Python找到主要因素

Question

So I need to find the prime factors of a number using a python program and I am able to find all the factors using:

def primeFactors(n):
    list = []
    for x in range(2,n//2):
       if n % x == 0:
           list.append(x)
return list

But I don't understand how I'm suppose to make the program ignore factors that are a multiple of the prime factors.

I found the following code:

def primes(n):
divisors = [ d for d in range(2,n//2+1) if n % d == 0 ]
return [ d for d in divisors if \
         all( d % od != 0 for od in divisors if od != d ) ]

But I don't actually understand what it does, and this is for an assignment, so I can't just copy and paste.

So I was wondering if someone could lead me in the right direction of what I'm suppose to do. Also I can't use any fancy functions it has to be done mainly using the built in stuff like loops and lists and basic math. And I'm using python 2.7 not 3.0.

Answer 1

Divisors gets all of the possible divisors, same as you have done in your code. The value returned by

[ d for d in divisors if \
         all( d % od != 0 for od in divisors if od != d ) ]

Keeps a divisor d if for all other possible divisors d is not divisible by any other divisor. all returns true if all values passed to it are true, and the expression inside all simply checks if a given divisor d is not a divisor of any other divisor od in your list. In this way the final list of values returned only includes the divisors that are not a multiple of the factors already present.

Answer 2

def primes(n):
    divisors = [ d for d in range(2,n//2+1) if n % d == 0 ]
    return [ d for d in divisors if \
         all( d % od != 0 for od in divisors if od != d ) ]

Use an example, say n = 12 divisors will be the following list [2, 3, 4, 6]

Now the return part is a bit tricky and for good reason, one shouldn't be abusing list comprehension in such a way.

So lets turn this part into an actual loop:

ans = []
for d in divisors:
    for od in divisors:
        if od != d:
            if d % od == 0:
                break
    else: ans.append(d)
return ans

Yes the else is in the right spot

Continuing with the example, what this part does is to search through the list of divisors and to see if any of them is divisible by any of the others. If this is so, we ignore that one, otherwise we keep it.

The end result is a list of prime factors.

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I also mentioned in my comment that the divisors list can also be generated using:

divisors = [2] + [ d for d in xrange(3,int(n**2+1), 2) if n % d == 0]

This method is useful because it avoids including even numbers except 2, which is the only even prime. Also the size of divisors is smaller because we only check up to the square root of a number which is most times smaller than the number divided by 2