我正在创建Eratosthenes的筛网,以便更有效地求和1到大数n之间的素数。我想要做的是创建一个从2到n的列表,然后删除2的倍数,然后删除3的倍数,再删除列表中下一个数字的倍数,依此类推。我创建的代码在时间上性能很慢,几乎就像通过检查每个条目是否为质数来创建列表一样。我想我要进行的操作数是有序的:n的平方根(第一个while循环)乘以n的平方根(小于第二个while循环)。因此,我不确定remove方法或其他方法是否会降低它的速度。
我的代码是这个:
def sieve_of_Eratosthenes(n):
L= list(range(2,n+1))
# creates a list from 2 to n
i=2
while i*i <=n: # going to remove multiples of i, starting at i^2
k=i # if j <i then ij already checked
while i*k <= max(L):
try:
L.remove(i*k) # there is an error if the element is not in
# the list so need to add these two lines
except ValueError:
pass # do nothing!
k=k+1
i=L[i-1] # list index starts at 0, want i to be next element in the list
# print(L)
return L
假设
问题是有关如何改善软件的运行时间,因为它的运行速度很慢。
执行以下两个代码更改以加快代码的速度
代码
def sieve_of_Eratosthenes2(n):
if n < 2:
return []
if n < 3:
return [2]
L = [True] * (n+1) # all numbers set as primes initially
# modifies prime flag in list for odd numbers
for i in range(3, n, 2): # Check odd numbers for prime (no need to check even numbers)
if L[i]: # A prime
L[i*i::i] = [False]*len(L[i*i::i]) # from i^2 in increments of i
# Report prime 2 + odd primes
return [2] + [i for i in range(3, n, 2) if L[i]] # Get odd numbers whose flag is
# still True
新代码
%timeit sieve_of_Eratosthenes2(100000)
16 ms ± 1.58 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
旧代码
sieve_of_Eratosthenes2(100000)
261.45 seconds (using time module)