Re: pythonize this!
On Jun 15, 1:49 pm, superpollo wrote: > goal (from e.c.m.): evaluate > 1^2+2^2+3^2-4^2-5^2+6^2+7^2+8^2-9^2-10^2+...-2010^2, where each three > consecutive + must be followed by two - (^ meaning ** in this context) My functional approach :) from operator import add from functools import reduce def square(l): return (v**2 for v in l) def sign(l, s): i = 0 for v in l: yield s[i % len(s)] * v i += 1 print reduce(add, sign(square(xrange(1, 2011)), [1, 1, 1, -1, -1])) (536926141) ~Aki -- http://mail.python.org/mailman/listinfo/python-list
Re: pythonize this!
On Jun 15, 2:37 pm, Peter Otten <__pete...@web.de> wrote: > >>> from itertools import cycle, izip > >>> sum(sign*i*i for sign, i in izip(cycle([1]*3+[-1]*2), range(1, 2011))) Wow!! :D I didn't knew cycle, great! With that i can reduce my solution (which isn't still elegant as your) to: print reduce(add,imap(mul, cycle([1, 1, 1, -1, -1]), (v**2 for v in xrange(1, 2011 (536926141) ~Aki -- http://mail.python.org/mailman/listinfo/python-list
