Re: [Tutor] declaring list
hello list I asked yesterday how we can "declare" an array in python. Thanks Brian van den Broek and Frank Schley for your responses. A[None] * n works for me. To answer Brian's question, I was writing the standard backtracking procedure to find the permutations of 1..n taken r at a time. def permutations (n, r): A, Used = [None] * (r+1), [False] * (n+1) def choose (m): if m > r: print A[1:], else: for j in range (1, n+1): if not Used [j]: Used [j] = True A[m] = j choose (m+1) Used [j] = False choose (1) Thanks Logesh Pillay ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] declaring list in python
Logesh Pillay said unto the world upon 10/01/06 11:28 PM: > Hello list > > I want to declare a list of a specific size as global to some nested > function like so Hi Logesh, what problem are you trying to solve by doing this? Knowing that will help generate more useful answers, I suspect. > def foo (n): > A[] (of size n) > def foo1 > ... > > The only way I can think to declare the list is to use dummy values: > A = [0] * n > > A = [] * n doesn't work. [] * n = [] A = [None] * n would be a better way to created an n-placed list of dummy values in Python, I think. > I'd prefer not to use dummy values I have no use for. Is there any way? This is why knowing your problem would be helpful. Built-in Python data structures don't have size limitations that are declared when an instance of the data structure is created. (Python's not C.) There is no way (that I know of) to create an n-placed list save creating a list with n objects. So, I think you will have to either give up on not employing dummy values or give up on creating a list of a fixed length. You could subclass list to create a class with a max. and/or min. length, but I think knowing more about what you want to do would be helpful before getting into that :-) Best, Brian vdB ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
