[英]using filter and generator to generator endless prime number in python
以下是我发现的使用Eratosthenes筛子查找素数的python程序。 它使用过滤器和生成器。 我听不懂
def _odd_iter():
n = 1
while True:
n = n + 2
yield n
def _not_divisible(n):
return lambda x: x % n > 0
def primes():
yield 2
it = _odd_iter()
while True:
n = next(it)
yield n
it = filter(_not_divisible(n), it)
for n in primes():
if n < 1000:
print(n)
else:
break
我不明白的是it = filter(_not_divisible(n), it)
。 例如,对于数字105,如何用单行代码排除它?
这不只是一行代码,它是被反复运行线,以不同的价值观n
。
基本上, it
是一个迭代器,它产生尚未被筛子排除的候选质数。 您首先要使所有奇数都成为候选数。
it = _odd_iter()
然后,您反复选择剩下的第一个候选人,
while True:
n = next(it)
删除该候选人倍数的所有数字,
filter(_not_divisible(n), it)
并用删除倍数后剩余的所有内容替换您的候选素数。
it = ...
如果您假装filter
返回一个数字列表,而不是一个可迭代的列表,并且还假装_odd_iter()
返回一个奇数列表而不是一个可迭代的列表,则可以遍历循环并确定列表中每个点的内容。 例如,运行后
it = _odd_iter()
你开始
it = 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, ...
然后跑
n = next(it) # 3
这将第一件物品从前面拉下来,让您
it = 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, ...
并运行
it = filter(_not_divisible(3), it)
过滤掉3的所有倍数,
it = 5, 7, 11, 13, 17, 19, 23, 25, ...
然后返回循环顶部,从前面拉出新的第一个数字
n = next(it) # 5
离开
it = 7, 11, 13, 17, 19, 23, 25, ...
然后滤除5的所有倍数
it = filter(_not_divisible(5), it)
这使
it = 7, 11, 13, 17, 19, 23, ...
等等。
实际上,由于filter()
返回的是迭代器,而不是列表,因此您最终会获得嵌套的迭代器序列。 特别是,您从
it = _odd_iter()
然后在循环的第一次迭代之后
it = filter(_non_divisible(3), _odd_iter())
除了从迭代器中取出3
之外,然后在循环的第二次迭代之后
it = filter(_non_divisible(5), filter(_non_divisible(3), _odd_iter()))
除了从迭代器中也取了5
,然后
it = filter(_non_divisible(7), filter(_non_divisible(5), filter(_non_divisible(3), _odd_iter())))
等等。
对于找到的每个素数,将一个filter
应用于可迭代对象,所使用的过滤器是一项功能,它排除素数的所有倍数。
因此,您的可迭代项被包裹在与找到质数一样多的过滤器中,例如,排除了数字105,因为它可以被3整除,并且在找到质数3时添加了3的所有倍数的过滤器。
如果添加一些print
语句,它将更加清晰(我希望):
def _odd_iter():
n = 1
while True:
n = n + 2
yield n
def _not_divisible(n):
print('add filter for all multiples of', n)
return lambda x: print('check if', x, 'is divisible by', n, 'result: ', not (x % n > 0)) or x % n > 0
def primes():
yield 2
it = _odd_iter()
while True:
n = next(it)
yield n
it = filter(_not_divisible(n), it)
for n in primes():
if n < 20:
print(n)
else:
break
打印:
2
3
add filter for all multiples of 3
check if 5 is divisible by 3 result: False
5
add filter for all multiples of 5
check if 7 is divisible by 3 result: False
check if 7 is divisible by 5 result: False
7
add filter for all multiples of 7
check if 9 is divisible by 3 result: True
check if 11 is divisible by 3 result: False
check if 11 is divisible by 5 result: False
check if 11 is divisible by 7 result: False
11
add filter for all multiples of 11
check if 13 is divisible by 3 result: False
check if 13 is divisible by 5 result: False
check if 13 is divisible by 7 result: False
check if 13 is divisible by 11 result: False
13
add filter for all multiples of 13
check if 15 is divisible by 3 result: True
check if 17 is divisible by 3 result: False
check if 17 is divisible by 5 result: False
check if 17 is divisible by 7 result: False
check if 17 is divisible by 11 result: False
check if 17 is divisible by 13 result: False
17
add filter for all multiples of 17
check if 19 is divisible by 3 result: False
check if 19 is divisible by 5 result: False
check if 19 is divisible by 7 result: False
check if 19 is divisible by 11 result: False
check if 19 is divisible by 13 result: False
check if 19 is divisible by 17 result: False
19
add filter for all multiples of 19
check if 21 is divisible by 3 result: True
check if 23 is divisible by 3 result: False
check if 23 is divisible by 5 result: False
check if 23 is divisible by 7 result: False
check if 23 is divisible by 11 result: False
check if 23 is divisible by 13 result: False
check if 23 is divisible by 17 result: False
check if 23 is divisible by 19 result: False
首先,对迭代器进行过滤会返回另一个迭代器。 即当我们做类似的事情时:
it = filter(_not_divisible(3), it)
it = filter(_not_divisible(5), it)
我们得到一个链式迭代器“奇数且不能被3整除,不能被5整除”。 它在某种程度上类似于链式装饰器,等效于:
# assuming we have decorator @not divisible
@not_divisible(2)
def iter():
return xrange(inf)
# then, at every subsequent prime we do something like:
iter = not_divisible(3)(iter)
# next prime is 5:
iter = not_divisible(5)(iter)
... 等等
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