I have a numpy array like,
nums = np.array([17, 18, 19, 20, 21, 22, 23])
如何以Pythonic的方式从数组中过滤出质数?我知道可以执行简单的筛选操作,例如:
nums[nums > 20] #array([21, 22, 23])
有没有一种方法可以传递lambda函数来进行过滤?
期望输出:array([17, 19, 23])
I have a numpy array like,
nums = np.array([17, 18, 19, 20, 21, 22, 23])
如何以Pythonic的方式从数组中过滤出质数?我知道可以执行简单的筛选操作,例如:
nums[nums > 20] #array([21, 22, 23])
我会使用gmpy或者一个开发了优秀素数测试算法的第三方库来完成。米勒-拉宾素数测试通常是非常安全(而且快速!)的选择。如果你只想用慢速的方式,可以这样做:
import numpy as np
import math
def is_prime(n):
if n % 2 == 0 and n > 2:
return False
return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))
a = np.arange(1, 10**3)
foo = np.vectorize(is_prime)
pbools = foo(a)
primes = np.extract(pbools, a)
primes # => Output below
array([ 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163,
167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233,
239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389,
397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563,
569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641,
643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727,
733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821,
823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907,
911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997])
如果您想要筛选掉质数,只需在pbools变量上调用np.invert。对于任何谓词,都可以采用相同的方法。您还可以将lambda传递给vectorize。例如,假设我们只想筛选出同时满足是质数且与5的除法余数为1的数字(不管出于什么原因)。import numpy as np
import math
def is_prime(n):
if n % 2 == 0 and n > 2:
return False
return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))
a = np.arange(1, 10**3)
foo = np.vectorize(lambda x: (not (x + 1) % 5 or not (x - 1) % 5) and is_prime(x))
primes = a[foo(a)] # => Shorthand.... Output below
array([ 1, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109,
131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 239, 241, 251,
269, 271, 281, 311, 331, 349, 359, 379, 389, 401, 409, 419, 421,
431, 439, 449, 461, 479, 491, 499, 509, 521, 541, 569, 571, 599,
601, 619, 631, 641, 659, 661, 691, 701, 709, 719, 739, 751, 761,
769, 809, 811, 821, 829, 839, 859, 881, 911, 919, 929, 941, 971, 991])
import numpy as np
def primesfrom2to(n):
# https://dev59.com/snI95IYBdhLWcg3w-DH0#3035188
""" Input n>=6, Returns a array of primes, 2 <= p < n """
sieve = np.ones(n//3 + (n%6==2), dtype=np.bool)
sieve[0] = False
for i in range(int(n**0.5)//3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)//3) ::2*k] = False
sieve[(k*k+4*k-2*k*(i&1))//3::2*k] = False
return np.r_[2,3,((3*np.nonzero(sieve)[0]+1)|1)]
# generate 100.000 random integers from 1 to 1.000.000.000
a1 = np.random.randint(1, 10**9, 100000)
# generate all primes that are equal or less than a1.max()
primes = primesfrom2to(a1.max())
# print result
print(np.intersect1d(primes, a1))
看起来你的问题并不是关于质数,而是关于如何将函数应用于 numpy
数组。我使用了简单的 is_odd
示例。也许你正在寻找 np.vectorize
:
In [34]: nums = np.array([17, 18, 19, 20, 21, 22, 23])
In [35]: def is_odd(n):
if n % 2 == 1:
return True
return False
....:
In [36]: is_odd_v = np.vectorize(is_odd)
In [37]: nums[is_odd_v(nums)]
Out[37]: array([17, 19, 21, 23]
如果我没记错的话,np.vectorize
主要用于方便起见,并且性能不是很好。
is_prime
函数,它将返回一个经过筛选的质数np.array
。当然,我的解决方案很朴素且未经优化。 - Akavall有这样的设置:
import numpy as np
import math
nums = np.array([17, 18, 19, 20, 21, 22, 23])
因此现在我们创建一个包含所有可能整数候选项的数组:
divisors = np.arange(2,int(math.sqrt(np.max(nums)))+1) # Numbers from 2 to sqrt(max(nums))
print(divisors)
# [2 3 4]
print(nums[:,None] % divisors[None,:]) # Modulo operation on each element (0 means divisible)
[[1 2 1] [0 0 2] [1 1 3] [0 2 0] [1 0 1] [0 1 2] [1 2 3]] 现在我们如何得到质数...我们检查是否有零结果的行:
print(np.min(nums[:,None] % divisors[None,:], axis=1)) # Minimum of the modulo for that element
# [1 0 1 0 0 0 1]
然后创建一个掩码来索引它们:
print(nums[np.min(nums[:,None] % divisors[None,:], axis=1) > 0]) # So index them
# [17 19 23]
所以最终你所需的就是:
nums = np.array([17, 18, 19, 20, 21, 22, 23])
divisors = np.arange(2,int(math.sqrt(np.max(nums)))+1)
nums[np.min(nums[:,None] % divisors[None,:], axis=1) > 0]
所有其他的东西只是为了说明每个步骤在做什么。
这并不是简单的操作,因为它使用了将1D数组广播到2D数组中,但方法应该很清晰。如果您有任何问题,请告诉我。
2
和sqrt(max(array))
之间的每个数字,但您不需要测试所有这些数字。如果您有一个返回该范围内所有质数的函数,那就足够了。例如,使用@MaxU答案中的primesfrom2to
,更快的可能性是:nums = np.array([17, 18, 19, 20, 21, 22, 23])
# All prime numbers in the range from 2 to sqrt(max(nums))
divisors = primesfrom2to(int(math.sqrt(np.max(nums)))+1)
nums[np.min(nums[:,None] % divisors[None,:], axis=1) > 0]
但它使用了与之前相同的机制,但速度更快了。:-)
nums[[i for i in range(len(nums)) if sum([nums[i]%val==0 for val in range(2,nums[i]-1)])==0]]
[i for i in range(len(nums)) if sum([nums[i]%val==0 for val in range(2,nums[i]-1)])==0]
这基本上会遍历每个值并检查它是否不能被比它自己小的任何值整除(忽略1)
[i for i in range(len(nums)) #for every index
if sum(#calculate sum of booleans
[nums[i]%val==0 for val in range(2,nums[i]-1)] # check if it is divisble by any value smaller than itself
)==0 #check if the number of divisors is zero