# List of prime factors with repetition This post is a wiki. Anyone with karma >750 is welcome to improve it.

Is there a simple command on Sage wich gives, in place of factor(720)= 2^4 * 3^2 * 5, the list [2, 2, 2, 2, 3, 3, 5] of prime factors with repetition?

edit retag close merge delete

Sort by » oldest newest most voted

Some potentially useful variations, building on previous answers and comments by @calc314 and @logomath.

The function factor returns a factorization.

sage: F = factor(2016)
sage: F
2^5 * 3^2 * 7


This secretly wraps a list of pairs (prime, multiplicity), and you can get this list.

sage: list(F)
[(2, 5), (3, 2), (7, 1)]


You can also iterate through this list of pairs directly.

sage: for pm in F: print pm
(2, 5)
(3, 2)
(7, 1)


This means you can use list comprehension in smart ways.

sage: [f for f in F for _ in range(f)]
[2, 2, 2, 2, 2, 3, 3, 7]


Same as above, but calling (p, m) the pairs.

sage: [p for (p, m) in F for _ in range(m)]
[2, 2, 2, 2, 2, 3, 3, 7]


Getting a list of repetitions of each prime.

sage: [([p] * m) for (p, m) in F]
[[2, 2, 2, 2, 2], [3, 3], ]


Flattening it to get the list of prime factors with multiplicity.

sage: flatten([([p] * m) for (p, m) in F])
[2, 2, 2, 2, 2, 3, 3, 7]


Using add to sum lists starting from the empty list (included as an optional parameter to add).

sage: add([[p] * m for (p, m) in F], [])
[2, 2, 2, 2, 2, 3, 3, 7]


(Edit.) Same but using an iterator instead of a list.

sage: add(([p] * m for (p, m) in F), [])
[2, 2, 2, 2, 2, 3, 3, 7]

more

Thank you, very interesting how sage is full of options. As I am using it for very large n, I remarked that the version " add([[p] ..." is nearly tenfold slower than the others (n near to 10000 digits).

I don't know of a direct command. Here is one option, though.

F= factor(2016)
result=[f for f in F for i in range(f)]
result


You may wish to look up the factor command for more details and examples.

more

I just add here a lambda-version of it for reference:

result = lambda n: [f for f in factor(n) for i in range(f)]; result(2016)