Question

我在这里缺少技术词，但这里的问题是将int更改为float或float to int。

def factorize(n):
    def isPrime(n):
        return not [x for x in range(2,int(math.sqrt(n)))
                    if n%x == 0]
    primes = []
    candidates = range(2,n+1)
    candidate = 2
    while not primes and candidate in candidates:
        if n%candidate == 0 and isPrime(candidate):

            # WHY ERROR?
            #I have tried here to add float(), int() but cannot understand why it returns err
            primes = primes + [float(candidate)] + float(factorize(n/candidate))
        candidate += 1
    return primes

错误 - 尝试使用int()和float()等功能进行修复，但仍然存在：

TypeError: 'float' object cannot be interpreted as an integer

Answer 1

这个表达式是您的直接问题：

float(factorize(n/candidate))

factorize返回一个列表，但float需要将其参数作为字符串或数字。

（你的代码有很多很多其他问题，但也许你最好自己发现它们......）

Answer 2

请注意，您返回list并在行中

primes = primes + [float(candidate)] + float(factorize(n/candidate))

但是float适用于数字或字符串，而非列表。

正确的解决方案是：

primes = primes + [float(candidate)] + [float(x) for x in factorize(n/candidate)]
# Converting every element to a float

Answer 3

无法理解Gareth对many, many other problems的意义，问题在于消毒！

def factorize(n):
    # now I won`t get floats
    n=int(n)

    def isPrime(n):
        return not [x for x in range(2,int(math.sqrt(n)))
                    if n%x == 0]

    primes = []
    candidates = range(2,n+1)
    candidate = 2
    while not primes and candidate in candidates:
        if n%candidate == 0 and isPrime(candidate):
            primes = primes + [candidate] + factorize(n/candidate)
        candidate += 1
    return primes


clearString = sys.argv[1]
obfuscated = 34532.334
factorized = factorize(obfuscated)

print("#OUTPUT "+factorized)


#OUTPUT [2, 2, 89, 97]

更好但是你可以做更简单或更少的行吗？

def factorize(n):
    """ returns factors to n """

    while(1):
            if n == 1:
                    break

            c = 2 

            while n % c != 0:
                    c +=1

            yield c
            n /= c

 print([x for x in factorize(10003)])

时间比较

$ time python3.1 sieve.py 
[100003]

real    0m0.086s
user    0m0.080s
sys 0m0.008s
$ time python3.1 bad.py 
^CTraceback (most recent call last):
  File "obfuscate128.py", line 25, in <module>
    print(factorize(1000003))
  File "obfuscate128.py", line 19, in factorize
    if n%candidate == 0 and isPrime(candidate):
KeyboardInterrupt

real    8m24.323s
user    8m24.320s
sys 0m0.016s

at least O(n)是一个很小的轻描淡写，大声笑，我可以从谷歌找到，让我们考虑大素数的糟糕结果。 10003至少10002!个子进程10003典当10002，因为每个子进程都会失败并且在评估每个子进程并且每个n子进程将进行评估之前无法对其进行评估有n-1子进程。很好的例子，如何不分解。

解释因子分解中的float-int问题

3 个答案: