What is the most efficient way to create list of the same number with n elements?
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Are you really asking for the "most efficient way", or would any correct way suffice?Robᵩ– Robᵩ2013-03-31 00:14:14 +00:00Commented Mar 31, 2013 at 0:14
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1 Answer
number = 1
elements = 1000
thelist = [number] * elements
>>> [1] * 10
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
NB: Don't try to duplicate mutable objects (notably lists of lists) like that, or this will happen:
In [23]: a = [[0]] * 10
In [24]: a
Out[24]: [[0], [0], [0], [0], [0], [0], [0], [0], [0], [0]]
In [25]: a[0][0] = 1
In [26]: a
Out[26]: [[1], [1], [1], [1], [1], [1], [1], [1], [1], [1]]
If you are using numpy, for multidimensional lists numpy.repeat is your best bet. It can repeat arrays of all shapes over separate axes.
4 Comments
DSM
The question is tagged
numpy too, so maybe mention numpy.repeat(1, 10)? (Although it'll be slower than multiplying a list until n is really big.)
Gareth Latty
Note that this may have unexpected results (depending on how much you know of how Python variables work) with mutable types - it produces a list of references to the same object.
Jaime
To honor the numpy tag,
a = np.empty((elements,), dtype=np.int); a.fill(number) is much faster than [number] * elements for higher values of elements. But the return is not a real list.
Warren Weckesser
@Jaime: Also note that
empty + fill is faster than repeat