我正在尝试对以下代码进行cythonize:
def my_func(vector_b):
vector_b = np.unpackbits(np.frombuffer(vector_b, dtype=np.uint8))
vector_b = (vector_b * _n_vector_ranks_only)
min_ab = np.sum(np.minimum(vector_a, vector_b))
max_ab = np.sum(np.maximum(vector_a, vector_b))
return min_ab / max_ab
_n_vector_ranks_only = np.arange(1023, -1, -1, dtype=np.uint16)
# vector_a data type is same of vector_b, is not contained in db, it is passed manually
vector_a = np.frombuffer(vector_a, dtype=np.uint8)
vector_a = (vector_a * _n_vector_ranks_only)
#fetch all vectors from DB
df = dd.read_sql_table('mydb', 'postgresql://user:passwordg@localhost/table1', npartitions=16, index_col='id', columns=['data'])
res = df.map_partitions(lambda df: df.apply( lambda x: my_func(x['data']), axis=1), meta=('result', 'double')).compute(scheduler='processes')
#data is a binary array saved with numpy packbits
目前我在这里:
from ruzi_cython import ruzicka
def my_func(vector_b):
vector_b = np.unpackbits(np.frombuffer(vector_b, dtype=np.uint8))
vector_b = (vector_b * _n_vector_ranks_only)
#min_ab = np.sum(np.minimum(vector_a, vector_b))
#max_ab = np.sum(np.maximum(vector_a, vector_b))
#return min_ab / max_ab
return ruzicka.run_old(vector_a, vector_b)
这是ruzicka.pyx:
# cython: profile=True
import numpy as np
cimport numpy as np
cimport cython
ctypedef np.uint16_t data_type_t
@cython.boundscheck(False)
@cython.wraparound(False)
@cython.overflowcheck(False)
@cython.initializedcheck(False)
cdef double ruzicka_old(data_type_t[:] a, data_type_t[:] b):
cdef int i
cdef float max_ab = 0
cdef float min_ab = 0
for i in range(1024):
if a[i] > b[i]:
max_ab += a[i]
min_ab += b[i]
else:
max_ab += b[i]
min_ab += a[i]
return min_ab / max_ab
def run_old(a, b):
return ruzicka_old(a, b)
在那里我获得了很多表演。 在对两个数组进行乘法运算的第一部分中,我仍然无法获得良好的结果。
这是我做乘法的方法:
cdef double ruzicka(data_type_16[:] a, data_type_8[:] b):
cdef int i
cdef float max_ab = 0
cdef float min_ab = 0
cdef data_type_16 tmp = 0
for i in range(1024):
tmp = b[i] * (1023-i)
if a[i] > tmp:
max_ab += a[i]
min_ab += tmp
else:
max_ab += tmp
min_ab += a[i]
return min_ab / max_ab
答案 0 :(得分:2)
您似乎正在努力获取数组的第n位(本质上是np.unpackbits
做的事情)。
第n位包含在n//8
字节内(我使用的是//
除法和舍入运算符)。您可以使用&
(通过1<<m
移一位)来按“按位与”(m
)的方式访问字节中的单个位。这将为您提供数字2**(m-1)
,而您实际上只是在乎它是否为0。
因此,假设vector_b
是np.int8_t
的内存视图,您可以执行以下操作:
byte_idx = n//8
bit_idx = n%8 # remainder operator
bitmask = 1<<bit_idx
bit_is_true = 1 if (vector_b[byte_idx]&bitmask) else 0
您需要将其放在循环中,并cdef
放置变量的类型。