无法弄清楚如何使用all（）在我的“代码”中工作

Question

m = range(1, 2000000, 2)
sum1 = 2
for x in xrange(1, 2000000, 2):
    for y in m:
        if x != y:
            if x%y == 0:
                m.remove(x)
            if all(x%y != 0):
                sum1 += x

That's what I've written. It's about a problem, trying to add all the primes bellow two million. My problem is in the all() statement. What I want to happen is to check if x is a prime; that is true only if every x%y gives a remainder.

Also if I use a can I use a statement (break?) to stop the loop if y > x/3 like so:

 m = range(1, 2000000, 2)
sum1 = 2
for x in xrange(1, 2000000, 2):
    for y in m:
        if y > x/3:
            break
        else:
            if x != y:
                if x%y == 0:
                    m.remove(x)
                if all(x%y != 0):
                    sum1 += x

Answer 1

You have to pass a sequence or iterable to all -- it just tests whether or not all the items passed to it evaluate as true. Here's the right way to use all :

>>> all([True, True, True])
True
>>> all([True, False, True])
False
>>> all([x > 5 for x in range(10)])
False
>>> all([x > 5 for x in range(6, 10)])
True
>>> all(x > 5 for x in range(6, 10))
True

That last one is the best, because it takes advantage of short-circuiting .

However, your call to all in your code is pointless. The idea behind your code, it seems to me, is to go through all the values in m and remove those that are divisible by any number between 2 and 2000000. Once you've done that, m will contain only prime numbers.

Of course, your code still won't work if you remove all . That's because you're actually testing whether each number in m is divisible by the numbers [1, 3, 5, 7...1999999] . (That's the sequence signified by xrange(1, 2000000, 2) . Because you start at 1 , and everything is divisible by 1 , your code will count nothing as prime. And then, once you remove 1 from that sequence, anything divisible by 2 will be counted as prime by your code! You should think more carefully about which numbers you actually have to test in your inner loop.

Finally, you should think about how many loops this code will complete. Even once you have this working, it will take a very long time to generate a result. You should test it on smaller numbers first; and then, you should think about how to reduce the number of loops. (And -- only after you've thought about it a bit -- read this .)

But once you have this working, all you have to do is call sum on your list of primes.

Answer 2

Your use of all is incorrect, if you look at the documentation for it, it takes an iterable.

What you may be trying to do is use a generator expression, something of the form:

sum(x**2 for x in range(10))

which is very similar to the list comprehension

[x**2 for x in range(10)]

However, use of all in this manner wouldn't suddenly stop the generator expression, if it found a divisor. Use of any and checking for x == 0 would stop sooner, but as the code is currently formatted, would check for many divisors before deeming something prime.

This would be more appropriate:

primes = []
MAX = 2000000

number = 2
while number < MAX:
    for prime in primes:
        if number % prime == 0:
            number += 1
            continue

    primes.append(number)
    number += 1

total = sum(primes)

Answer 3

all() takes an iterable for an argument. In your situation, you would use it like this:

all(x%y for y in m)

where x%y for y in m is a generator expression. If I have an iterable

[item1, item2, item3, item4...]

all(iterable) is equivalent to:

item1 and item2 and item3 and item4...'