Got it !! @Alan, @DaveA: Hontoni Arigato :)
On Fri, Oct 30, 2009 at 9:55 PM, Dave Angel <da...@ieee.org> wrote: > Shashwat Anand wrote: > >> Shashwat Anand to Bangalore >> show details 5:31 AM (2 minutes ago) >> >> <snip> >> >> I wrote an LCM function of mine as follows: >> >> import fractions >> def lcm(mylist): >> # lcm by defination is Lowest Common Multiple >> # lcm (a*b) = a*b / gcd(a*b) >> # lcm (a, b, c) = lcm(lcm(a, b), c) >> # the function lcm() returns lcm of numbers in a list >> # for the special case of two numbers, pass the argument as lcm([a, b]) >> sol = 1 >> for i in mylist: >> sol = sol * i / fractions.gcd(sol, i) >> return sol >> >> print lcm(range(1, 11)) #gives lcm of numbers (1, 2, 3....,9 ,10) >> print lcm([2, 3]) #gives lcm of two numbers, a special case >> print lcm([2, 5, 6, 10]) #gives lcm of a random list >> >> >> However I also did a dirty hack as an alternate approach : >> >> import fractions >> l = [1] >> print max( ( l[i-2], l.append(l[-1] * i / fractions.gcd(l[-1], i ) ) ) for >> i >> in range(2, 12) )[0] >> # prints the LCM of list (1, 10) >> >> However to shorten the code i made it totally unpythonic. >> Can reduce( ) or any other function be of help ? >> >> Let me take a test-case as an example: >> I want to multiple all the content of a given list, >> now say my list is lst = [2, 3, 9] >> I can do: >> >> sol = 1 >> for i in lst: >> sol *= i >> print sol >> >> However this can also be done as: >> >> >> >>> reduce( operator.mul, lst) >>>>> >>>>> >>>> >> Can it be done via List Comprehension or we have to do dirty hacks as >> mentioned above :( >> Can the LCM function be reduced ? >> >> *The point is if we insert a variable( sol here in both test case and LCM >> case), which changes its values dynamically depending upon the next values >> of the list, how can I calculate so without going through the long way of >> using for-construct*, which makes me feel as though i'm coding in C. >> reduce( >> ) or even sum( ) does helps in some cases but it constrains me as per my >> requirement. Any pointers ? >> >> >> > Starting where Alan left off, > > def lcm(mylist): > def lcm2(a, b): > return a*b / fractions.gcd(a, b) > return reduce(lcm2, mylist) > > or, if you feel the need to get rid of the extra name, > > def lcm(mylist): > return reduce(lambda a,b : a*b/fractions.gcd(a,b), mylist) > > DaveA >
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