On Tue, 11 Oct 2005 11:22:39 +0200, Lasse Vågsæther Karlsen wrote: > This begs a different question along the same lines.
Er, no it doesn't. "Begs the question" does _not_ mean "asks the question" or "suggests the question". It means "assumes the truth of that which needs to be proven". http://en.wikipedia.org/wiki/Begging_the_question http://www.worldwidewords.org/qa/qa-beg1.htm (Both of these sources are far more forgiving of the modern mis-usage than I am. Obviously.) > If I have a generator or other iterable producing a vast number of > items, and use it like this: > > s = [k for k in iterable] > > if I know beforehand how many items iterable would possibly yield, would > a construct like this be faster and "use" less memory? > > s = [0] * len(iterable) > for i in xrange(len(iterable)): > s[i] = iterable.next() Faster? Maybe. Only testing can tell -- but I doubt it. But as for less memory, look at the two situations. In the first, you create a list of N objects. In the second, you end up with the same list of N objects, plus an xrange object, which may be bigger or smaller than an ordinary list of N integers, depending on how large N is. So it won't use *less* memory -- at best, it will use just slightly more. Is there a way from within Python to find out how much memory a single object uses, and how much memory Python is using in total? -- Steven. -- http://mail.python.org/mailman/listinfo/python-list
