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  • User input in perl - Issue with running script in KomodoEdit

    - by golwalkar.rohan
    i wrote this tiny code on gedit and ran it :- #/usr/bin/perl print "Enter the radius of circle: \n"; $radius = <>; chomp $radius; print "radius is: $radius\n"; $circumference = (2*3.141592654) * $radius; print "Circumference of circle with radius : $radius = $circumference\n"; Runs fine using command line.Ran the same code on Komodo Edit: facing an issue i expect first line as output as :- Enter the radius of circle: whearas it waits on the screen i.e waiting for an input and after that runs everything in sequence -- can someone tell me why it runs fine with command line but not Komodo?

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  • str is not callable error in python .

    - by mekasperasky
    import sys import md5 from TOSSIM import * from RadioCountMsg import * t = Tossim([]) #The Tossim object is defined here m = t.mac()#The mac layer is defined here , in which the communication takes place r = t.radio()#The radio communication link object is defined here , as the communication needs Rf frequency to transfer t.addChannel("RadioCountToLedsC", sys.stdout)# The various channels through which communication will take place t.addChannel("LedsC", sys.stdout) #The no of nodes that would be required in the simulation has to be entered here print("enter the no of nodes you want ") n=input() for i in range(0, n): m = t.getNode(i) m.bootAtTime((31 + t.ticksPerSecond() / 10) * i + 1) #The booting time is defined so that the time at which the node would be booted is given f = open("topo.txt", "r") #The topography is defined in topo.txt so that the RF frequencies of the transmission between nodes are are set lines = f.readlines() for line in lines: s = line.split() if (len(s) > 0): if (s[0] == "gain"): r.add(int(s[1]), int(s[2]), float(s[3])) #The topogrography is added to the radio object noise = open("meyer-heavy.txt", "r") #The noise model is defined for the nodes lines = noise.readlines() for line in lines: str = line.strip() if (str != ""): val = int(str) for i in range(0, 4): t.getNode(i).addNoiseTraceReading(val) for i in range (0, n): t.getNode(i).createNoiseModel() #The noise model is created for each node for i in range(0,n): t.runNextEvent() fk=open("key.txt","w") for i in range(0,n): if i ==0 : key=raw_input() fk.write(key) ak=key key=md5.new() key.update(str(ak)) ak=key.digest() fk.write(ak) fk.close() fk=open("key.txt","w") plaint=open("pt.txt") for i in range(0,n): msg = RadioCountMsg() msg.set_counter(7) pkt = t.newPacket()#A packet is defined according to a certain format print("enter message to be transported") ms=raw_input()#The message to be transported is taken as input #The RC5 encryption has to be done here plaint.write(ms) pkt.setData(msg.data) pkt.setType(msg.get_amType()) pkt.setDestination(i+1)#The destination to which the packet will be sent is set print "Delivering " + " to" ,i+1 pkt.deliver(i+1, t.time() + 3) fk.close() print "the key to be displayed" ki=raw_input() fk=open("key.txt") for i in range(0,n): if i==ki: ms=fk.readline() for i in range(0,n): msg=RadioCountMsg() msg.set_counter(7) pkt=t.newPacket() msg.data=ms pkt.setData(msg.data) pkt.setType(msg.get_amType()) pkt.setDestination(i+1) pkt.deliver(i+1,t.time()+3) #The key has to be broadcasted here so that the decryption can take place for i in range(0, n): t.runNextEvent(); this code gives me error here key.update(str(ak)) . when i run a similar code on the python terminal there is no such error but this code pops up an error . why so?

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  • Python unit-testing with nose: Making sequential tests

    - by cool-RR
    I am just learning how to do unit-testing. I'm on Python / nose / Wing IDE. (The project that I'm writing tests for is a simulations framework, and among other things it lets you run simulations both synchronously and asynchronously, and the results of the simulation should be the same in both.) The thing is, I want some of my tests to use simulation results that were created in other tests. For example, synchronous_test calculates a certain simulation in synchronous mode, but then I want to calculate it in asynchronous mode, and check that the results came out the same. How do I structure this? Do I put them all in one test function, or make a separate asynchronous_test? Do I pass these objects from one test function to another? Also, keep in mind that all these tests will run through a test generator, so I can do the tests for each of the simulation packages included with my program.

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  • Creating a spider using Scrapy, Spider generation error.

    - by Nacari
    I just downloaded Scrapy (web crawler) on Windows 32 and have just created a new project folder using the "scrapy-ctl.py startproject dmoz" command in dos. I then proceeded to created the first spider using the command: scrapy-ctl.py genspider myspider myspdier-domain.com but it did not work and returns the error: Error running: scrapy-ctl.py genspider, Cannot find project settings module in python path: scrapy_settings. I know I have the path set right (to python26/scripts), but I am having difficulty figuring out what the problem is. I am new to both scrapy and python so there is a good possibility that I have failled to do something important. Also, I have been using eclipse with the Pydev plugin to edit the code if that might cause some problems.

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  • Throw of a die in Java

    - by Arkapravo
    The throw of a die is a popular program in Java, public class Die { /* This program simulates rolling a die */ public static void main(String[] args) { int die; // The number on the die. die = (int)(Math.random()*6 + 1); System.out.println (die); } // end main() } // end class What I wish to do is make it repeat, 500 times. I have not been able to put this program into a loop of 500. I usually program in Python, thus I guess my Java has rusted ! Any help is most welcome !

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  • Using EclipseLink

    - by Ross Peoples
    I am still new to Java and Eclipse and I'm trying to get my application to connect to a database. I think I want to use EclipseLink, but all of the documentation on the matter assumes you already know everything there is to know about everything. I keep getting linked back to this tutorial: http://www.vogella.de/articles/JavaPersistenceAPI/article.html But it's basically useless because it doesn't tell you HOW to do anything. For the Installation section, it tells you to download EclipseLink and gives you a link to the download page, but doesn't tell you what to do with it after you download. The download page doesn't either. I used the "Install new software" option in Eclipse to install EclipseLink into Eclipse, but it gave me like 4 different options, none of which are explained anywhere. It gave me options JPA, MOXy, SDO, etc, but I don't know which one I need. I just installed them all. Everything on the web assumes you are already a Java guru and things that are second nature to Java devs are never explained, so it's very frustrating for someone trying to learn. So how do I install and USE EclipseLink in my project and what do I need to do to connect it to a Microsoft SQL server? Again, I am new to all of this so I have no clue what to do. Thanks for the help.

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  • Running an allocation simulation repeatedly breaks after the first run.

    - by Az
    Background I have a bunch of students, their desired projects and the supervisors for the respective projects. I'm running a battery of simulations to see which projects the students end up with, which will allow me to get some useful statistics required for feedback. So, this is essentially a Monte-Carlo simulation where I'm randomising the list of students and then iterating through it, allocating projects until I hit the end of the list. Then the process is repeated again. Note that, within a single session, after each successful allocation of a project the following take place: + the project is set to allocated and cannot be given to another student + the supervisor has a fixed quota of students he can supervise. This is decremented by 1 + Once the quota hits 0, all the projects from that supervisor become blocked and this has the same effect as a project being allocated Code def resetData(): for student in students.itervalues(): student.allocated_project = None for supervisor in supervisors.itervalues(): supervisor.quota = 0 for project in projects.itervalues(): project.allocated = False project.blocked = False The role of resetData() is to "reset" certain bits of the data. For example, when a project is successfully allocated, project.allocated for that project is flipped to True. While that's useful for a single run, for the next run I need to be deallocated. Above I'm iterating through thee three dictionaries - one each for students, projects and supervisors - where the information is stored. The next bit is the "Monte-Carlo" simulation for the allocation algorithm. sesh_id = 1 for trial in range(50): for id in randomiseStudents(1): stud_id = id student = students[id] if not student.preferences: # Ignoring the students who've not entered any preferences for rank in ranks: temp_proj = random.choice(list(student.preferences[rank])) if not (temp_proj.allocated or temp_proj.blocked): alloc_proj = student.allocated_proj_ref = temp_proj.proj_id alloc_proj_rank = student.allocated_rank = rank successActions(temp_proj) temp_alloc = Allocated(sesh_id, stud_id, alloc_proj, alloc_proj_rank) print temp_alloc # Explained break sesh_id += 1 resetData() # Refer to def resetData() above All randomiseStudents(1) does is randomise the order of students. Allocated is a class defined as such: class Allocated(object): def __init__(self, sesh_id, stud_id, alloc_proj, alloc_proj_rank): self.sesh_id = sesh_id self.stud_id = stud_id self.alloc_proj = alloc_proj self.alloc_proj_rank = alloc_proj_rank def __repr__(self): return str(self) def __str__(self): return "%s - Student: %s (Project: %s - Rank: %s)" %(self.sesh_id, self.stud_id, self.alloc_proj, self.alloc_proj_rank) Output and problem Now if I run this I get an output such as this (truncated): 1 - Student: 7720 (Project: 1100241 - Rank: 1) 1 - Student: 7832 (Project: 1100339 - Rank: 1) 1 - Student: 7743 (Project: 1100359 - Rank: 1) 1 - Student: 7820 (Project: 1100261 - Rank: 2) 1 - Student: 7829 (Project: 1100270 - Rank: 1) . . . 1 - Student: 7822 (Project: 1100280 - Rank: 1) 1 - Student: 7792 (Project: 1100141 - Rank: 7) 2 - Student: 7739 (Project: 1100267 - Rank: 1) 3 - Student: 7806 (Project: 1100272 - Rank: 1) . . . 45 - Student: 7806 (Project: 1100272 - Rank: 1) 46 - Student: 7714 (Project: 1100317 - Rank: 1) 47 - Student: 7930 (Project: 1100343 - Rank: 1) 48 - Student: 7757 (Project: 1100358 - Rank: 1) 49 - Student: 7759 (Project: 1100269 - Rank: 1) 50 - Student: 7778 (Project: 1100301 - Rank: 1) Basically, it works perfectly for the first run, but on subsequent runs leading upto the nth run, in this case 50, only a single student-project allocation pair is returned. Thus, the main issue I'm having trouble with is figuring out what is causing this anomalous behaviour especially since the first run works smoothly. Thanks in advance, Az

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  • Convert string to decimal retaining the exact input format

    - by Brett
    Hi All, I'm sure this is a piece of cake but I'm really struggling with something that seems trivial. I need to check the inputted text of a textbox on the form submit and check to see if it's within a desired range (I've tried a Range Validator but it doesn't work for some reason so I'm trying to do this server-side). What I want to do is: Get the value inputted (eg. 0.02), replace the commas and periods, convert that to a decimal (or double or equivalent) and check to see if it's between 0.10 and 35000.00. Here's what I have so far: string s = txtTransactionValue.Text.Replace(",", string.Empty).Replace(".", string.Empty); decimal transactionValue = Decimal.Parse(s); if (transactionValue >= 0.10M && transactionValue <= 35000.00M) // do something If I pass 0.02 into the above, transactionValue is 2. I want to retain the value as 0.02 (ie. do no format changes to it - 100.00 is 100.00, 999.99 is 999.99) Any ideas? Thanks in advance, Brett

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  • La première beta de Ubuntu 10.04 LTS est sortie : nouveau design, nouveau kernel, nouveaux logiciels

    La beta 1 de Ubuntu 10.04 LTS est sortie Nouveau design, nouveau kernel, nouveaux logiciels Fini le marron dans Ubuntu. Le choc est rude, certes. Mais la première beta de la mouture 10.04 de l'OS consolera les nostalgique de l'ancien design de la distribution Linux (certainement) la plus connue au monde. Bien qu'il ne s'agisse que de la première beta, Lucid Lynx propose en effet déjà plusieurs nouveautés majeures. Sous le capot, on trouve par exemple le kernel 2.6.32 de Linux, Gnome 2.30 et une prise en charge améliorée des cartes graphiques NVIDIA. Coté logiciel, Ubuntu 10.04 propose notamment OpenOffice.org 3.2.0 (la suite bureautique de Su...

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  • Replace values in a dataframe based on another factor which contains NA's in R

    - by PaulHurleyuk
    I have a dataframe which contains (among other things) a numeric column with a concentration, and a factor column with a status flag. This status flag contains NA's. Here's an example df<-structure(list(conc = c(101.769, 1.734, 62.944, 92.697, 25.091, 27.377, 24.343, 55.084, 0.335, 23.280), status = structure(c(NA, NA, NA, NA, NA, NA, 2L, NA, 1L, NA), .Label = c("<LLOQ", "NR"), class = "factor")), .Names = c("conc", "status"), row.names = c(NA, -10L), class = "data.frame") I want to replace the concentration column with a string for some values of the flag column, or with the concentration value formatted to a certain number of significant digits. When I try this ifelse(df$status=="NR","NR",df$conc) The NA's in the status flag don't trigger either the true or false condition (and return NA) - as the documentation suggests it will. I could loop over the rows and use IF then else on each one but this seems inefficient. Am I missing something ? I've tried as.character(df$status) as well which doesn't work. My mojo must be getting low....

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  • Ubuntu représente 40% du marché desktop Linux, avec près de 12 millions d'utilisateurs

    Ubuntu représente 40% du marché desktop Linux, avec près de 12 millions d'utilisateurs Coup de tonnerre dans le monde des distributions Linux. Canonical vient de dévoiler qu'il y aurait aujourd'hui environ 12 millions d'utilisateurs d'Ubuntu de par le monde. C'est un chiffre colossal, comparé aux autres distrbutions. Par exemple, Fedora ne totalise "que" 4.5 millions d'utilisateurs. En 2008, on comptait seulement 8 millions d'utilisateurs d'Ubuntu. C'est donc plus de 50 % de croissance qui viennent d'être enregistrés. De plus, selon le très sérieux Linux Counter, il y aurait actuellement 29 millions d'utilisateurs de Linux sur notre planète. Donc, près de 40 % d'entre eux auraient choisis de ...

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  • Python learner needs help spotting an error

    - by Protean
    This piece of code gives a syntax error at the colon of "elif process.loop(i, len(list_i) != 'repeat':" and I can't seem to figure out why. class process: def loop(v1, v2): if v1 < v2 - 1: return 'repeat' def isel(chr_i, list_i): for i in range(len(list_i)): if chr_i == list_i[i]: return list_i[i] elif process.loop(i, len(list_i) != 'repeat': return 'error'()

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  • convert bitset to string ??

    - by mr.bio
    Hi .. What is wrong with this code ? set<string> nk ; bitset<3> bs1(string("100")); nk.insert(bs1.to_string()); error: no matching function for call to `std::bitset<3u::to_string()' why?!

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  • Awk: how to have backclashes in printf?

    - by HH
    Works: awk '{print $$1"\t&\t"$$2"\t\\\\"}' .file > file.tex Does not work, why? awk '{printf %.2f"\t&\t"\.2f"\t\\\\",$$1,$$2}' .file > file.tex Error: awk: {printf %.2f"\t&\t"\.2f"\t\\\\",$1,$2} awk: ^ backslash not last character on line

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  • Using a handle to collect output from CreateProcess()

    - by Stef
    Hi I am using CreateProcess() to run an external console application in Windows from my GUI application. I would like to somehow gather the output to know whether there were errors. Now I know I have to do something with hStdOutput, but I fail to understand what. I am new to c++ and an inexperienced programmer and I actually don't know what to do with a handle or how to light a pipe. How do I get the output to some kind of variable (or file)? This is what I have a the moment: void email::run(string path,string cmd){ WCHAR * ppath=new(nothrow) WCHAR[path.length()*2]; memset(ppath,' ',path.length()*2); WCHAR * pcmd= new(nothrow) WCHAR[cmd.length()*2]; memset(pcmd,' ',cmd.length()*2); string tempstr; ToWCHAR(path,ppath); //creates WCHAR from my std::string ToWCHAR(cmd,pcmd); STARTUPINFO info={sizeof(info)}; info.dwFlags = STARTF_USESHOWWINDOW; //hide process PROCESS_INFORMATION processInfo; if (CreateProcess(ppath,pcmd, NULL, NULL, FALSE, 0, NULL, NULL, &info, &processInfo)) { ::WaitForSingleObject(processInfo.hProcess, INFINITE); CloseHandle(processInfo.hProcess); CloseHandle(processInfo.hThread); } delete[](ppath); delete[](pcmd); } This code probably makes any decent programmer scream, but (I shouldn't even say it:) It works ;-) The Question: How do I use hStdOutput to read the output to a file (for instance)?

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  • Open Source Utilization Questions: How do you lone wold programmers best take advantage of open sour

    - by Funkyeah
    For Clarity: So you come up with an idea for a new program and want to start hacking, but you also happen to be a one-man army. How do you programming dynamos best find and utilize existing open-source software to give you the highest jumping off point possible when diving into your new project? When you do jump in where the shit do you start from? Any imaginary scenarios would be welcome, e.g. a shitty example might be utilizing a open-source database with an open-source IM client as a starting off point to a make a new client where you could tag and store conversations and query those tags at a later time.

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  • Why is my Ubuntu system not using the correct kernel?

    - by Brooks Moses
    We're having a bit of confusion on a Ubuntu remote system -- /boot/grub/menu.lst suggests the system should boot into kernel 2.6.35-30-generic, but it is actually running kernel 2.6.32-27-generic. Where should I look to start figuring out why this is happening and how to fix it? Specifically, /boot/grub/menu.lst has default 0 and the first entry is title Ubuntu 10.10, kernel 2.6.35-30-generic uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.35-30-generic root=UUID=67717ee3-cbf9-45d2- ae97-820256f4c4fd ro quiet splash initrd /boot/initrd.img-2.6.35-30-generic Further, I've confirmed that /boot/vmlinuz-2.6.35-30-generic and /boot/initrd.img-2.6.35-30-generic exist and have appropriate permissions. Meanwhile, uname -a returns: $ uname -a Linux cuda2 2.6.32-27-generic #49-Ubuntu SMP Thu Dec 2 00:51:09 UTC 2010 x86_64 GNU/Linux Edit: I've also tried re-running update-grub, and rebooting; no luck. Here's the full menu.lst, as requested by a commenter: # menu.lst - See: grub(8), info grub, update-grub(8) # grub-install(8), grub-floppy(8), # grub-md5-crypt, /usr/share/doc/grub # and /usr/share/doc/grub-legacy-doc/. ## default num # Set the default entry to the entry number NUM. Numbering starts from 0, and # the entry number 0 is the default if the command is not used. # # You can specify 'saved' instead of a number. In this case, the default entry # is the entry saved with the command 'savedefault'. # WARNING: If you are using dmraid do not use 'savedefault' or your # array will desync and will not let you boot your system. default 0 ## timeout sec # Set a timeout, in SEC seconds, before automatically booting the default entry # (normally the first entry defined). timeout 3 ## hiddenmenu # Hides the menu by default (press ESC to see the menu) hiddenmenu # Pretty colours #color cyan/blue white/blue ## password ['--md5'] passwd # If used in the first section of a menu file, disable all interactive editing # control (menu entry editor and command-line) and entries protected by the # command 'lock' # e.g. password topsecret # password --md5 $1$gLhU0/$aW78kHK1QfV3P2b2znUoe/ # password topsecret # # examples # # title Windows 95/98/NT/2000 # root (hd0,0) # makeactive # chainloader +1 # # title Linux # root (hd0,1) # kernel /vmlinuz root=/dev/hda2 ro # # # Put static boot stanzas before and/or after AUTOMAGIC KERNEL LIST ### BEGIN AUTOMAGIC KERNELS LIST ## lines between the AUTOMAGIC KERNELS LIST markers will be modified ## by the debian update-grub script except for the default options below ## DO NOT UNCOMMENT THEM, Just edit them to your needs ## ## Start Default Options ## ## default kernel options ## default kernel options for automagic boot options ## If you want special options for specific kernels use kopt_x_y_z ## where x.y.z is kernel version. Minor versions can be omitted. ## e.g. kopt=root=/dev/hda1 ro ## kopt_2_6_8=root=/dev/hdc1 ro ## kopt_2_6_8_2_686=root=/dev/hdc2 ro # kopt=root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro ## default grub root device ## e.g. groot=(hd0,0) # groot=67717ee3-cbf9-45d2-ae97-820256f4c4fd ## should update-grub create alternative automagic boot options ## e.g. alternative=true ## alternative=false # alternative=true ## should update-grub lock alternative automagic boot options ## e.g. lockalternative=true ## lockalternative=false # lockalternative=false ## additional options to use with the default boot option, but not with the ## alternatives ## e.g. defoptions=vga=791 resume=/dev/hda5 # defoptions=quiet splash ## should update-grub lock old automagic boot options ## e.g. lockold=false ## lockold=true # lockold=false ## Xen hypervisor options to use with the default Xen boot option # xenhopt= ## Xen Linux kernel options to use with the default Xen boot option # xenkopt=console=tty0 ## altoption boot targets option ## multiple altoptions lines are allowed ## e.g. altoptions=(extra menu suffix) extra boot options ## altoptions=(recovery) single # altoptions=(recovery mode) single ## controls how many kernels should be put into the menu.lst ## only counts the first occurence of a kernel, not the ## alternative kernel options ## e.g. howmany=all ## howmany=7 # howmany=all ## specify if running in Xen domU or have grub detect automatically ## update-grub will ignore non-xen kernels when running in domU and vice versa ## e.g. indomU=detect ## indomU=true ## indomU=false # indomU=detect ## should update-grub create memtest86 boot option ## e.g. memtest86=true ## memtest86=false # memtest86=true ## should update-grub adjust the value of the default booted system ## can be true or false # updatedefaultentry=false ## should update-grub add savedefault to the default options ## can be true or false # savedefault=false ## ## End Default Options ## title Ubuntu 10.10, kernel 2.6.35-30-generic uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.35-30-generic root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro quiet splash initrd /boot/initrd.img-2.6.35-30-generic title Ubuntu 10.10, kernel 2.6.35-30-generic (recovery mode) uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.35-30-generic root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro single initrd /boot/initrd.img-2.6.35-30-generic title Ubuntu 10.10, kernel 2.6.32-32-server uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.32-32-server root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro quiet splash initrd /boot/initrd.img-2.6.32-32-server title Ubuntu 10.10, kernel 2.6.32-32-server (recovery mode) uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.32-32-server root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro single initrd /boot/initrd.img-2.6.32-32-server title Ubuntu 10.10, kernel 2.6.32-27-generic uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.32-27-generic root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro quiet splash initrd /boot/initrd.img-2.6.32-27-generic title Ubuntu 10.10, kernel 2.6.32-27-generic (recovery mode) uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/vmlinuz-2.6.32-27-generic root=UUID=67717ee3-cbf9-45d2-ae97-820256f4c4fd ro single initrd /boot/initrd.img-2.6.32-27-generic title Chainload into GRUB 2 root 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/grub/core.img title Ubuntu 10.10, memtest86+ uuid 67717ee3-cbf9-45d2-ae97-820256f4c4fd kernel /boot/memtest86+.bin ### END DEBIAN AUTOMAGIC KERNELS LIST To add complication and joy to my life, this is a desktop machine in a remote datacenter; we don't have either local access or serial-console access. Suggestions?

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  • Implementing a linear, binary SVM (support vector machine)

    - by static_rtti
    I want to implement a simple SVM classifier, in the case of high-dimensional binary data (text), for which I think a simple linear SVM is best. The reason for implementing it myself is basically that I want to learn how it works, so using a library is not what I want. The problem is that most tutorials go up to an equation that can be solved as a "quadratic problem", but they never show an actual algorithm! So could you point me either to a very simple implementation I could study, or (better) to a tutorial that goes all the way to the implementation details? Thanks a lot!

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  • Python: how to enclose strings in a list with < and >

    - by Michael Konietzny
    Hello, i would like to enclose strings inside of list into < (formatted like <%s). The current code does the following: def create_worker (general_logger, general_config): arguments = ["worker_name", "worker_module", "worker_class"] __check_arguments(arguments) def __check_arguments(arguments): if len(sys.argv) < 2 + len(arguments): print "Usage: %s delete-project %s" % (__file__," ".join(arguments)) sys.exit(10) The current output looks like this: Usage: ...\handler_scripts.py delete-project worker_name worker_module worker_class and should look like this: Usage: ...\handler_scripts.py delete-project <worker_name> <worker_module> <worker_class> Is there any short way to do this ? Greetings, Michael

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  • Allocation algorithm help, using Python.

    - by Az
    Hi there, I've been working on this general allocation algorithm for students. The pseudocode for it (a Python implementation) is: for a student in a dictionary of students: for student's preference in a set of preferences (ordered from 1 to 10): let temp_project be the first preferred project check if temp_project is available if so, allocate it to them and make the project UNavailable to others Quite simply this will try to allocate projects by starting from their most preferred. The way it works, out of a set of say 100 projects, you list 10 you would want to do. So the 10th project wouldn't be the "least preferred overall" but rather the least preferred in their chosen set, which isn't so bad. Obviously if it can't allocate a project, a student just reverts to the base case which is an allocation of None, with a rank of 11. What I'm doing is calculating the allocation "quality" based on a weighted sum of the ranks. So the lower the numbers (i.e. more highly preferred projects), the better the allocation quality (i.e. more students have highly preferred projects). That's basically what I've currently got. Simple and it works. Now I'm working on this algorithm that tries to minimise the allocation weight locally (this pseudocode is a bit messy, sorry). The only reason this will probably work is because my "search space" as it is, isn't particularly large (just a very general, anecdotal observation, mind you). Since the project is only specific to my Department, we have their own limits imposed. So the number of students can't exceed 100 and the number of preferences won't exceed 10. for student in a dictionary/list/whatever of students: where i = 0 take the (i)st student, (i+1)nd student for their ranks: allocate the projects and set local_weighting to be sum(student_i.alloc_proj_rank, student_i+1.alloc_proj_rank) these are the cases: if local_weighting is 2 (i.e. both ranks are 1): then i += 1 and and continue above if local weighting is = N>2 (i.e. one or more ranks are greater than 1): let temp_local_weighting be N: pick student with lowest rank and then move him to his next rank and pick the other student and reallocate his project after this if temp_local_weighting is < N: then allocate those projects to the students move student with lowest rank to the next rank and reallocate other if temp_local_weighting < previous_temp_allocation: let these be the new allocated projects try moving for the lowest rank and reallocate other else: if this weighting => previous_weighting let these be the allocated projects i += 1 and move on for the rest of the students So, questions: This is sort of a modification of simulated annealing, but any sort of comments on this would be appreciated. How would I keep track of which student is (i) and which student is (i+1) If my overall list of students is 100, then the thing would mess up on (i+1) = 101 since there is none. How can I circumvent that? Any immediate flaws that can be spotted? Extra info: My students dictionary is designed as such: students[student_id] = Student(student_id, student_name, alloc_proj, alloc_proj_rank, preferences) where preferences is in the form of a dictionary such that preferences[rank] = {project_id}

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