I wonder if there is a more elegant way to do the following. For example with list comprehension.
Consider a simple list :
l = ["a", "b", "c", "d", "e"]
I want to duplicate each elements n times. Thus I did the following :
n = 3
duplic = list()
for li in l:
duplic += [li for i in range(n)]
At the end duplic is :
['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'e', 'e', 'e']
You can use
duplic = [li for li in l for _ in range(n)]
This does the same as your code. It adds each element of l (li for li in l) n times (for _ in range n).
You can use:
l = ["a", "b", "c", "d", "e"]
n=3
duplic = [ li for li in l for i in range(n)]
Everytime in python that you write
duplic = list()
for li in l:
duplic +=
there is a good chance that it can be done with a list comprehension.
Try this:
l = ["a", "b", "c", "d", "e"]
print sorted(l * 3)
Output:
['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'e', 'e', 'e']
sorted function will mess up everything.
from itertools import chain
n = 4
>>>list(chain.from_iterable(map(lambda x: [x]*n,l)))
['a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'c', 'd', 'd', 'd', 'd', 'e', 'e', 'e', 'e']
In [12]: l
Out[12]: ['a', 'b', 'c', 'd', 'e']
In [13]: n
Out[13]: 3
In [14]: sum((n*[item] for item in l), [])
Out[14]: ['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'e', 'e', 'e']
