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R� 4 t|R� t}]P� ! 4 PT tR� t�R� t�R� t�R� t�R� t�R� t�R�R/R� lt�R# ]4 d ]2t3 EL�i ; i ]7 d RA t6 EL�i ; i ]7 d RCt: EL�i ; i ]7 d ]ItJ ELUi ; i)�a Imported from the recipes section of the itertools documentation.
All functions taken from the recipes section of the itertools library docs
[1]_.
Some backward-compatible usability improvements have been made.
.. [1] http://docs.python.org/library/itertools.html#recipes
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triplewise�unique�unique_everseen�unique_justseenT��strict)�sumprodc � � \ W4 # �N)r, )�x�ys &&�8/usr/lib/python3/dist-packages/more_itertools/recipes.py�<lambda>ra l s � �J�q�,� )�heappush_max�heappushpop_maxFc �* � \ \ W4 4 # )z�Return first *n* items of the *iterable* as a list.
>>> take(3, range(10))
[0, 1, 2]
If there are fewer than *n* items in the iterable, all of them are
returned.
>>> take(10, range(3))
[0, 1, 2]
)�listr )�n�iterables &&r` rR rR x s � � ��x�#�$�$rb c �, � \ V \ V4 4 # )a� Return an iterator over the results of ``func(start)``,
``func(start + 1)``, ``func(start + 2)``...
*func* should be a function that accepts one integer argument.
If *start* is not specified it defaults to 0. It will be incremented each
time the iterator is advanced.
>>> square = lambda x: x ** 2
>>> iterator = tabulate(square, -3)
>>> take(4, iterator)
[9, 4, 1, 0]
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4 R4 # \ d \ \ YR7 4 u # i ; i)zsReturn an iterator over the last *n* items of *iterable*.
>>> t = tail(3, 'ABCDEFG')
>>> list(t)
['E', 'F', 'G']
N��maxlen)�lenr �max� TypeError�iterr )rg rh �sizes && r` rQ rQ � sK � �8��8�}�� �h��A�t�x� 0�$�7�7�� � /��E�(�-�.�.�/�s �* � A
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c �X � Vf \ V ^ R7 R# \ \ WV4 R4 R# )a� Advance *iterable* by *n* steps. If *n* is ``None``, consume it
entirely.
Efficiently exhausts an iterator without returning values. Defaults to
consuming the whole iterator, but an optional second argument may be
provided to limit consumption.
>>> i = (x for x in range(10))
>>> next(i)
0
>>> consume(i, 3)
>>> next(i)
4
>>> consume(i)
>>> next(i)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
If the iterator has fewer items remaining than the provided limit, the
whole iterator will be consumed.
>>> i = (x for x in range(3))
>>> consume(i, 5)
>>> next(i)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
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