在 python 列表推导中解包元组（不能使用 *-operator）

Question

I am trying to create a list based on another list, with the same values repeated 3 times consecutively.

At the moment, I am using:

>>> my_list = [ 1, 2 ]
>>> three_times = []
>>> for i in range( len( my_list ) ):
...   for j in range( 3 ):
...     three_times.append( my_list[ i ] )
...
>>> print three_times
[1, 1, 1, 2, 2, 2]

But I would like to do it using a more Pythonic way, such as:

>>> my_list = [ 1, 2 ]
>>> three_times = []
>>> three_times = [ (value,) * 3 for value in my_list ]
>>> print three_times
[(1, 1, 1), (2, 2, 2)]

However, I cannot find a way to unpack the tuples.

Something like three_times = [ *( (value,) * 3 ) for value in my_list ] would be perfect for unpacking the tuples but this is not a correct syntax.

Answer 1

You can't use * iterable unpacking in a list comprehension, that syntax is only available in calls, and in Python 3, when using assignments.

If you want to use a list comprehension, just put your for loops in series; you do want to access the values from my_list directly rather than generate indices though:

[v for v in my_list for _ in range(3)]

Answer 2

Accepted answer is correct, but I made some efficiency tests, so sharing it for passers-by.

Summary: Use chain.from_iterable for ~x2 speed improvement over list comprehension. And use np.repeat for ~x6 speed improvement if you don't mind importing numpy , but don't use np.repeat if eventually converting back to list .

In [1]: from itertools import chain
   ...: import numpy as np
   ...: 
   ...: def nested_list_comprehension(seq, repeats):
   ...:     return [v for v in seq for _ in range(repeats)]
   ...: 
   ...: def chain_from_iterable_tuple(seq, repeats):
   ...:     return list(chain.from_iterable((v,) * repeats for v in seq))
   ...: 
   ...: def chain_from_iterable_list(seq, repeats):
   ...:     return list(chain.from_iterable([v] * repeats for v in seq))
   ...: 
   ...: def numpy_repeat_list(seq, repeats):
   ...:     return list(np.repeat(seq, repeats))
   ...: 
   ...: def numpy_repeat(seq, repeats):
   ...:     return np.repeat(seq, repeats)

In [2]: seq = list(range(1000))
   ...: repeats = 100

In [3]: assert (
   ...:     nested_list_comprehension(seq, repeats)
   ...:     == chain_from_iterable_tuple(seq, repeats)
   ...:     == chain_from_iterable_list(seq, repeats)
   ...:     == numpy_repeat_list(seq, repeats)
   ...: )

In [4]: %timeit nested_list_comprehension(seq, repeats)
   ...: %timeit chain_from_iterable_tuple(seq, repeats)
   ...: %timeit chain_from_iterable_list(seq, repeats)
   ...: %timeit numpy_repeat_list(seq, repeats)
   ...: %timeit numpy_repeat(seq, repeats)
1.53 ms ± 2.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
814 µs ± 3.79 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
842 µs ± 2.02 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
3.65 ms ± 22.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
268 µs ± 1.44 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)