The search algorithm for all divisors of the number

Day! When deciding on the network, the algorithm for finding all the dividers of the number. Well, that is, for eight it is necessary to issue [1,2,4,8], and not [2.2,2] - a list of divisors. I rewrote this Nanovo algorithm, I ask the "senior comrades" "to tell how to improve. If there is time))

   def   divisors  ( n ):   from  collections  import  counter ls = get_ls (n) # for n = 1568 -& gt; ls = [2, 2, 2, 2, 2, 7, 7]   Pairs =  dict  (couter (ls)) # {2: 5, 7: 2}    from  itertools  Import  product, Starmap  from  oparator  Import  mul  from  funstools  Import  reduce # list of all different divisions of the number   bass = [b  for  b  in  pairs.keys ()] # [2, 7]   # list of unique divisors to obtain the number   Powers = [I  for  i  in   range  (k+ 1 )]  for  k  in  pairs.values ​​()]  # generating all sets for obtaining dividers of the original number   multi = product (*powers)   zip  (bases, power)  for  power  in  multi)  # sets of numbers, sets which you need to change to obtain a divider   rezzz = (Starmap ( for  row  in  wrk)  # retu List of all dividers    retu  [reduce (mul, i)  in  reziz]  

for example Divisors (1568) Retus [1, 7, 49, 2, 14, 98, 4, 28, 196, 8, 56, 392, 16, 112, 784, 32, 224, 1568]

get_ls (n) gives a list of decomposition of the number to the work of dividers

for example:

   deg   get_ls  ( n ):  " "" "" "" "decompose the number" "" "" "  #result = [1]    Result = []  i =  2    when  i*i & lt; = n:   if  n % i ==  0 :  n n n n n n n n n n n. // = i result.ppend (i)   else :  i +=  1    if  n & gt;  1 :  result.ppend (n)   retu  result    
 what can be improved? ACCUMELATE from Itertools. Well, in general, according to the algorithm. 
  
 it is clear that the improvements of the type 
  
   retu  [red, i)  for  i  in  (Starmap ( pow , row)  for  row  in  ( zip  (bases, power)  for  power  in  product (*powers)))]  
  
 is not interesting.