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
[Tutor] declaring list in python
Hello list I want to declare a list of a specific size as global to some nested function like so 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 = [] I'd prefer not to use dummy values I have no use for. Is there any way? Thanks Logesh Pillay ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] permutations using list comprehensions
Dear list I was really impressed by this version of quicksort:- def qsort (t): if len (t) == 0: return [] else: return qsort([x for x in t[1:] if x = t[0]]) + [t[0]] + qsort([x for x in t[1:] if x t[0]]) I'm trying to produce something of a similar structure to generate the permutations of a sequence by translating the ffg bit of Haskell:- perms [] = [[]] perms xs = [ x : ps | x - xs , ps - perms ( xs\\[x]) ] '\\' applies to lists means elements of the first list excluding any elements in common with the second list. The best I've been able to do is pretty obvious:- def perms (t): if len (t) == 0: return [] else: return [[x] + perms ([y for y in t if y x]) for x in t] Needless to say, it doesn't work. It does produce a list of lists but they are so horribly nested that I've no idea whether I am getting close. I've also tried adding and removing square brackets in various combinations. Any ideas? Logesh ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] variable scope in nested functions
Thanks Kent for your reply. You said This is a limitation of Python's nested scopes. You can't assign to a variable in an enclosing scope. One way to work around this is to use a mutable value like a list in the enclosing scope: def foo (n): counter = [0] def choose (i): if (solution found): counter[0] += 1 print counter[0], (solution) else choose (i+1) choose (1) I can't help thinking that this is a little unsatisfactory and suggests (with all due respect) an unfinished quality to Python. Consider a program I've just translated into python from my version in C. def erdos(n): #Explores Erdos theorem that any found, r, s = False, 1, [0]*1000#integer can be expressed as +-1.1 def choose (d, k): #+-2.2...+-r.r for some r and some for i in -1, 1: #combinations of +'s and -'s s[d] = i if d r: choose (d+1, k - i*d*d) elif k == i*d*d: found = True printout () def printout (): print n, '=', for i in range (1, r+1): if s[i] == 1: print '+', i,'.',i, else: print '-',i,'.',i, print '\n' while not found: choose(1, n) r += 1 The program is supposed to return to the python prompt as soon as it finds solution(s) at the smallest width r. I entered it exactly as it is expecting the same sort of error message for the boolean variable 'found'. There was none. Instead python simply fails to perform 'found = True' in choose() and it just keeps running (incidentally demonstrating the other part of this Erdos theorem that the no. can be so expressed in an infinite number of ways). Why the inconsistency in handling enclosing scope variables of type 'int' and 'bool'? Also, it can't be desirable that the interpreter effectively ignores a line of code without warning. Logesh ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] variable scope in nested functions
Hello list I am having trouble with a variable to act as a counter in a nested recursive function which will only occasionally find an answer. Something like a static variable in C. Why does this sort of thing not work? def foo (n): counter = 0 def choose (i): if (solution found): counter += 1 print counter, (solution) else choose (i+1) choose (1) I get an error message that counter is referenced before being declared. Declaring counter as global does not help either Incidentally how does one nest comments in python. I want to comment out part of a line. Is it just Kate my editor which won't allow it Thanks Logesh ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] quicksort using list comprehension
I'm trying to program quicksort using list comprehension. The following gives me a type mismatch error for +. def qsort (t): if len (t) 1: return qsort([x for x in t[1:] if x = t[0]]) + [t[0]] + qsort([x for x in t[1:] if x t[0]]) I know this sounds strange but I have and idea it (or something similar) worked when I last tried Python a yr or so ago. An earlier version of Python? Logesh Pillay ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
