【378】python any() and all()

Reference:

[1] Python all() - Python Standard Library

[2] Python any() - Python Standard Library

all() and any() 函数主要用于需要判断某个数组是不是都满足了某种条件，设置一个跟数组一样的 bool 数组，判断 bool 数组是否都为 True，或者有没有 True 等等。

Q:

Insert your code into consecutive_primes.py so as to find all sequences of 6 consecutive prime 5-digit numbers, say (a,b,c,d,e,f), with b=a+2, c=b+4, d=c+6, e=d+8, and f=e+10. So a, b, c, d, e and f are all 5-digit prime numbers and no number between a and b, between b and c, between c and d, between d and e, and between e and f is prime.

If you are stuck, but only when you are stuck, then use consecutive_primes_scaffold.py.

A: by McDelfino，在这个程序中使用了 all() 函数用来判断是否对应的数为质数，如果都返回 True 才会通过。

from math import sqrt

def is_prime(n):
    if n%2 == 0: return False
    for i in range(3,round(sqrt(n))+1, 2):
        if n%i == 0:
            return False
    return True

print('The solutions are:\n')
# The list of all even i's such that a + i is one of a, b, c, d, e, f.
good_leaps = tuple(sum(range(0, k, 2)) for k in range(2, 13, 2))

# Write a loop that generates all odd numbers a between 10_000 and 99_999 and tests whether
# for all i = 0, 2, 4, ..., 30, i is in good_leaps iff a + i is prime.
for a in range(10_001, 100_000-5, 2):
    if all([is_prime(a+i) for i in good_leaps]):
        tmp = [a+i for i in good_leaps]
        count = 0
        for j in range(tmp[1]+2, tmp[5], 2):
            if is_prime(j): count+=1
        if count == 3:
            print(*[a+i for i in good_leaps])

The solutions are:

13901 13903 13907 13913 13921 13931
21557 21559 21563 21569 21577 21587
28277 28279 28283 28289 28297 28307
55661 55663 55667 55673 55681 55691
68897 68899 68903 68909 68917 68927