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  • python add two list and createing a new list

    - by Adam G.
    lst1 = ['company1,AAA,7381.0 ', 'company1,BBB,-8333.0 ', 'company1,CCC, 3079.999 ', 'company1,DDD,5699.0 ', 'company1,EEE,1640.0 ', 'company1,FFF,-600.0 ', 'company1,GGG,3822.0 ', 'company1,HHH,-600.0 ', 'company1,JJJ,-4631.0 ', 'company1,KKK,-400.0 '] lst2 =['company1,AAA,-4805.0 ', 'company1,ZZZ,-2576.0 ', 'company1,BBB,1674.0 ', 'company1,CCC,3600.0 ', 'company1,DDD,1743.998 '] output I need == ['company1,AAA,2576.0','company1,ZZZ,-2576.0 ','company1,KKK,-400.0 ' etc etc] I need to add it similar product number in each list and move it to a new list. I also need any symbol not being added together to be added to that new list. I am having problems with moving through each list. This is what I have: h = [] z = [] a = [] for g in hhl: spl1 = g.split(",") h.append(spl1[1]) for j in c_hhl: spl2 = j.split(",") **if spl2[1] in h: converted_num =(float(spl2[2]) +float(spl1[2])) pos=('{0},{1},{2}'.format(spl2[0],spl2[1],converted_num)) z.append(pos)** else: pos=('{0},{1},{2}'.format(spl2[0],spl2[1],spl2[2])) z.append(pos) for f in z: spl3 = f.split(",") a.append(spl3[1]) for n in hhl[:]: spl4 = n.split(",") if spl4[1] in a: got = (spl4[0],spl4[1],spl4[2]) hhl.remove(n) smash = hhl+z #for i in smash: for i in smash: print(i) I am having problem iterating through the list to make sure I get all of the simliar product to a new list,(bold) and any product not in list 1 but in lst2 to the new list and vice versa. I am sure there is a much easier way.

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  • Python 3.0 - Dynamic Class Instance Naming

    - by Jon
    I want to use a while loop to initialize class objects with a simple incremented naming convention. The goal is to be able to scale the number of class objects at will and have the program generate the names automatically. (ex. h1...h100...h1000...) Each h1,h2,h3... being its own instance. Here is my first attempt... have been unable to find a good example. class Korker(object): def __init__(self,ident,roo): self.ident = ident self.roo = roo b = 1 hwinit = 'h' hwstart = 0 while b <= 10: showit = 'h' + str(b) print(showit) #showit seems to generate just fine as demonstrated by print str(showit) == Korker("test",2) #this is the line that fails b += 1 The errors I get range from a string error to a cannot use function type error.... Any help would be greatly appreciated.

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  • Python Speeding Up Retrieving data from extremely large string

    - by Burninghelix123
    I have a list I converted to a very very long string as I am trying to edit it, as you can gather it's called tempString. It works as of now it just takes way to long to operate, probably because it is several different regex subs. They are as follow: tempString = ','.join(str(n) for n in coords) tempString = re.sub(',{2,6}', '_', tempString) tempString = re.sub("[^0-9\-\.\_]", ",", tempString) tempString = re.sub(',+', ',', tempString) clean1 = re.findall(('[-+]?[0-9]*\.?[0-9]+,[-+]?[0-9]*\.?[0-9]+,' '[-+]?[0-9]*\.?[0-9]+'), tempString) tempString = '_'.join(str(n) for n in clean1) tempString = re.sub(',', ' ', tempString) Basically it's a long string containing commas and about 1-5 million sets of 4 floats/ints (mixture of both possible),: -5.65500020981,6.88999986649,-0.454999923706,1,,,-5.65500020981,6.95499992371,-0.454999923706,1,,, The 4th number in each set I don't need/want, i'm essentially just trying to split the string into a list with 3 floats in each separated by a space. The above code works flawlessly but as you can imagine is quite time consuming on large strings. I have done a lot of research on here for a solution but they all seem geared towards words, i.e. swapping out one word for another. EDIT: Ok so this is the solution i'm currently using: def getValues(s): output = [] while s: # get the three values you want, discard the 3 commas, and the # remainder of the string v1, v2, v3, _, _, _, s = s.split(',', 6) output.append("%s %s %s" % (v1.strip(), v2.strip(), v3.strip())) return output coords = getValues(tempString) Anyone have any advice to speed this up even farther? After running some tests It still takes much longer than i'm hoping for. I've been glancing at numPy, but I honestly have absolutely no idea how to the above with it, I understand that after the above has been done and the values are cleaned up i could use them more efficiently with numPy, but not sure how NumPy could apply to the above. The above to clean through 50k sets takes around 20 minutes, I cant imagine how long it would be on my full string of 1 million sets. I'ts just surprising that the program that originally exported the data took only around 30 secs for the 1 million sets

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  • Adding a decorator that converts strings to lowercase in Python

    - by user2905382
    So I am new to learning decorators and I have gone through countless tutorials and while I understand and can mostly follow all of the examples, I think the best way to learn, would be to implement a decorator myself. So I am going to use this example below. I realize a decorator is not at all necessary to do this, but for the sake of learning, I would like to add a decorator that filters the strings like dog name and breed and turns them into lowercase. Any ideas or pointers in the right direction would be appreciated. class Dogs: totalDogs = 0 dogList=[] def __init__(self, breed, color, age): self.breed=breed self.color=color self.age=age Dogs.dogList.append(self.breed) Dogs.totalDogs += 1 def displayDogs(self): print "breed: ", self.breed print "color: ",self.color print "age: ",self.age print "list of breeds:", Dogs.dogList print "total dogs: ", Dogs.totalDogs def somedecorator(*args): #now what terrier=Dogs("TeRrIer", "white", 5) terrier.displayDogs() retriever=Dogs("goldenRETRIEVER", "brown", 10) retriever.displayDogs()

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  • Sorting a meta-list by first element of children lists in Python

    - by thismachinechills
    I have a list, root, of lists, root[child0], root[child1], etc. I want to sort the children of the root list by the first value in the child list, root[child0][0], which is an int. Example: import random children = 10 root = [[random.randint(0, children), "some value"] for child in children] I want to sort root from greatest to least by the first element of each of it's children. I've taken a look at some previous entries that used sorted() and a lamda function I'm entirely unfamiliar with, so I'm unsure of how to apply that to my problem. Appreciate any direction that can by given Thanks

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  • Python 3: Recursivley find if number is even

    - by pythonhack
    I am writing a program that must find if a number is even or not. It needs to follow this template. I can get it to find if a number is even or not recursively (call function and subtract 2, base case zero), but I am having a hard time following this template, based on how the isEven function is called in the main function. Any help would be greatly appreciated. Write a recursive function called isEven that finds whether a number is even or not: def isEven() #recursivley determine whether number is even or not def main(): number=int(input(“Enter a number : “)) if (isEven(number)): print(“Number is even”) else: print(“Number is not even”) main() Thank you! Appreciate it.

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  • threading in Python taking up too much CPU

    - by KevinShaffer
    I wrote a chat program and have a GUI running using Tkinter, and to go and check when new messages have arrived, I create a new thread so Tkinter keeps doing its thing without locking up while the new thread goes and grabs what I need and updates the Tkinter window. This however becomes a huge CPU hog, and my guess is that it has to do somehow with the fact that the Thread is started and never really released when the function is done. Here's the relevant code (it's ugly and not optimized at the moment, but it gets the job done, and itself does not use too much processing power, as when I run it not threaded, it doesn't take up much CPU but it locks up Tkinter) Note: This is inside of a class, hence the extra tab. def interim(self): threading.Thread(target=self.readLog).start() self.after(5000,self.interim) def readLog(self): print 'reading' try: length = len(str(self.readNumber)) f = open('chatlog'+str(myport),'r') temp = f.readline().replace('\n','') while (temp[:length] != str(self.readNumber)) or temp[0] == '<': temp = f.readline().replace('\n','') while temp: if temp[0] != '<': self.updateChat(temp[length:]) self.readNumber +=1 else: self.updateChat(temp) temp = f.readline().replace('\n','') f.close() Is there a way to better manage the threading so I don't consume 100% of the CPU very quickly?

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  • Python 3 order of testing undetermined

    - by user578598
    string='a' p=0 while (p <len(string)) & (string[p]!='c') : p +=1 print ('the end but the process already died ') while (p <1) & (string[p]!='c') : IndexError: string index out of range I want to test a condition up to the end of a string (example string length=1) why are both parts of the and executed is the condition is already false! as long as p < len(string). the second part does not even need executing. if it does a lot of performance can be lost

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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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  • 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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  • 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 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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  • 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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  • 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 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\ 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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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  • 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 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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  • 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 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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  • 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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