On Wednesday, September 21, 2016 at 4:14:10 PM UTC+2, Daiyue Weng wrote: > Hi, first of all, let me rephase the problem. > > For an arbitrary list of integers (the integers in the list are not > necessary to be sequential), e.g. [1,2,3,6,8,9,10,11,13], > > if a set of consecutive integers having a difference of 1 between them, put > them in a list, i.e. there are two such lists in this example, > > [1,2,3], > > [8,9,10,11], > > and then put such lists in another list (i.e. [[1,2,3], [8,9,10,11]]). Put > the rest integers (non-sequential) in a separated list, i.e. > > `[6, 13]`. > > Note that, the problem only considers sequential integers with step_size = > 1. > > I tried to use itertools.groupby and itertools.count, > > from itertools import groupby, count > lst = [1,2,3,6,8,9,10,11,13] > c = count() > result = [list(g) for i, g in groupby(lst, key=lambda x: x-next(c))] > > but the result is close to the requirement shown as a list of lists, > > [[1, 2, 3], [6], [8, 9, 10, 11], [13]] > > but it didn't separate sequential lists from non-sequential ones. > Also, I couldn't find a way to put [6] and [13] in a single list. > > I have also tried to separate sequential lists from non-sequential ones, > > result = [list(g) for i, g in groupby(lst, key=lambda x: x-next(c) == 1)] # > tried to extract [1,2,3] and [8,9,10,11] from the list > > or > > result = [list(g) for i, g in groupby(lst, key=lambda x: x-next(c) > 1)] # > tried to extract [6] and [13] from the list > > but they all ended up with > > [[1, 2, 3], [6, 8, 9, 10, 11, 13]] > > So two questions here, > > 1. How does itertools.groupby key function work in my case? > > 2. How to improve the program to achieve my goals? > > > Many thanks > > On 21 September 2016 at 00:19, John Pote <johnp...@jptechnical.co.uk> wrote: > > > On 20/09/2016 12:52, Daiyue Weng wrote: > > > > Hi, I have a list numbers say, > >> > >> [1,2,3,4,6,8,9,10,11] > >> > >> First, I want to calculate the sum of the differences between the numbers > >> in the list. > >> > > At least for this first part a little pencil, paper and algebra yields a > > simple formula of constant and minimal calculation time. I had an intuitive > > guess and wrote down the sum of differences for a couple of examples, > > [1, 2, 5] => 4 > > [9, 11, 17, 19] => 10 > > It works for negative differences as well, > > [1, 2, 5, 1] => 0 > > The trick is to spot the relation between the sum of differences and the > > numbers in the list. A few lines of algebra then gave a very simple formula. > > > > As for the rest it's down to code as others have hinted at. > > > > Second, if a sequence of numbers having a difference of 1, put them in a > >> list, i.e. there are two such lists, > >> > >> [1,2,3] > >> > >> [8,9,10,11] > >> > >> and also put the rest numbers in another list, i.e. there is only one such > >> list in the example, > >> > >> [6]. > >> > >> Third, get the lists with the max/min sizes from above, i.e. in this > >> example, the max list is, > >> > >> [8,9,10,11] > >> > >> min list is > >> > >> [1,2,3]. > >> > >> What's the best way to implement this? > >> > >> cheers > >> > > > > -- > > https://mail.python.org/mailman/listinfo/python-list > > On Wednesday, September 21, 2016 at 4:14:10 PM UTC+2, Daiyue Weng wrote: > Hi, first of all, let me rephase the problem. > > For an arbitrary list of integers (the integers in the list are not > necessary to be sequential), e.g. [1,2,3,6,8,9,10,11,13], > > if a set of consecutive integers having a difference of 1 between them, put > them in a list, i.e. there are two such lists in this example, > > [1,2,3], > > [8,9,10,11], > > and then put such lists in another list (i.e. [[1,2,3], [8,9,10,11]]). Put > the rest integers (non-sequential) in a separated list, i.e. > > `[6, 13]`. > > Note that, the problem only considers sequential integers with step_size = > 1. > > I tried to use itertools.groupby and itertools.count, > > from itertools import groupby, count > lst = [1,2,3,6,8,9,10,11,13] > c = count() > result = [list(g) for i, g in groupby(lst, key=lambda x: x-next(c))] > > but the result is close to the requirement shown as a list of lists, > > [[1, 2, 3], [6], [8, 9, 10, 11], [13]] > > but it didn't separate sequential lists from non-sequential ones. > Also, I couldn't find a way to put [6] and [13] in a single list. > > I have also tried to separate sequential lists from non-sequential ones, > > result = [list(g) for i, g in groupby(lst, key=lambda x: x-next(c) == 1)] # > tried to extract [1,2,3] and [8,9,10,11] from the list > > or > > result = [list(g) for i, g in groupby(lst, key=lambda x: x-next(c) > 1)] # > tried to extract [6] and [13] from the list > > but they all ended up with > > [[1, 2, 3], [6, 8, 9, 10, 11, 13]] > > So two questions here, > > 1. How does itertools.groupby key function work in my case? > > 2. How to improve the program to achieve my goals? > > > Many thanks > > On 21 September 2016 at 00:19, John Pote <johnp...@jptechnical.co.uk> wrote: > > > On 20/09/2016 12:52, Daiyue Weng wrote: > > > > Hi, I have a list numbers say, > >> > >> [1,2,3,4,6,8,9,10,11] > >> > >> First, I want to calculate the sum of the differences between the numbers > >> in the list. > >> > > At least for this first part a little pencil, paper and algebra yields a > > simple formula of constant and minimal calculation time. I had an intuitive > > guess and wrote down the sum of differences for a couple of examples, > > [1, 2, 5] => 4 > > [9, 11, 17, 19] => 10 > > It works for negative differences as well, > > [1, 2, 5, 1] => 0 > > The trick is to spot the relation between the sum of differences and the > > numbers in the list. A few lines of algebra then gave a very simple formula. > > > > As for the rest it's down to code as others have hinted at. > > > > Second, if a sequence of numbers having a difference of 1, put them in a > >> list, i.e. there are two such lists, > >> > >> [1,2,3] > >> > >> [8,9,10,11] > >> > >> and also put the rest numbers in another list, i.e. there is only one such > >> list in the example, > >> > >> [6]. > >> > >> Third, get the lists with the max/min sizes from above, i.e. in this > >> example, the max list is, > >> > >> [8,9,10,11] > >> > >> min list is > >> > >> [1,2,3]. > >> > >> What's the best way to implement this? > >> > >> cheers > >> > > > > -- > > https://mail.python.org/mailman/listinfo/python-list > >

## Advertising

And here is a straightforward one without any helpers: l = [1, 2, 3, 6, 8, 9, 10, 11, 13, 17, 19, 20] n = len(l) seq_list = None non_seq_list = [] all_seq_list = [] diff = [] for i, li in enumerate(l): # Handle edge case if i == n - 1: if l[-1] - l[-2] == 1: seq_list.append(l[-1]) all_seq_list.append(seq_list) else: non_seq_list.append(l[-1]) break d = l[i+1] - li diff.append(d) if d == 1: if seq_list is None: seq_list = [] seq_list.append(li) else: if seq_list is not None: seq_list.append(li) all_seq_list.append(seq_list) seq_list = None else: non_seq_list.append(li) print('input', l) print('sum of differences', sum(diff)) print('sequential lists', all_seq_list) print('non-sequential entries', non_seq_list) length_seq_list = sorted([(len(s), s) for s in all_seq_list]) print('shortest', length_seq_list[0]) print('longest', length_seq_list[-1]) This gives: input [1, 2, 3, 6, 8, 9, 10, 11, 13, 17, 19, 20] sum of differences 19 sequential lists [[1, 2, 3], [8, 9, 10, 11], [19, 20]] non-sequential entries [6, 13, 17] shortest (2, [19, 20]) longest (4, [8, 9, 10, 11]) -- https://mail.python.org/mailman/listinfo/python-list