重复整数列表

Question

What is the most pythonic way to create a list of given size N= l*k where l is the number of different symbols (integers for simplicity) and k is the subsequence length like this:

N=12 , l=4 , k=3

[ 0,0,0, 1,1,1, 2,2,2, 3,3,3 ]

or this for example N=15 l=3 , k=5 :

[ 0,0,0,0,0, 1,1,1,1,1, 2,2,2,2,2 ]

this function should be called very often so speed is desirable.

Answer 1

Using numpy you can do this:

In [23]: import numpy as np

In [26]: a=np.arange(3).repeat(5)

In [27]: a
Out[27]: array([0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2])

or python builtin:

In [29]: [l for l in range(3) for k in range(5)]
Out[29]: [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2]

Answer 2

>>> l=3
>>> k=5
>>> [e for i in range(l) for e in [i]*k]
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2]

Answer 3

I like this lazy version (it returns an iterator, not a list, and you can generate values out of it when you want).

l, k = 3, 5
itertools.chain.from_iterable(itertools.repeat(i, k) for i in xrange(l))

It outputs this:

list(itertools.chain.from_iterable(itertools.repeat(i, k) for i in xrange(l)))
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2]

Answer 4

Perhaps with itertools.repeat and itertools.chain.from_iterable ?

>>> from itertools.import repeat, chain
>>> k = 3
>>> l = 4
>>> list(chain.from_iterable(list(repeat(x, k)) for x in xrange(l))
[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3]

Answer 5

>>> sum([[x]*3 for x in xrange(4)], [])
[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3]

Putting it into a function:

def combine(l, k):
   return sum([[x]*k for x in xrange(l)], [])

Answer 6

import itertools

l = 3
k = 5

print(list(itertools.chain(*[[i] * k for i in range(l)])))

[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2]