On Tue, Jul 19, 2011 at 11:08 AM, Robert Kern <robert.k...@gmail.com> wrote:
> On Tue, Jul 19, 2011 at 07:38, Andrea Cimatoribus
> <g.plantagen...@gmail.com> wrote:
>> Dear all,
>> I would like to avoid the use of a boolean array (mask) in the following
>> statement:
>>
>> mask = (A != 0.)
>> B = A[mask]
>>
>> in order to be able to move this bit of code in a cython script (boolean
>> arrays are not yet implemented there, and they slow down execution a lot as
>> they can't be defined explicitely).
>> Any idea of an efficient alternative?
>
> You will have to count the number of True values, create the B array
> with the right size, then run a simple loop to assign into it where A
> != 0. This makes you do the comparisons twice.
>
> Or you can allocate a B array the same size as A, run your loop to
> assign into it when A != 0 and incrementing the index into B, then
> slice out or memcpy out the portion that you assigned.
According to my calculations, the last method is the fastest, though
the savings aren't considerable.
In cython, defining some test mask functions (saved as cython_mask.pyx):
import numpy as N
cimport numpy as N
def mask1(N.ndarray[N.int32_t, ndim=1] A):
cdef N.ndarray[N.int32_t, ndim=1] B
B = A[A != 0]
return B
def mask2(N.ndarray[N.int32_t, ndim=1] A):
cdef int i
cdef int count = 0
for i in range(len(A)):
if A[i] == 0: continue
count += 1
cdef N.ndarray[N.int32_t, ndim=1] B = N.empty(count, dtype=int)
count = 0
for i in range(len(A)):
if A[i] == 0: continue
B[count] = A[i]
count += 1
return B
def mask3(N.ndarray[N.int32_t, ndim=1] A):
cdef N.ndarray[N.int32_t, ndim=1] B = N.empty(len(A), dtype=int)
cdef int i
cdef int count = 0
for i in range(len(A)):
if A[i] == 0: continue
B[count] = A[i]
count += 1
return B[:count]
In [1]: import numpy as N
In [2]: import timeit
In [3]: from cython_mask import *
In [4]: A = N.random.randint(0, 2, 10000)
In [5]: def mask4(A):
...: return A[A != 0]
...:
In [6]: %timeit mask1(A)
10000 loops, best of 3: 195 us per loop
In [7]: %timeit mask2(A)
10000 loops, best of 3: 136 us per loop
In [8]: %timeit mask3(A)
10000 loops, best of 3: 117 us per loop
In [9]: %timeit mask4(A)
10000 loops, best of 3: 193 us per loop
~Brett
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