1

我尝试从 gmpy2.mpz 转换为 numpy 布尔数组，但不能完全正确。（gmpy2：https ://gmpy2.readthedocs.io ）

import gmpy2
import numpy as np

x = gmpy2.mpz(int('1'*1000,2))

print("wrong conversion 1")
y = np.fromstring(gmpy2.to_binary(x), dtype=bool) # this is wrong
print(np.sum(y)) # this returns 127 instead of 1000

print("wrong conversion 2")
y = np.fromstring(gmpy2.to_binary(x), dtype=np.uint8)
print(y) # array([  1,   1, 255 ... 255], dtype=uint8)
y_bool = np.unpackbits(y)
slow_popcount = np.sum(y_bool, dtype=int)
print(slow_popcount) # 1002. should be 1000

print("Fudging an answer. This is wrong as well.")
y = np.fromstring(gmpy2.to_binary(x)[2:], dtype=np.uint8)
# is that slicing [2:] a slow operation?
y_bool = np.unpackbits(y)
print np.sum(y_bool, dtype=int) # 1000

更多测试：

np.fromstring(gmpy2.to_binary(gmpy2.mpz(int('1'*64,2))), dtype=np.uint8)
# array([  1,   1, 255, 255, 255, 255, 255, 255, 255, 255], dtype=uint8)
np.fromstring(gmpy2.to_binary(gmpy2.mpz(int('1'*65,2))), dtype=np.uint8)
# array([  1,   1, 255, 255, 255, 255, 255, 255, 255, 255,   1], dtype=uint8
np.fromstring(gmpy2.to_binary(gmpy2.mpz(int('1'*66,2))), dtype=np.uint8)
# array([  1,   1, 255, 255, 255, 255, 255, 255, 255, 255,   3], dtype=uint8)
np.fromstring(gmpy2.to_binary(gmpy2.mpz(int('1'*1024,2))), dtype=np.uint8)
# array([  1,   1, 255 ... 255], dtype=uint8)

顺便说一句，我实际上想快速获取 gmpy2.mpz 的所有设置位的索引列表、数组或 numpy 数组。我尝试转换的实际 4,777,000 gmpy2.mpz 每个有 760,000 位，其中大约 2,000 位为 1。计算机上的 gmp 库是用 intel icc 编译的。

谢谢

4

2 回答 2

1

有几个选项。该函数gmpy2.bit_scan1(x, n)将返回索引 >= n 的设置的第一位的索引。

>>> x = gmpy2.mpz(123456)
>>> bin(x)
'0b11110001001000000'
>>> n = 0
>>> while True:
...     n = gmpy2.bit_scan1(x, n)
...     if n is None:
...         break
...     print(n)
...     n = n + 1
... 
6
9
13
14
15
16

gmpy2还支持称为 的整数类型xmpz。它是该mpz类型的实验版本。主要区别在于xmpz类型是可变的 - 就地操作将直接修改值而不创建副本。这使得该xmpz类型对于位操作非常有用。例如，您可以使用切片表示法提取和修改位位置。

该xmpz类型还支持称为iter_set、iter_clear和的方法iter_bits。

>>> x_str='1'*8+'01'
>>> x_int=gmpy2.xmpz(x_str, 2)
>>> list(x_int.iter_set())
[0, 2, 3, 4, 5, 6, 7, 8, 9]
>>> list(x_int.iter_clear())
[1]
>>> list(x_int.iter_bits())
[True, False, True, True, True, True, True, True, True, True]

我最初编写该xmpz类型是为了评估优化就地操作的任何性能改进。位操作看到了最大的好处。这是埃拉托色尼筛法的简短而快速的实现。

def sieve(limit=1000000):
    '''Returns a generator that yields the prime numbers up to limit.'''

    sieve_limit = gmpy2.isqrt(limit) + 1
    limit += 1
    # Mark bit positions 0 and 1 as not prime.
    bitmap = gmpy2.xmpz(3)
    # Process 2 separately. This allows us to use p+p for the step size
    # when sieving the remaining primes.
    bitmap[4 : limit : 2] = -1
    # Sieve the remaining primes.
    for p in bitmap.iter_clear(3, sieve_limit):
        bitmap[p*p : limit : p+p] = -1
    return bitmap.iter_clear(2, limit)
于 2016-05-30T04:49:31.887 回答
0

这有效，但速度很慢。

import gmpy2
import numpy as np
x_str = '1'*8+'01'
print(x_str)
x_int = int(x_str,2)
x_mpz = gmpy2.mpz(x_int)
x_01 = bin(int(x_mpz))[2:] # get rid of '0b'
x_bin = x_01.replace('1','\x01').replace('0','\x00')
x_np_bool = np.fromstring(x_bin, dtype = bool)
x_1_index = np.where(x_np_bool)[0]
print(x_1_index)
于 2016-05-30T03:52:08.203 回答