence.Entity; import play.data.Form; import play.data.validation.Constraints.Required; import play.db.ebean.Model.Finder; import play.mvc.Result; import controllers.routes; @Entity public class Task { public Long id; @Required public String label; // search public static Finder<Long,Task> find = new Finder( Long.class, Task.class); // display tasks public static List<Task> all() { return find.all(); } // create task public static void create(Task task) { task.create(task); } // delete task public static void delete(Long id) { find.ref(id).delete(id); // find.ref(id).delete(); } // create new task public static Result newTask() { Form<Task> filledForm = taskForm.bindFromRequest(); if(filledForm.hasErrors()) { return badRequest( views.html.index.render(Task.all(), filledForm) ); } else { Task.create(filledForm.get()); return redirect(routes.Application.tasks()); } } } I get a compile error on Task.java on the line static Form<Task> taskForm = form(Task.class); As I'm working on eclipse (the project is eclipsified before import), it's telling me that taskForm cannot be resolved and it also underlines every play 2 command e.g. "render(), redirect(), bindFromRequest()" asking me to create a method for it. Any ideas how to solve the compilations error and also how to get Eclipse to recognize the play2 commands? EDIT: updated Application.java package controllers; import models.Task; import play.data.Form; import play.mvc.Controller; import play.mvc.Result; public class Application extends Controller { // create new task public static Result newTask() { Form<Task> filledForm = form(Task.class).bindFromRequest(); if(filledForm.hasErrors()) { return badRequest( views.html.index.render(Task.all(), filledForm) ); } else { Task.newTask(filledForm.get()); return redirect(routes.Application.tasks()); } } public static Result index() { return redirect(routes.Application.tasks()); } public static Result tasks() { return ok( views.html.index.render(Task.all(), taskForm)); } public static Result deleteTask(Long id) { return TODO; } } Updated task.java package models; import java.util.List; import javax.persistence.Entity; import play.data.Form; import play.data.validation.Constraints.Required; import play.db.ebean.Model; import play.db.ebean.Model.Finder; import play.mvc.Result; import controllers.routes; @Entity public class Task extends Model { public Long id; @Required public String label; // Define a taskForm static Form<Task> taskForm = form(Task.class); // search public static Finder<Long,Task> find = new Finder( Long.class, Task.class); // display tasks public static List<Task> all() { return find.all(); } // create new task public static Result newTask(Task newTask) { save(task); } // delete task public static void delete(Long id) { find.ref(id).delete(id); // find.ref(id).delete(); } }

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  • How/When/Where to Extend Gem Classes (via class_eval and Modules) in Rails 3?

    - by viatropos
    What is the recommended way to extend class behavior, via class_eval and modules (not by inheritance) if I want to extend a class buried in a Gem from a Rails 3 app? An example is this: I want to add the ability to create permalinks for tags and categories (through the ActsAsTaggableOn and ActsAsCategory gems). They have defined Tag and Category models. I want to basically do this: Category.class_eval do has_friendly_id :title end Tag.class_eval do has_friendly_id :title end Even if there are other ways of adding this functionality that might be specific to the gem, what is the recommended way to add behavior to classes in a Rails 3 application like this? I have a few other gems I've created that I want to do this to, such as a Configuration model and an Asset model. I would like to be able to add create an app/models/configuration.rb model class to my app, and it would act as if I just did class_eval. Anyways, how is this supposed to work? I can't find anything that covers this from any of the current Rails 3 blogs/docs/gists.

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  • Automatic images translation to 3d model

    - by farrakhov-bulat
    I'm quite interested in automatic images translation to 3d models. Not really for commercial product, but from the point of possible academic research and implementation. What I'd like to achieve is almost transparent for user process of transformation series of images (fewer is better) to 3d model which might be shown in flash/silverlight/javafx or similar. Consider online furniture store with 3d models of all items in stock. Kinda cool to have ability to see the product in 3d before purchasing it. I managed to find a few pieces of software, like insight3d, but it couldn't be used in my case I guess. So, are there any similar projects or tips for me? If it would require to write that piece of software - I'd really love to dig into research on this field.

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  • Indexing a method return (depending on Internationalization)

    - by Hedde
    Consider a django model with an IntegerField with some choices, e.g. COLORS = ( (0, _(u"Blue"), (1, _(u"Red"), (2, _(u"Yellow"), ) class Foo(models.Model): # ...other fields... color = models.PositiveIntegerField(choices=COLOR, verbose_name=_(u"color")) My current (haystack) index: class FooIndex(SearchIndex): text = CharField(document=True, use_template=True) color = CharField(model_attr='color') def prepare_color(self, obj): return obj.get_color_display() site.register(Product, ProductIndex) This obviously only works for keyword "yellow", but not for any (available) translations. Question: What's would be a good way to solve this problem? (indexing method returns based on the active language) What I have tried: I created a function that runs a loop over every available language (from settings) appending any translation to a list, evaluating this against the query, pre search. If any colors are matched it converts them backwards into their numeric representation to evaluate against obj.color, but this feels wrong.

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  • rails: best way to store comments in mysql

    - by ciss
    Hello. Okay i have two models: posts and comments. as you can think comments has column :post_id. My models Comments belongs_to :post Post has_many :comments So, this is pretty simple association but i have some problems with ordering comments. at first time, when i create my comments migration file i just add column :position. This column indicate comment position in the post. But now i think what where is more good way to do this. so i can't make my choise: 1) uses t.column :datatime :created_at, :default = Time.now() 2) or use timestamps? this is undiscovered for me, please tell me about your exp.

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  • Hibernate not Loading a class

    - by Noor
    Hi, I have a class Auction that contains a Class Item and Users but when I am getting the class, the class item and Users are not being loaded. Auction Class Mapping File: <?xml version="1.0"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> <!-- Generated Dec 28, 2010 9:14:12 PM by Hibernate Tools 3.4.0.Beta1 --> <hibernate-mapping> <class name="com.BiddingSystem.Models.Auction" table="AUCTION"> <id name="AuctionId" type="long"> <column name="AUCTIONID" /> <generator class="native" /> </id> <property name="StartTime" type="java.util.Date"> <column name="STARTTIME" /> </property> <property name="EndTime" type="java.util.Date"> <column name="ENDTIME" /> </property> <property name="StartingBid" type="long"> <column name="STARTINGBID" /> </property> <property name="MinIncrement" type="long"> <column name="MININCREMENT" /> </property> <many-to-one name="CurrentItem" class="com.BiddingSystem.Models.Item" fetch="join" cascade="all"> <column name="ItemId" /> </many-to-one> <property name="AuctionStatus" type="java.lang.String"> <column name="AUCTIONSTATUS" /> </property> <property name="BestBid" type="long"> <column name="BESTBID" /> </property> <many-to-one name="User" class="com.BiddingSystem.Models.Users" fetch="join"> <column name="UserId" /> </many-to-one> </class> </hibernate-mapping> When I am doing this: Query query=session.createQuery("from Auction where UserId="+UserId); List <Auction> AllAuctions= new LinkedList<Auction>(query.list()); The Users and Item are null

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  • Running [R] on a Netbook

    - by Thomas
    I am interested in purchasing a netbook to do field research in another country. My hardware specifications for the nebtook are fairly basic: Be rugged enough to survive a bit of wear and tear Fairly fast processing (the ability to upgrade from 1GB of RAM to 2GB) A battery life of longer than 6 hours At least a 10 inch screen A decent camera for Skyping However, I am mainly concerned about being able to do basic statistical analysis in conjunction with R Be able run a Spreadsheet program to do basic data input (like Excel or Open Office) Use R to do basic data analysis (Regression, some simulation (nothing crazy), data cleaning, and some of the functionality) Word Processing (Word or Open Office) Do you have any suggestions on which models or brands my fit my needs? Some of the models I am considering: Samsung NB-30 Toshiba NB 305 Asus Eee PC 1005HA Lenovo S10-2 Does anyone use R on a netbook, and if so do you have any recommendations on how best to optimize it? This article from Lifehacker mentions some OS. Anybody use these in conjunction with R? Any help would be much appreciated.

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