Re: [Tutor] reduce with comprehension
Hi John, Everything is possible with list comprehensions! [x for z in a for y in z for x in y] [1, 2, 3, 31, 32, 4, 5, 6, 7, 71, 72, 8, 9] impressive! But in the general case can you do the same job as reduce in a list comprehension? ie apply an operation to the first two elements of a sequence and replace them with the result, then repeat until the list becomes a single value? I cannot think of a way to do that in the generaln case with a comprehension. I'd be interested to see if there is a recipe for that. Alan G ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
On 22/11/05, Alan Gauld [EMAIL PROTECTED] wrote: But in the general case can you do the same job as reduce in a list comprehension? ie apply an operation to the first two elements of a sequence and replace them with the result, then repeat until the list becomes a single value? Well ... probably not. I may have been exaggerating ever-so-slightly in my praise of list comprehensions :-) Although you may be able to cheat: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/204297 (but I'm pretty certain that this feature is on Guido's hitlist) Hmm... def f(x, y): ... return x + y ... arr = range(10) sum(arr) # Our target 45 tmp = [0] [f(x, y) for x in arr for y in [tmp[-1]] if tmp.append(f(x, y)) or True][-1] 45 Let's try some more... def f(x, y): ... return x*y ... arr = range(1, 10)# Don't want to include 0! reduce(f, arr) # Our target 362880 tmp = [1] [f(x, y) for x in arr for y in [tmp[-1]] if tmp.append(f(x, y)) or True][-1] 362880 print 'Magic!' Magic! -- John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
But in the general case can you do the same job as reduce in a list comprehension? def f(x, y): ... return x + y ... tmp = [0] [f(x, y) for x in arr for y in [tmp[-1]] if tmp.append(f(x, y)) or True][-1] 45 Let's try some more... def f(x, y): ... return x*y ... tmp = [1] [f(x, y) for x in arr for y in [tmp[-1]] if tmp.append(f(x, y)) or True][-1] 362880 OK, you've nearly convinced me that its possible but I think I still prefer reduce()! :-) Alan G. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
János Juhász said unto the world upon 2005-11-21 01:20: Hi, I can't imagine how this could be made with list comprehension. import operator a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) reduce(operator.add, a) # it makes a long list now ([1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]) When I make list comprehension, the list hierarchy is allways the same or deeper than the original hierarchy. But now it should be reduced with one level. [item for item in a] # the deepnes is the same [([1], [2], [3, 31, 32], [4]), ([5], [6], [7, 71, 72]), ([8], [9])] [(item, item) for item in a] # it is deeper with one level Is it possible to substitute reduce with comprehension anyway ? Yours sincerely, __ János Juhász Hi János, I think it is. a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) b = [] [b.extend(x) for x in a] [None, None, None] b [[1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]] But don't do that :-) Seems very poor form to me, as the desired operation is a side effect of the creation of the list via the comprehension. If, however, you are just trying to avoid reduce: b=[] for x in a: b.extend(x) b [[1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]] Notice, too, that these ways result in a list of lists, whereas yours provided a tuple of lists. HTH, Brian vdB ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) reduce(operator.add, a) # it makes a long list now ([1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]) When I make list comprehension, the list hierarchy is allways the same or [item for item in a] # the deepnes is the same [([1], [2], [3, 31, 32], [4]), ([5], [6], [7, 71, 72]), ([8], [9])] [(item, item) for item in a] # it is deeper with one level You are not using operator add anywhere in there. So you won't reduce anything! But reduce is tricky to reproduce using comprehensions because it involves combining two elements using an operation. And one of the elements is the running result. So count = 0 [count + item for item in alist] won't work because we never change count - and assignments aren't allowed inside a comprehension. There may be a clever way of doing it but I can't think of it and certainly not a way that would be as readable as reduce()! Do you have any particular need to avoid reduce? Or is it just curiosity? Alan g ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
János Juhász wrote: Hi, I can't imagine how this could be made with list comprehension. import operator a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) reduce(operator.add, a) # it makes a long list now Is it possible to substitute reduce with comprehension anyway ? Not that I know. Why are you trying to eliminate reduce()? If it is for compatibility with Python 3.0, PEP 3000 recommends replacing reduce() with an explicit loop. http://www.python.org/peps/pep-3000.html#built-in-namespace Kent -- http://www.kentsjohnson.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
On 21/11/05, János Juhász [EMAIL PROTECTED] wrote: I can't imagine how this could be made with list comprehension. import operator a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) reduce(operator.add, a) # it makes a long list now ([1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]) Everything is possible with list comprehensions! a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) [x for y in a for x in y] [[1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]] We can even go a level deeper! [x for z in a for y in z for x in y] [1, 2, 3, 31, 32, 4, 5, 6, 7, 71, 72, 8, 9] Just make sure your depth is consistent throughout. And remember that that single-line expression is hiding nested FOR loops! -- John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] reduce with comprehension
Hi John, thanks it. It is great. I looked for it, but I couldn't made it. I have tried it with wrong order: # I have tried it so [x for x in y for y in a] [[8], [8], [8], [9], [9], [9]] # that is wrong, # Instead of that you wrote [x for y in a for x in y] [[1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]] Yours sincerely, __ János Juhász [EMAIL PROTECTED] wrote on 2005.11.21 23:26:03: On 21/11/05, János Juhász [EMAIL PROTECTED] wrote: I can't imagine how this could be made with list comprehension. import operator a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) reduce(operator.add, a) # it makes a long list now ([1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]) Everything is possible with list comprehensions! a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9])) [x for y in a for x in y] [[1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9]] We can even go a level deeper! [x for z in a for y in z for x in y] [1, 2, 3, 31, 32, 4, 5, 6, 7, 71, 72, 8, 9] Just make sure your depth is consistent throughout. And remember that that single-line expression is hiding nested FOR loops! -- John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
