Ken G. wrote: > I have been working on this problem for several days and I am not making > any progress. I have a group of 18 number, in ascending order, within a > list. They ranged from 1 to 39. Some numbers are duplicated as much as > three times or as few as none. > > I started with one list containing the numbers. For example, they are > listed as like below: > > a = [1, 2, 3, 3, 4] > > I started off with using a loop: > > for j in range (0, 5): > x = a[0] # for example, 1 > > How would I compare '1' with 2, 3, 3, 4? > > Do I need another duplicated list such as b = a and compare a[0] with > either b[0], b[1], b[2], b[3], b[4]? > > Or do I compare a[0] with a[1], a[2], a[3], a[4]? > > In any event, if a number is listed more than once, I would like to know > how many times, such as 2 or 3 times. For example, '3' is listed twice > within a list.
I haven't seen a solution that takes advantage of the fact that the numbers are sorted, so here's one: >>> items = [1, 2, 3, 3, 4] >>> value = items[0] >>> n = 0 >>> result = [] >>> for item in items: ... if value != item: ... print value, "-->", n ... value = item ... n = 1 ... else: ... n += 1 ... 1 --> 1 2 --> 1 3 --> 2 >>> print value, "-->", n 4 --> 1 Python has a cool feature called generator that allows you to encapsulate the above nicely: >>> def count_runs(items): ... it = iter(items) ... value = next(it) ... n = 1 # next consumes one item ... for item in it: # the for loop starts with the second item ... if value != item: ... yield value, n ... value = item ... n = 1 ... else: ... n += 1 ... yield value, n ... >>> for v, n in count_runs([1, 1, 1, 2, 2, 3, 4, 4, 4, 4]): ... print v, "-->", n ... 1 --> 3 2 --> 2 3 --> 1 4 --> 4 Finally, there is itertools.groupby() which allows you to simplify the implementation count_runs(): >>> from itertools import groupby >>> def count_runs2(items): ... for key, group in groupby(items): ... yield key, sum(1 for _ in group) ... >>> for v, n in count_runs2([1, 1, 1, 2, 2, 3, 4, 4, 4, 4]): ... print v, "-->", n ... 1 --> 3 2 --> 2 3 --> 1 4 --> 4 Peter _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
