The aim is to find groups of increasing/monotonic numbers given a list of integers. Each item in the resulting group must be of a +1 increment from the previous item
Given an input:
x = [7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5]
I need to find groups of increasing numbers and achieve:
increasing_numbers = [(7,8,9,10), (0,1,2,3,4,5)]
And eventually also the number of increasing numbers:
len(list(chain(*increasing_numbers)))
And also the len of the groups:
increasing_num_groups_length = [len(i) for i in increasing_numbers]
I have tried the following to get the number of increasing numbers:
>>> from itertools import tee, chain
>>> def pairwise(iterable):
... a, b = tee(iterable)
... next(b, None)
... return zip(a, b)
...
>>> x = [8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6]
>>> set(list(chain(*[(i,j) for i,j in pairwise(x) if j-1==i])))
set([1, 2, 3, 4, 5, 6, 8, 9, 10, 11])
>>> len(set(list(chain(*[(i,j) for i,j in pairwise(x) if j-1==i]))))
10
But I'm unable to keep the order and the groups of increasing numbers.
How can I achieve the increasing_numbers
groups of integer tuples and also the increasing_num_groups_length
?
Also, is there a name for such/similar problem?
EDITED
I've came up with this solution but it's super verbose and I'm sure there's an easier way to achieve the increasing_numbers
output:
>>> from itertools import tee, chain
>>> def pairwise(iterable):
... a, b = tee(iterable)
... next(b, None)
... return zip(a, b)
...
>>> x = [8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6]
>>> boundary = iter([0] + [i+1 for i, (j,k) in enumerate(pairwise(x)) if j+1!=k] + [len(x)])
>>> [tuple(x[i:next(boundary)]) for i in boundary]
[(8, 9, 10, 11), (1, 2, 3, 4, 5, 6)]
Is there a more pythonic / less verbose way to do this?
Another input/output example:
[in]:
[17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35, 36, 40]
[out]:
[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14),
(28, 29, 30, 31, 32, 33, 34, 35, 36)]
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