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
此时此刻,我正站在这一点上:
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)
在哪里鲁齐卡.pyx这是:
# 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)
在那里我得到了很多表演。 我仍然不能用很好的结果来计算第一部分,在这里我将两个数组相乘。你知道吗
我就是这样做乘法的:
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
看起来您正在努力获取数组的第n位(实际上是在做
np.unpackbits
所做的事情)。你知道吗第n位包含在
n//8
字节中(我使用的是//
除法取整运算符)。您可以使用1<<m
(一个位被m
移位)执行“按位与”(&
)访问字节中的单个位。这会给你一个数字2**(m-1)
,你真的只关心它是不是0。你知道吗因此假设
vector_b
是np.int8_t
内存视图,您可以执行以下操作:你需要把它放在一个循环中,
cdef
变量的类型。你知道吗相关问题 更多 >
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