CODEWITHSUNDEEP
python
user3378649
user3378649

Reputation: 5354

List of Lists as result in list comprehension

I am trying to find a solution for a problem where I have to loop in a list for every element.

This is not the real problem that I am trying to solve, but I am using a simple example to illustrate the problem and what I want to understand. 

aList= [3, 4, 5, 6, 8, 9, 10,12]

I should regroup number divisible by each others.

result should give: 

result = [[3], [4], [5], [6,3], [4,8],[3,9], [5,10], [3,4,6,12]]

I use this function:

def divisible(x,y):
 if x%y== 0: 
   return True
 else: 
   return False

well, to solve this problem using two loops, we can use: 

globaList= []

for x in aList: 
  internalist=[]
  internalist.append(x)
  for y in aList:
    if divisible(x,y): 
       internalist.append(y)
    globaList.append(internalist)

I tried to write this double-loops in list comprehension, but didn't know how to make better way. 

result= [[x for x in aList ] for y in aList if divisible(x,y) ]

Upvotes: 3

Views: 72

Answers (2)

Joohwan
Joohwan

Reputation: 2512

You don't really need a helper function divisible:

aList= [3, 4, 5, 6, 8, 9, 10,12]
print [[x for x in aList if y % x == 0] for y in aList]

If you really want to use a helper function you can make that more succinct by going: 

def divisible(x,y):
    return  x % y == 0

Upvotes: 1

Cory Kramer
Cory Kramer

Reputation: 117971

def divisible(x,y):
    if x%y== 0: 
        return True
    else: 
        return False

aList= [3, 4, 5, 6, 8, 9, 10,12]

>>> [[x for x in aList if divisible(y,x)] for y in aList]
[[3], [4], [5], [3, 6], [4, 8], [3, 9], [5, 10], [3, 4, 6, 12]]

Upvotes: 3

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