from collections import deque
n = 10000
%timeit sum(1 for _ in range(n))                      # 2.5
%timeit deque(enumerate(range(n), 1), maxlen=1)[0][0] # 2.6
>>>
1000 loops, best of 3: 439 µs per loop
1000 loops, best of 3: 289 µs per loop

from more_itertools import divide, always_iterable, adjacent, groupby_transform, context
[tuple(i) for i in divide(3, range(10))]
>>>
[(0, 1, 2, 3), (4, 5, 6), (7, 8, 9)]

always_iterable(None)
>>>
()

always_iterable(1)
>>>
(1,)

always_iterable([1,3,5])
>>>
[1, 3, 5]

list(adjacent(lambda x: x==3, range(7)))
>>>
[(False, 0),
 (False, 1),
 (True, 2),
 (True, 3),
 (True, 4),
 (False, 5),
 (False, 6)]

list(adjacent(lambda x: x==3, range(7), distance=2)) #Bis zu 2 Unterschiede vorher und nachher
>>>
[(False, 0),
 (True, 1),
 (True, 2),
 (True, 3),
 (True, 4),
 (True, 5),
 (False, 6)]

[(i,list(j)) for i,j in groupby_transform([1,1,3,2,2,2],
                                          lambda x: f'<{x}>', #Schlüsselkonvertierung
                                          lambda x: x*10)]    #Wertumwandlung
>>>
[('<1>', [10, 10]), ('<3>', [30]), ('<2>', [20, 20, 20])]

with open(Dateiname) as fp:
    print(fp.read())

↓ Sie können so schreiben

print(*[fp.read() for fp in context(open(Dateiname))])

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