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
EDIT:
Here's a code-golf solution (142 characters):
def f(x):s=[0]+[i for i in range(1,len(x)) if x[i]!=x[i-1]+1]+[len(x)];return [x[j:k] for j,k in [s[i:i+2] for i in range(len(s)-1)] if k-j>1]
Expanded version:
def igroups(x):
s = [0] + [i for i in range(1, len(x)) if x[i] != x[i-1] + 1] + [len(x)]
return [x[j:k] for j, k in [s[i:i+2] for i in range(len(s)-1)] if k - j > 1]
Commented version:
def igroups(x):
# find the boundaries where numbers are not consecutive
boundaries = [i for i in range(1, len(x)) if x[i] != x[i-1] + 1]
# add the start and end boundaries
boundaries = [0] + boundaries + [len(x)]
# take the boundaries as pairwise slices
slices = [boundaries[i:i + 2] for i in range(len(boundaries) - 1)]
# extract all sequences with length greater than one
return [x[start:end] for start, end in slices if end - start > 1]
Original solution:
Not sure whether this counts as "pythonic" or "not too verbose":
def igroups(iterable):
items = iter(iterable)
a, b = None, next(items, None)
result = [b]
while b is not None:
a, b = b, next(items, None)
if b is not None and a + 1 == b:
result.append(b)
else:
if len(result) > 1:
yield tuple(result)
result = [b]
print(list(igroups([])))
print(list(igroups([0, 0, 0])))
print(list(igroups([7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5])))
print(list(igroups([8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6])))
print(list(igroups([9, 1, 2, 3, 1, 1, 2, 3, 5])))
Output:
[]
[]
[(7, 8, 9, 10), (0, 1, 2, 3, 4, 5)]
[(8, 9, 10, 11), (1, 2, 3, 4, 5, 6)]
[(1, 2, 3), (1, 2, 3)]