Repeat a sequence of a list [closed]

Question

Problem

Given the following list:

l=[1,2,3,4,5,6]

I want to repeat the n first elements of the list r times, and after that, do the same for the next n elements of the list, and so on...

For example, for n=2 and r=2:

result=[1,2,1,2,3,4,3,4,5,6,5,6]

And for n=3 and r=2:

result=[1,2,3,1,2,3,4,5,6,4,5,6]

My clumsy attempt

l=[1,2,3,4,5,6]
n=3
r=2
l_new=[]
result=[]
for i in range(0,len(l),n):
    reduced=l[i:(n+i)]
    l_new.append(reduced)

for elem in l_new:
    elem_new=elem*r
    result+=(elem_new)

How can I improve this code? What is a more clean/efficient/good practice way of doing this?

Answer 1

def flatten(arr):
    return [item for sublist in arr for item in sublist]

def flatmap(f, arr):
    return flatten(map(f, arr))

def chunk(arr, r):
    return [arr[i:i+r] for i in range(0,len(arr), r)]

def repeat_n_r_times(arr, n, r):
  return flatmap(lambda x: x * n, chunk(arr, r))

Answer 2

you can achieve it cleanly as following.

l=[1,2,3,4,5,6]

def repeat_n_r_times(l, n, r):
    list_breakdown = [l[i:i+n] for i in range(0, len(l), n)]
    #make copies r times 
    lst = [lst*r for lst in list_breakdown]
    # merge all sublist
    return [item for sublist in lst for item in sublist]

print(repeat_n_r_times(l, 3, 2))
# OUTPUT: [1,2,3,1,2,3,4,5,6,4,5,6]
print(repeat_n_r_times(l, 2, 2))
# OUTPUT: result=[1,2,1,2,3,4,3,4,5,6,5,6]