Rasika Sapate via Tutor wrote: > Dear Python group, > I had written following code. > > super = [] > sub = [""]*3 > other = ["a","b","c","d"] > sub[0] = "hi" > sub[1] = "hello" > for item in other: > sub[2] = item > super.append(sub) > for item in super: > print item > > > Output : > ['hi', 'hello', 'd'] > ['hi', 'hello', 'd'] > ['hi', 'hello', 'd'] > ['hi', 'hello', 'd'] > > > Expected output: > ['hi', 'hello', 'a] > ['hi', 'hello', 'b'] > ['hi', 'hello', 'c'] > ['hi', 'hello', 'd'] > > > Is there anything wrong in this code or any feature of python?
When you replace an item in a list you don't magically copy that list, so in a for loop >>> outer = [] >>> inner = [1, 2, 3] >>> for item in "abc": ... inner[2] = item ... outer.append(inner) ... >>> You append the same list three time to outer >>> outer[0] is inner True >>> outer[2] is inner True and inner[2] holds the last value you assigned to it. What you want is a new list on each iteration, and one way to get that when the inner list is long is list concatenation with the + operator: >>> outer = [] >>> inner_start = [1, 2] >>> for item in "abc": ... outer.append(inner_start + [item]) ... >>> outer[0] is outer[2] False >>> outer [[1, 2, 'a'], [1, 2, 'b'], [1, 2, 'c']] In your case with only three items I recommend that you use a list literal: >>> outer = [] >>> for item in "abc": ... outer.append([1, 2, item]) ... >>> outer [[1, 2, 'a'], [1, 2, 'b'], [1, 2, 'c']] Python has a concise way to spell those append-only loops which is called list comprehension and looks like this: >>> [[1, 2, item] for item in "abc"] [[1, 2, 'a'], [1, 2, 'b'], [1, 2, 'c']] _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
