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• Devise routes to only use custom sign in page

  - by eblume

  I am trying to learn Devise and seem to have hit a problem that is nearly the same as many, many questions here on SO but just slightly different enough that I am not getting much traction following those questions' threads. I have a very simple application. Users can sign in and sign out (and also the devise 'forgot password' stuff is enabled - I am not concerned about it at this time) and that is it. They can't register, they can't edit their profile other than to change their password, they can't do anything except sign in and sign out. (Account creation will, for now, be handled manually at the command line.) I would vastly prefer that the only page that users can log in from was "/" (root_path). I already have it working where you can log in from "/", and that is great. The problem I am having is that if you type in your user/password combination incorrectly from the root_path login page, it automatically sends you to the Devise sign-in page to continue trying to sign in. How can I make Devise redirect to root_path on sign-in failure? Here are my routes - they are probably not optimal or correctly configured and I would appreciate pointers on fixing them: root to: "core_pages#splash" devise_for :users, skip: [:sessions] as :user do # get 'signin' => 'devise/sessions#new', as: :new_user_session post 'signin' => 'devise/sessions#create', as: :user_session delete 'signout' => 'devise/sessions#destroy', as: :destroy_user_session, via: Devise.mappings[:user].sign_out_via end match '/home' => 'core_pages#home' Note the commented-out 'get signin' line. The system works without this line but, surprisingly (to me), "GET /signin" results in a HTTP 400 (OK) response and renders the Devise login template. I would prefer it return some sort of 'invalid request' or just silently redirect the user to root_path. Running rake routes on these routes gives: root / core_pages#splash user_password POST /users/password(.:format) devise/passwords#create new_user_password GET /users/password/new(.:format) devise/passwords#new edit_user_password GET /users/password/edit(.:format) devise/passwords#edit PUT /users/password(.:format) devise/passwords#update user_session POST /signin(.:format) devise/sessions#create destroy_user_session DELETE /signout(.:format) devise/sessions#destroy home /home(.:format) core_pages#home Thanks!

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• Efficient file buffering & scanning methods for large files in python

  - by eblume

  The description of the problem I am having is a bit complicated, and I will err on the side of providing more complete information. For the impatient, here is the briefest way I can summarize it: What is the fastest (least execution time) way to split a text file in to ALL (overlapping) substrings of size N (bound N, eg 36) while throwing out newline characters. I am writing a module which parses files in the FASTA ascii-based genome format. These files comprise what is known as the 'hg18' human reference genome, which you can download from the UCSC genome browser (go slugs!) if you like. As you will notice, the genome files are composed of chr[1..22].fa and chr[XY].fa, as well as a set of other small files which are not used in this module. Several modules already exist for parsing FASTA files, such as BioPython's SeqIO. (Sorry, I'd post a link, but I don't have the points to do so yet.) Unfortunately, every module I've been able to find doesn't do the specific operation I am trying to do. My module needs to split the genome data ('CAGTACGTCAGACTATACGGAGCTA' could be a line, for instance) in to every single overlapping N-length substring. Let me give an example using a very small file (the actual chromosome files are between 355 and 20 million characters long) and N=8 import cStringIO example_file = cStringIO.StringIO("""\ header CAGTcag TFgcACF """) for read in parse(example_file): ... print read ... CAGTCAGTF AGTCAGTFG GTCAGTFGC TCAGTFGCA CAGTFGCAC AGTFGCACF The function that I found had the absolute best performance from the methods I could think of is this: def parse(file): size = 8 # of course in my code this is a function argument file.readline() # skip past the header buffer = '' for line in file: buffer += line.rstrip().upper() while len(buffer) = size: yield buffer[:size] buffer = buffer[1:] This works, but unfortunately it still takes about 1.5 hours (see note below) to parse the human genome this way. Perhaps this is the very best I am going to see with this method (a complete code refactor might be in order, but I'd like to avoid it as this approach has some very specific advantages in other areas of the code), but I thought I would turn this over to the community. Thanks! Note, this time includes a lot of extra calculation, such as computing the opposing strand read and doing hashtable lookups on a hash of approximately 5G in size. Post-answer conclusion: It turns out that using fileobj.read() and then manipulating the resulting string (string.replace(), etc.) took relatively little time and memory compared to the remainder of the program, and so I used that approach. Thanks everyone!

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