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  • Ubuntu 12.04 Preseed LDAP Config

    - by Arturo
    I'm trying to deploy Ubuntu 12.04 via xCAT, everything works except the automatic configuration of LDAP, the preseed file is read but the file /etc/nsswitch is not written properly. My Preseed File: [...] ### LDAP Setup nslcd nslcd/ldap-bindpw password ldap-auth-config ldap-auth-config/bindpw password ldap-auth-config ldap-auth-config/rootbindpw password ldap-auth-config ldap-auth-config/binddn string cn=proxyuser,dc=example,dc=net libpam-runtime libpam-runtime/profiles multiselect unix, ldap, gnome-keyring, consolekit, capability ldap-auth-config ldap-auth-config/dbrootlogin boolean false ldap-auth-config ldap-auth-config/rootbinddn string cn=manager,dc=xcat-domain,dc=com nslcd nslcd/ldap-starttls boolean false nslcd nslcd/ldap-base string dc=xcat-domain,dc=com ldap-auth-config ldap-auth-config/pam_password select md5 ldap-auth-config ldap-auth-config/move-to-debconf boolean true ldap-auth-config ldap-auth-config/ldapns/ldap-server string ldap://192.168.32.42 ldap-auth-config ldap-auth-config/ldapns/base-dn string dc=xcat-domain,dc=com ldap-auth-config ldap-auth-config/override boolean true libnss-ldapd libnss-ldapd/clean_nsswitch boolean false libnss-ldapd libnss-ldapd/nsswitch multiselect passwd,group,shadow nslcd nslcd/ldap-reqcert select ldap-auth-config ldap-auth-config/ldapns/ldap_version select 3 ldap-auth-config ldap-auth-config/dblogin boolean false nslcd nslcd/ldap-uris string ldap://192.168.32.42 nslcd nslcd/ldap-binddn string [...] After the installation, nsswitch.conf rimains unchanged. Has someone an idea?? Thanks!

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  • Solaris ldap Authentication

    - by Tman
    Iv been having a trouble trying to get my Solaris 10 server to authenticate against an eDir server.im managed to Set up my linux(RHeL,SLES) servers to authenticate against the ldap Server.which works fine. Here is my configuration Files. ldapclient list: NS_LDAP_FILE_VERSION= 2.0 NS_LDAP_BINDDN= cn=proxyuser,o=AEDev NS_LDAP_BINDPASSWD= {NS1}ecfa88f3a945c22222233 NS_LDAP_SERVERS= 192.168.0.19 NS_LDAP_SEARCH_BASEDN= ou=auth,o=AEDev NS_LDAP_AUTH= simple NS_LDAP_SEARCH_SCOPE= sub NS_LDAP_CACHETTL= 0 NS_LDAP_CREDENTIAL_LEVEL= anonymous NS_LDAP_SERVICE_SEARCH_DESC= group:ou=Groups,ou=auth,o=AEDev NS_LDAP_SERVICE_SEARCH_DESC= shadow:ou=users,ou=auth,o=AEDev?sub?objectClass=shadowAccount NS_LDAP_SERVICE_SEARCH_DESC= passwd:ou=auth,o=AEDev?sub?objectClass=posixAccount NS_LDAP_BIND_TIME= 10 NS_LDAP_SERVICE_AUTH_METHOD= pam_ldap:simple getent passwd works fine: root:x:0:0:Super-User:/:/sbin/sh daemon:x:1:1::/: bin:x:2:2::/usr/bin: sys:x:3:3::/: adm:x:4:4:Admin:/var/adm: lp:x:71:8:Line Printer Admin:/usr/spool/lp: uucp:x:5:5:uucp Admin:/usr/lib/uucp: nuucp:x:9:9:uucp Admin:/var/spool/uucppublic:/usr/lib/uucp/uucico smmsp:x:25:25:SendMail Message Submission Program:/: listen:x:37:4:Network Admin:/usr/net/nls: gdm:x:50:50:GDM Reserved UID:/: webservd:x:80:80:WebServer Reserved UID:/: postgres:x:90:90:PostgreSQL Reserved UID:/:/usr/bin/pfksh svctag:x:95:12:Service Tag UID:/: nobody:x:60001:60001:NFS Anonymous Access User:/: noaccess:x:60002:60002:No Access User:/: nobody4:x:65534:65534:SunOS 4.x NFS Anonymous Access User:/: tlla:x:2012:100::/home/tlla: test:x:2011:100::/home/test: thato:x:2010:100::/home/thato: pam.conf login auth sufficient pam_unix_auth.so.1 #server_policy login auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass login auth required pam_dial_auth.so.1 rlogin auth sufficient pam_rhosts_auth.so.1 rlogin auth requisite pam_authtok_get.so.1 rlogin auth required pam_dhkeys.so.1 rlogin auth required pam_unix_cred.so.1 rlogin auth sufficient pam_unix_auth.so.1 rlogin auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass rsh auth sufficient pam_rhosts_auth.so.1 rsh auth required pam_unix_cred.so.1 rsh auth sufficient pam_unix_auth.so.1 #server_policy rsh auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass other auth requisite pam_authtok_get.so.1 other auth required pam_dhkeys.so.1 other auth required pam_unix_cred.so.1 other auth sufficient pam_unix_auth.so.1 other auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass passwd auth required pam_passwd_auth.so.1 passwd auth sufficient pam_unix_auth.so.1 ssh account sufficient pam_unix.so.1 ssh account sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass other account requisite pam_roles.so.1 other account sufficient pam_unix_account.so.1 other account sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass other password required pam_dhkeys.so.1 other password requisite pam_authtok_get.so.1 other password requisite pam_authtok_check.so.1 other password required pam_authtok_store.so.1 other password sufficient pam_unix.so.1 other password sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass Local Authentication Works But LDAP Authentication Doesn't Work.

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  • Oracle’s Vision for the Social-Enabled Enterprise

    - by Richard Lefebvre
    2 years ago, Social was a nice to have. Now it’s a must-have’- Mark Hurd .Do you agree? Check out  the on demand version of the Oracle’s Vision for the Social-Enabled Enterprise Exclusive Webcast in a 30' video HERE  Smart companies are developing social media strategies to engage customers, gain brand insights, and transform employee collaboration and recruitment. Join Oracle President Mark Hurd and senior Oracle executives to learn more about Oracle's vision for the social-enabled enterprise

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  • Google I/O 2010 - Google Buzz, location, and social gaming

    Google I/O 2010 - Google Buzz, location, and social gaming Google I/O 2010 - Surf the stream: Google Buzz, location, and social gaming Social Web 201 Bob Aman, Timothy Jordan Google Buzz has a feature-rich API that allows you to do all kinds of interesting things with conversations and location. In this session we'll build a Buzz-tastic mobile game using App Engine, HTML5, and the Buzz API for social awesomeness. For all I/O 2010 sessions, please go to code.google.com From: GoogleDevelopers Views: 2 0 ratings Time: 31:18 More in Science & Technology

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  • SEO, Social Media & Skinny Jeans

    No longer is social media just the "cool thing" to do - like rocking an 80's inspired flat top hairdo in 2010. Social networking now has a definitive effect on your SEO efforts, and as such, it must be monitored closely. While previous social networking efforts were generally tied to connecting with a business's customer base, Google is now including real-time social media into their search results - resulting in increased (and practically effortless) SEO opportunities.

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  • Project Euler 10: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 10.  As always, any feedback is welcome. # Euler 10 # http://projecteuler.net/index.php?section=problems&id=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. import time start = time.time() def primes_to_max(max): primes, number = [2], 3 while number < max: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes primes = primes_to_max(2000000) print sum(primes) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • What's wrong with relative imports in Python?

    - by Oddthinking
    I recently upgraded versions of pylint, a popular Python style-checker. It has gone ballistic throughout my code, pointing out places where I import modules in the same package, without specifying the full package path. The new error message is W0403. W0403: Relative import %r, should be %r Used when an import relative to the package directory is detected. Example For example, if my packages are structured like this: /cake /__init__.py /icing.py /sponge.py /drink and in the sponge package I write: import icing instead of import cake.icing I will get this error. While I understand that not all Pylint messages are of equal importance, and I am not afraid to dismiss them, I don't understand why such a practice is considered a poor idea. I was hoping someone could explain the pitfalls, so I could improve my coding style rather than (as I currently plan to do) turning off this apparently spurious warning.

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  • Project Euler 15: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 15.  As always, any feedback is welcome. # Euler 15 # http://projecteuler.net/index.php?section=problems&id=15 # Starting in the top left corner of a 2x2 grid, there # are 6 routes (without backtracking) to the bottom right # corner. How many routes are their in a 20x20 grid? import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) rows, cols = 20, 20 print factorial(rows+cols) / (factorial(rows) * factorial(cols)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 9: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 9.  As always, any feedback is welcome. # Euler 9 # http://projecteuler.net/index.php?section=problems&id=9 # A Pythagorean triplet is a set of three natural numbers, # a b c, for which, # a2 + b2 = c2 # For example, 32 + 42 = 9 + 16 = 25 = 52. # There exists exactly one Pythagorean triplet for which # a + b + c = 1000. Find the product abc. import time start = time.time() product = 0 def pythagorean_triplet(): for a in range(1,501): for b in xrange(a+1,501): c = 1000 - a - b if (a*a + b*b == c*c): return a*b*c print pythagorean_triplet() print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Compound assignment operators in Python's Numpy library

    - by Leonard
    The "vectorizing" of fancy indexing by Python's numpy library sometimes gives unexpected results. For example: import numpy a = numpy.zeros((1000,4), dtype='uint32') b = numpy.zeros((1000,4), dtype='uint32') i = numpy.random.random_integers(0,999,1000) j = numpy.random.random_integers(0,3,1000) a[i,j] += 1 for k in xrange(1000): b[i[k],j[k]] += 1 Gives different results in the arrays 'a' and 'b' (i.e. the appearance of tuple (i,j) appears as 1 in 'a' regardless of repeats, whereas repeats are counted in 'b'). This is easily verified as follows: numpy.sum(a) 883 numpy.sum(b) 1000 It is also notable that the fancy indexing version is almost two orders of magnitude faster than the for loop. My question is: "Is there an efficient way for numpy to compute the repeat counts as implemented using the for loop in the provided example?"

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  • Project Euler 5: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 5.  As always, any feedback is welcome. # Euler 5 # http://projecteuler.net/index.php?section=problems&id=5 # 2520 is the smallest number that can be divided by each # of the numbers from 1 to 10 without any remainder. # What is the smallest positive number that is evenly # divisible by all of the numbers from 1 to 20? import time start = time.time() def gcd(a, b): while b: a, b = b, a % b return a def lcm(a, b): return a * b // gcd(a, b) print reduce(lcm, range(1, 20)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 8: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Python or C server hosting for indie development

    - by Richard Fabian
    I've written a lot of the game, but it's singleplayer. Now we want to join up and play together. I want to host it like an MMO, but haven't got any personal ability to host (no static IPs or direct access to a reasonable router that will allow me to port forward) so I wondered if there were any free / very cheap hosting solutions for people developing games that need to develop their MMO side. In my case it's a world server for a 2D game where the world map can be changed by the players. So, GAE sounds expensive, as there would be quite a few updates per second (I heard they bill for data updates but not for download, but can't find refernce to billing anywhere on the FAQs) I'd prefer to be able to write the server in python as that's what the game is written in (with pygame), but C is fine, and maybe even better as it might prompt me to write some more performant world generator code ;)

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  • Project Euler 2: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 2.  As always, any feedback is welcome. # Euler 2 # http://projecteuler.net/index.php?section=problems&id=2 # Find the sum of all the even-valued terms in the # Fibonacci sequence which do not exceed four million. # Each new term in the Fibonacci sequence is generated # by adding the previous two terms. By starting with 1 # and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # Find the sum of all the even-valued terms in the # sequence which do not exceed four million. import time start = time.time() total = 0 previous = 0 i = 1 while i <= 4000000: if i % 2 == 0: total +=i # variable swapping removes the need for a temp variable i, previous = previous, previous + i print total print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • PHP developer wanting to learn python

    - by dclowd9901
    I'm pretty familiar at this point with PHP (Javascript, too), up to the point of OOP in PHP, and am looking to branch out my knowledge. I'm looking at Python next, but a lot of it is a bit alien to me as a PHP developer. I'm less concerned about learning the language itself. I'm positive there's plenty of good resources, documentation and libraries to help me get the code down. I'm less sure about the technical aspects of how to set up a dev environment, unit testing and other more mundane details that are very important, aid in rapid development, but aren't as widely covered. Are there any good resources out there for this?

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  • Project Euler 16: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 16.  As always, any feedback is welcome. # Euler 16 # http://projecteuler.net/index.php?section=problems&id=16 # 2^15 = 32768 and the sum of its digits is # 3 + 2 + 7 + 6 + 8 = 26. # What is the sum of the digits of the number 2^1000? import time start = time.time() print sum([int(i) for i in str(2**1000)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 7: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Finding the html tag value with Python [on hold]

    - by MrWho
    Consider a html page, which contains a line like below: file: 'http://google.com/video.mp4' I want to search for google.com/video.mp4 in that file and save it in a variable.I want to code it with python. Shortly, I want to elicit a link from a html page, so I need to get the link by using regular expressions or the other techniques in which I'm asking about. PS: What should I exactly try to clarify?it's really annoying that the administrators don't even say what is exactly unclear about the question, they've just learned to close or on hold the topics!

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  • Project Euler 4: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 4.  As always, any feedback is welcome. # Euler 4 # http://projecteuler.net/index.php?section=problems&id=4 # Find the largest palindrome made from the product of # two 3-digit numbers. A palindromic number reads the # same both ways. The largest palindrome made from the # product of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of # two 3-digit numbers. import time start = time.time() def isPalindrome(s): return s == s[::-1] max = 0 for i in xrange(100, 999): for j in xrange(i, 999): n = i * j; if (isPalindrome(str(n))): if (n > max): max = n print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 13: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 6: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 6.  As always, any feedback is welcome. # Euler 6 # http://projecteuler.net/index.php?section=problems&id=6 # Find the difference between the sum of the squares of # the first one hundred natural numbers and the square # of the sum. import time start = time.time() square_of_sums = sum(range(1,101)) ** 2 sum_of_squares = reduce(lambda agg, i: agg+i**2, range(1,101)) print square_of_sums - sum_of_squares print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • updater stuck on downloading python files

    - by Tim
    However i am a java programmer is consider myself as a linux noob. So i could use a little help here... I am trying to update my ubuntu from version 10.04 to 12.04.1 (LTS). The downloading start and runs at around 10MB/s untill i am somewhere near 26% and the speed just drops to 0. I figured i could use the command "do-release-update" so i could see what it was doing. Again the same occured: the downloading stucked on 26% at "http://nl.archive.ubuntu.com/ubunutu/ precise/main python-qt4 1386 4.9.1-2ubuntu1". It says it downloaded 41% of that file/package. It also tries another wget on the same file every X seconds. Help? Greetings Tim, Holland.

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  • Project Euler 20: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 20.  As always, any feedback is welcome. # Euler 20 # http://projecteuler.net/index.php?section=problems&id=20 # n! means n x (n - 1) x ... x 3 x 2 x 1 # Find the sum of digits in 100! import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) print sum([int(i) for i in str(factorial(100))]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 1: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 1.  As always, any feedback is welcome. # Euler 1 # http://projecteuler.net/index.php?section=problems&amp;id=1 # If we list all the natural numbers below 10 that are # multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of # these multiples is 23. Find the sum of all the multiples # of 3 or 5 below 1000. import time start = time.time() print sum([x for x in range(1000) if x % 3== 0 or x % 5== 0]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue') # Also cool def constraint(x): return x % 3 == 0 or x % 5 == 0 print sum(filter(constraint, range(1000)))

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