素数分解程序中赋值错误前引用的变量？

Question

我正在写一些代码，用来寻找数的素因子分解。主函数通过数字递增；我这样做是因为我想使用代码进行计时实验。我不介意它不是超级高效，对我来说，项目的一部分将是让它更高效自己。它还没有完全完成。你知道吗

import math
import time

primfac=[]

def primcheck(n):
    for x in xrange(2, int(n**0.5)+1):
        if n % x == 0:
            return False
    return True

def primes(n):
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]


def factfind(lsp,n): #finds factors of n among primes
    for i in lsp:
        if n%i==0:
            primfac.append(i)
        else:
            i+=1

def simplify(lsp, n):
    x = 1
    for i in lsp:
        x=i*x
    if x != n:
        print "needs exponent, computing"
        for i in lsp:
            y=n/i
            if primcheck(y) == True:
                lsp.append(y)
            else:
                lsp.append(factfind(primes,y))

def primfacfind(n1,n2):
    while n1 <= n2:
        time_start = time.clock()
        if primcheck(n1) == True:
            print "prime"
            time_elapsed = time.clock() - time_start
            print "time:", time_elapsed
            n1+=1
        else:
            n = n1
            print "starting #", n

            factfind(primes(n),n)
            print primfac

            del primfac
            primfac[:] = []
            simplify(primfac, n)

            time_elapsed = time.clock() - time_start
            print "time:", time_elapsed

            n1+=1

primfacfind(6,15)

当我运行代码时，它会给出以下错误消息：

Traceback (most recent call last):
  File "python", line 65, in <module>
  File "python", line 54, in primfacfind
UnboundLocalError: local variable 'primfac' referenced before assignment

我不明白，因为我已经测试了除了simplify函数之外的所有函数，即使这样，唯一的新部分还是print之后的行。你知道吗

Answer 1

python中的函数只能以只读方式访问在函数范围之外声明的变量，而不使用global关键字。你知道吗

import math 
import time

primfac=[]

def primcheck(n):
    for x in xrange(2, int(n**0.5)+1):
        if n % x == 0:
            return False
    return True

def primes(n):
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]


def factfind(lsp,n): #finds factors of n among primes
    for i in lsp:
        if n%i==0:
            primfac.append(i)
        else:
            i+=1

def simplify(lsp, n):
    x = 1
    for i in lsp:
        x=i*x
    if x != n:
        print "needs exponent, computing"
        for i in lsp:
            y=n/i
            if primcheck(y) == True:
                lsp.append(y)
            else:
                lsp.append(factfind(primes,y))

def primfacfind(n1,n2):
    global primfac
    while n1 <= n2:
        time_start = time.clock()
        if primcheck(n1) == True:
            print "prime"
            time_elapsed = time.clock() - time_start
            print "time:", time_elapsed
            n1+=1
        else:
            n = n1
            print "starting #", n

            factfind(primes(n),n)
            print primfac

            del primfac
            primfac = []
            simplify(primfac, n)

            time_elapsed = time.clock() - time_start
            print "time:", time_elapsed

            n1+=1

primfacfind(6,15)